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Chapter 2, Problem 1P
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Two particles of identical mass m are acted on only by the gravitational force of one upon the
other. If the distance d between the particles is constant, what is the angular velocity of the line
joining them? Use Newton’s second law with the center of mass of the system as the origin of the
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inertial frame.
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Step 1 of 2
Angular velocity of an object can be defined as the time rate of change of its angular position
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relative to the origin.
___
Use the following equation to calculate the angular velocity.
Here, v is the linear velocity and r is the radius of orbit.
Comment
Step 2 of 2
Calculate the angular velocity of the line joining the two identical masses.
The gravitational force between the two identical masses of m separated at a distance of d is,
The two masses are identical, so their center of mass should be equal to the half of their distance
d, that is,
So, the two bodies can move in circular path with the above radius.
The equation for the centripetal force is,
Here, m is the mass of the object,
is the angular velocity and r is the radius of the orbital path.
The gravitational force must be equal to the centripetal force to move the two bodies in circular
path.
Substitute
for r in above equation.
Therefore, the angular velocity of the line joining the two identical masses is
.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / advanced physics / advanced physics questions and answers / at a given instant t0, a 1000-kg earth-orbiting satellite has the …
Question: At a given instant t0, a 1000-kg earth-orbiting satellite has the i…
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At a given instant t0, a 1000-kg earth-orbiting satellite has the inertial position and velocity vectors r0=
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3207i +5459j +2714k (km) and v0 = -6.532i +0.7835j +6.142k (km/s)
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reached by the satellite and the time at which it occurs.
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Q: At a given instant t0, a 1000-kg earth-orbiting satellite has the inertial position and velocity vectors r0= 3207i +5459j
+2714k (km) and v0 = -6.532i +0.7835j +6.142k (km/s) Solve the Two-Body Equation of Relative Motion numerically in
order to nd the maximum altitude reached by the satellite and the time at which it occurs. (NOTE: The book says that
the answer is 9670 km for the...
A: See answer
Questions viewed by other students
Q: At a given instant t0, a 1000-kg earth-orbiting satellite has the inertial position and velocity vectors r0= 3207i +5459j
+2714k (km) and v0 = -6.532i +0.7835j +6.142k (km/s) Solve the Two-Body Equation of Relative Motion numerically in
order to nd the maximum altitude reached by the satellite and the time at which it occurs. (NOTE: The book says that
the answer is 9670 km for the...
A: See answer
Q: At a given instant t0, a 1000 kg earth-orbiting satellite has the inertial position and velocity vectors r0 = 436 i + 6083 j +
2529 k (km) and v0 = −7.340 i − 0.5125 j + 2.497 k (km/s). After 89 minutes, a rocket motor with Isp = 300 s and T = 0 kN
thrust aligned with the velocity vector ignites and burns for 120 s. Use numerical integration to nd the maximum altitude
reached by...
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Chapter 2, Problem 3P
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Consider the two-body problem illustrated in Figure 2.1. If a force T (such as rocket thrust) acts
on m2 in addition to the mutual force of gravitation F21, show that
(a) Equation (2.22) becomes
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(b) If the thrust vector T has a magnitude T and is aligned with the velocity vector v, then
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Step-by-step solution
Step 1 of 5
The expression for the gravitational force between two objects of masses
is as
follows:
Here, G is the universal gravitational constant and r is the distance between
The vector form of the equation
is as follows:
Comment
Step 2 of 5
The free body diagrams of the two masses
which are in inertial frame is shown in the
following figure:
Figure: Free body diagrams.
In the figure,
and ,
and
are the position vectors,
is gravitational force on
is gravitational force on
due to
,
.
Comment
Step 3 of 5
(a)
From Newton’s third law of motion,
If a thrust T acts on
mass
in addition to mutual gravitational force, then the net force acting the
is as follows:
From Newton’s second law of motion, the net force acting on the mass
Rearrange the equation
The gravitational force on
Substitute
for
is,
for a.
due to
is as follows:
in the equation
and solve for a.
Replace acceleration vector with the second derivative of position vector and rewrite the equation
as follows:
Here,
is the second order derivative of position vector.
Replace
as its gravitational parameter
in the equation
and solve for
:
Hence, the equation
is proved.
Comment
Step 4 of 5
(b)
The unit vector of the thrust
is as follows:
Comment
Step 5 of 5
Here, T is the magnitude of thrust and
is the unit vector.
A unit vector can be formed by taking a vector and dividing it by its length.
The velocity vector and the thrust vector are in the same direction. Hence, the unit vector of
velocity and the unit vector of thrust are the same.
Here, v is the magnitude of velocity vector.
Substitute
for
in the equation
and solve for T.
Therefore, the expression for the trust vector is
.
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition
Orbital Mechanics for Engineering Students
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Three particles of identical mass m are acted on only by their mutual gravitational attraction.
They are located at the vertices of an equilateral triangle with sides of length d. Consider the
motion of any one of the particles about the system center of mass G and, using G as the origin
of the inertial frame, employ Newton’s second law to determine the angular velocity ω required
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for d to remain constant.
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Consider the following figure.
Consider the origin of the system at point C, as shown in the figure.
The location of center of mass system
is,
Find the gravitational force on the mass at the point C due to the other two masses.
Write the expression for the acceleration of the mass
Substitute
for
situated at the point.
.
The magnitude of tangential acceleration is calculated as,
Comments (1)
Step 2 of 2
Here, point G is considered as the center of rotation of the entire system.
Then from the geometry of the system of the masses.
Write the expression of the tangential acceleration in terms of the angular velocity.
Rearrange the equation for the angular velocity.
Substitute
for
and
for
,
Hence, the required angular velocity is
.
Comments (1)
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / advanced physics / advanced physics questions and answers / 2.5 at a given instant, a 1000-kg earth-orbiting satellite has the…
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Eqn 2.22:
where
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Q: At a given instant, a 1000-kg earth-orbiting satellite has the inertial position and velocity vectors 6600i (km) and vo =
12j(km/s). Solve Eqn (2.22) numerically to nd the distance of the spacecraft from the center of the earth and its speed 24
h later. ro =
A: See answer
Q: At a given instant t0, a 1000-kg earth-orbiting satellite has the inertial position and velocity vectors r0= 3207i +5459j
+2714k (km) and v0 = -6.532i +0.7835j +6.142k (km/s) Solve the Two-Body Equation of Relative Motion numerically in
order to nd the maximum altitude reached by the satellite and the time at which it occurs.
A: See answer
Show more
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If r, in meters, is given by
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, where t is the time in seconds,
calculate (a)
and (b) ∥ṙ∥ at t = 2 s.
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(a)
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Express the relation for position vector
Here,
.
is the vector.
Substitute
for
to find r.
Differentiate with respect to t and substitute 2s for t.
Therefore, value of
is
.
Comments (4)
Step 2 of 3
(b)
Express the relation for derivative of position vector.
Substitute
for
and differentiate with respect to t.
Comment
Step 3 of 3
Express the relation for length of the vector to find
.
Substitute
for
and
for t.
Therefore, the value of
is
.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 7p
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Starting with Eqn (2.35a), prove that ṙ = v·ûr and interpret this result.
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Start with the equation relating the dot product of the vector, r, of length r, and its first time
derivative with the product of their magnitudes.
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The derivative of the vector r is the velocity vector, v.
Now, divide both sides by the vector length, r.
Comment
Step 2 of 2
The vector r divided by its length, r, is a unit vector in the direction of the position
The conclusion is that
.
is the component of the velocity that points in the same direction as the
r vector. Hence, the equation
is proved.
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 8p
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Show that ûr·dûr/dt = 0, where ûr = r/r. Use only the fact that ûr is a unit vector. Interpret this
result.
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Step-by-step solution
Step 1 of 1
The unit vector
has a length of 1 and points in the direction of the position vector, r. It is
found by taking the vector, r, and dividing it by its length.
The derivative of the unit vector
must be perpendicular to
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can be found directly, but the direction of the derivative
because
cannot change in length. The dot product of two
perpendicular vectors is zero; therefore, the equation
is proved.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 9p
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Starting with Eqn (2.38), show that ûr dûr/dt = 0.
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Step 1 of 2
Start with equation 2.38 relating position, r, and angular momentum, h.
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The vector r divided by its length is the same as a unit vector in the direction of position
Substitute
for
in the equation
.
.
Comment
Step 2 of 2
Take a dot product on each side of the equation with
The cross product of the unit vector
.
, and angular momentum, h, produces a vector that is
perpendicular to both. The dot product between two perpendicular vectors is zero. So, the above
equation becomes,
Therefore, the equation
is proved.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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Chapter 2, Problem 10P
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for any orbit.
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For orbital motion the components of velocity are given as follow.
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And radial component of velocity,
Here,
is the gravitational parameter,
eccentricity and
is the specific relative angular momentum,
is the
is the angle between e and r.
Comment
Step 2 of 2
The resultant velocity in orbit is,
Substitute,
for
and
for
,
Solve further,
Substitute, 1 for
,
Hence, it is proved
.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 11p
Orbital Mechanics for Engineering Students
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Relative to a nonrotating, earth-centered Cartesian coordinate system, the position and velocity
vectors of a spacecraft are
and
.
Calculate the orbit’s (a) eccentricity vector and (b) the true anomaly.
My Textbook Solutions
Step-by-step solution
Step 1 of 5
(a)
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Express the relation for angular momentum in the cross product of position and velocity.
Use vectors of
for position and
Thus, the cross product
for velocity to express the cross product in terms of matrix.
is
.
Comments (6)
Step 2 of 5
Express the relation for eccentricity.
Here, e is the eccentricity C is the Laplace vector, μ is gravitational parameter of the Earth.
Expression for the Laplace vector is,
Use Laplace vector to rewrite eccentricity vector.
Comments (1)
Step 3 of 5
Express cross product of
Thus, cross product
.
is
.
Comments (1)
Step 4 of 5
Use Laplace vector to find eccentricity.
Therefore, the eccentricity vector is
.
Comments (3)
Step 5 of 5
(b)
Express the relation for true anomaly.
Here,
is the angle between the position vector and the eccentricity vector, e is the eccentricity
and r is the position vector.
Rewrite the expression in terms of
Therefore, true anomaly is
.
.
Comments (2)
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 12p
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Show that the eccentricity is 1 for rectilinear orbits (h = 0).
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Step 1 of 2
The magnitude of the eccentricity, e, can be found by taking the square root of the dot product of
the eccentricity vector, e, with itself.
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The eccentricity vector, e, can be found from the position vector, r, the angular momentum
vector, h, velocity vector, v, and the gravitation parameter of the object being orbited, μ.
Comment
Step 2 of 2
For rectilinear orbits, the angular momentum, h is 0. This lets the eccentricity vector, e, be
rewritten in terms of a unit vector in the direction of the position, ur.
Now substitute the eccentricity vector, e, into the formula for the magnitude of the eccentricity, e.
Therefore, the eccentricity is 1 for rectilinear orbits.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 13p
Orbital Mechanics for Engineering Students
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Relative to a nonrotating, earth-centered Cartesian coordinate system, the velocity of a
spacecraft is
and the unit vector in the direction of the radius is
. Calculate (a) the radial component of velocity vr, (b) the
My Textbook Solutions
azimuth component of velocity v⊥, and (c) the flight path angle γ.
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Step 1 of 3
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(a)
The expression for the radial component of the velocity is,
Here, v is the velocity vector and
is the unit vector in the direction of the position.
Substitute
and
for
for
Therefore, the radial component of velocity is
.
.
Comment
Step 2 of 3
(b)
To calculate tangential velocity, use the Pythagorean Theorem of a right triangle, with the
magnitude of the velocity as the hypotenuse, and the radial velocity as one leg.
Substitute
for
and
for vr
Therefore, the azimuthal component of velocity is
.
Comment
Step 3 of 3
(c)
The flight path angle,
, can be found by taking the arctangent of the ratio of the radial velocity,
, and the tangential velocity,
Substitute
for
.
and
Therefore, the flight path angle is
for
.
.
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 14p
Orbital Mechanics for Engineering Students
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If the specific energy ε of the two-body problem is negative, show that m2 cannot move outside a
sphere of radius µ/|ε| centered at m1.
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The orbital specific energy
is,
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Here,
is the speed of satellite,
is the gravitational parameter of the object and
is the
distance from the center of the orbit to the satellite.
Comment
Step 2 of 2
If the specific energy of the orbit is negative then,
Substitute,
for
,
The specific kinetic energy
is,
Or
Substitute,
for
Thus the radius of
in
,
is always less than
when specific energy is negative
.
Comments (4)
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 15p
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Chapter 2, Problem 15P
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Relative to a nonrotating Cartesian coordinate frame with the origin at the center O of the earth, a
spacecraft in a rectilinear trajectory has the velocity
when its distance
from O is 10,000 km. Find the position vector r when the spacecraft comes to rest.
My Textbook Solutions
Step-by-step solution
Step 1 of 2
Use the specific energy of the orbit and the magnitude of position vector to find the position
vector.
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The specific energy of the orbit can be determined from the magnitude of position vector, r, and
the magnitude velocity vector, v.
The magnitude of the velocity vector is,
Substitute
for
The specific energy is,
Substitute
for
,
for
, and 10000 km for
.
Comment
Step 2 of 2
The specific energy of the orbit is constant. Solve the specific energy equation for position, r for
speed 0 km/s.
Rearrange equation for
.
To calculate the position vector, calculate the new magnitude by the unit vector vr.
Substitute
for
,
for v, and
Therefore, the position vector is
for v,
.
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 16p
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Chapter 2, Problem 16P
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The specific angular momentum of a satellite in circular earth orbit is 60,000 km2/s. Calculate the
period.
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The position r of an object revolving in an orbit of eccentricity e can be given as follows:
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Here, h is the angular momentum,
is the gravitational parameter, and
is the angle between
the eccentricity vector and the position vector.
The relation between the time period and position vector of an object revolving in an orbit can be
given as follows:
Here, r is the position and
is the gravitational parameter.
Comment
Step 2 of 4
The gravitational parameter can be given as follows:
Here, G is the gravitational constant, M is the object around which the object of mass m is
orbiting.
Comment
Step 3 of 4
The satellite is orbiting in a circular orbit such that the eccentricity of the orbit is 0.
Substitute 0 for e in the expression
.
Now,
The time period of the satellite can be given as follows:
Substitute
for r in the above expression.
Comment
Step 4 of 4
The gravitational parameter can be given as follows:
The mass of the satellite is very less as compared to the mass of the Earth so that it can be
neglected.
Thus, the gravitational parameter can be written as follows:
Substitute
for G and
Substitute
for h and
Therefore, the period is
for M in the above expression.
for
in the expression
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.
.
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 17p
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A spacecraft is in a circular orbit of Mars at an altitude of 200 km. Calculate its speed and its
period.
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Use the radius of the satellite’s orbit, velocity in a circular orbit, and the gravitational parameter to
find the orbital period.
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The radius of the satellite’s orbit is,
Here, z is the height of the orbit,
Substitute
for
is the radius of the Mars.
and
for
.
The velocity in a circular orbit is related to the radius; r, and gravitational parameter, µ is,
Substitute
for G,
for M, and
for
.
Comment
Step 2 of 2
Kepler’s third law states the square of the period of the revolutions is directly proportional to cube
of the semi major axis. For a satellite orbiting an asteroid having orbital radius
period
Here,
, and orbital
, Kepler’s third law is given as:
is universal gravitational constant, and
is the mass of the asteroid.
Rewrite the equation for T.
Substitute
for G,
for M, and
for
.
Convert the orbital period from seconds to hours.
Therefore, the orbital period is
.
Comments (2)
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 19p
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Chapter 2, Problem 19P
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Determine the true anomaly θ of the point(s) on an elliptical orbit at which the speed equals the
speed of a circular orbit with the same radius, that is, vellipse = vcircle.
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Step-by-step solution
Step 1 of 3
The velocity of a satellite in a circular orbit is,
Here,
is gravitational parameter and
is the radius of circle.
And the specific energy of an elliptical orbit
,
Also
Here,
is the semi major axis of ellipse and
is the radius of rotation as shown in figure.
Comment
Step 2 of 3
Equate the both expressions of specific energy
,
As given in problem,
Substitute,
for
and
for
,
Comment
Step 3 of 3
Consider the following formula,
Here,
is eccentricity.
Substitute,
for
,
Solve further,
Thus the value of true anomaly
is
.
Comments (2)
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 19p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 19P
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Problem
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Determine the true anomaly θ of the point(s) on an elliptical orbit at which the speed equals the
speed of a circular orbit with the same radius, that is, vellipse = vcircle.
My Textbook Solutions
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Step-by-step solution
Step 1 of 3
The velocity of a satellite in a circular orbit is,
Here,
is gravitational parameter and
is the radius of circle.
And the specific energy of an elliptical orbit
,
Also
Here,
is the semi major axis of ellipse and
is the radius of rotation as shown in figure.
Comment
Step 2 of 3
Equate the both expressions of specific energy
,
As given in problem,
Substitute,
for
and
for
,
Comment
Step 3 of 3
Consider the following formula,
Here,
is eccentricity.
Substitute,
for
,
Solve further,
Thus the value of true anomaly
is
.
Comments (2)
Was this solution helpful?
40
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 20p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 20P
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Calculate the flight path angle at the locations found in Problem 2.19.
My Textbook Solutions
Step-by-step solution
Step 1 of 4
The velocity of a satellite in a circular orbit,
is a function of the gravitation parameter of the
body being orbited, µ, and the radius of the orbit, r.
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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The specific energy of an elliptical orbit, e, is negative and is a function of the gravitational
parameter, µ, and the semi-major axis.
Substituting this value into the specific energy equation allows the velocity at any point in the
orbit,
to be calculated from the semi-major axis, the distance from the focus, r, and the
gravitational parameter of the body being orbited.
Comment
Step 2 of 4
To find the radius where the velocity in a circular orbit,
elliptical orbit,
is the same as the velocity in an
set the two equations equal to each other.
Now, solve for r.
Comment
Step 3 of 4
The radius, r, at any true anomaly, ?, is based on the semi-major axis of the ellipse, a, and the
eccentricity of the orbit, e.
When the velocity of an elliptical orbit,
is the same as the velocity of a circular orbit,
the radius r is equal to the semi-major axis.
Now solve for ?.
Comment
Step 4 of 4
The tangent of the flight path angle ? is related to the eccentricity, e, and the true anomaly, ?.
Take the inverse tangent of each side to solve for the flight path angle, ?.
Comment
Was this solution helpful?
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 21p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 21P
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An unmanned satellite orbits the earth with a perigee radius of 10,000 km and an apogee radius
of 100,000 km. Calculate (a) the eccentricity of the orbit; (b) the semimajor axis of the orbit
(kilometers); (c) the period of the orbit (hours); (d) the specific energy of the orbit (kilometers
squared per seconds squared); (e) the true anomaly at which the altitude is 10,000 km (degrees);
My Textbook Solutions
(f) vr and v⊥ at the points found in part (e) (kilometers per second); and (g) the speed at perigee
and apogee (kilometers per second).
Step-by-step solution
Orbital
Mechanics...
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Step 1 of 9
(a)
Express the relation for eccentricity of elliptical orbit.
Here, e is the eccentricity,
Substitute
is the radius of perigee and
for
and
for
is radius of apogee.
to find e.
Therefore, the eccentricity of elliptical orbit is
Comment
Step 2 of 9
(b)
Express the relation for semi major axis of the orbit.
Here, a is the semi major axis of the orbit
Substitute
for
is the radius of perigee and
and
for
is radius of apogee
to find a.
Therefore, the semi major axis of the orbit is
.
Comment
Step 3 of 9
(c)
Express the relation for time period of orbit
Here,
is the gravitational parameter of the orbit.
Substitute
for a and
for
Therefore, period of orbit is
to find T.
.
Comment
Step 4 of 9
(d)
Express the relation for Specific energy of the orbit.
Here,
is the specific energy of the orbit, a is the semi major axis of the orbit,
is the
gravitational parameter of the orbit.
Substitute
for
and
for a to find
Therefore, the specific energy of the orbit is
.
.
Comment
Step 5 of 9
(e)
Express the relation for the distance to the focus.
Here, r is the distance of the focus, a is the semi major axis of the orbit, e is the eccentricity of
the orbit and
is the true anomaly
Rewrite the expression for true anomaly.
Substitute
for a,
for e, and
Therefore, true anomaly at
for r to find
is
.
.
Comments (4)
Step 6 of 9
(f)
Express the relation for radial velocity
Here,
is the radial velocity and
is the gravitational parameter.
Rewrite the expression for h.
Use the relation for h to rewrite the radial velocity expression.
Substitute
for
,
for
Therefore, the radial velocity is
,
for
, and
for e to find
.
Comments (6)
Step 7 of 9
Express the relation for azimuthal velocity.
Substitute
for
,
for
Therefore, the azimuthal velocity is
,
for
, and
for e.
.
Comments (2)
Step 8 of 9
(g)
Express the relation for the velocity at perigee
Substitute
for
,
for e, and
for
to find
.
Therefore, the speed at perigee is
Comment
Step 9 of 9
Express the relation for the velocity at apogee
Here, e is the eccentricity,
Substitute
to find
for
is the gravitational parameter and
,
for
,
is the radius at apogee.
for e, and
for
.
Therefore, the speed at apogee is
.
Comment
Was this solution helpful?
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16
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 22p
Orbital Mechanics for Engineering Students
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homework questions
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Chapter 2, Problem 22P
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A spacecraft is in a 400-km by 600-km LEO. How long (in minutes) does it take to coast from the
perigee to the apogee?
My Textbook Solutions
Step-by-step solution
Step 1 of 4
The expression for the radius at perigee is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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2nd Edition
6th Edition
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Here,
is the minimum altitude,
Substitute 400 km for
is the radius of the Earth.
and
for
,
The radius at apogee is,
Here,
is the maximum altitude,
Substitute 600 km for
is the radius of the Earth.
and
for
,
Comment
Step 2 of 4
The semi-major axis of the orbit is,
Substitute
for
and
for
Comment
Step 3 of 4
The expression for the period of the orbit is,
Here, µ is the gravitational parameter of the Earth.
Substitute
for
and
for
.
Comment
Step 4 of 4
The time it takes to get from perigee to apogee
apogee to perigee
, is the same as the time it takes to get from
.
The total orbital period is equal to the sum of the time it takes to get from perigee to apogee and
the time it takes to return.
Rewrite the equation for time from perigee to apogee,
Substitute
for T.
Therefore, the time from perigee to apogee,
is
.
Comment
Was this solution helpful?
15
3
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 23p
Orbital Mechanics for Engineering Students
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homework questions
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Chapter 2, Problem 23P
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The altitude of a satellite in an elliptical orbit around the earth is 2000 km at apogee and 500 km
at perigee. Determine (a) the eccentricity of the orbit; (b) the orbital speeds at perigee and
apogee; and (c) the period of the orbit.
My Textbook Solutions
Step-by-step solution
Step 1 of 5
(a)
Orbital
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s of...
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6th Edition
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The expression for the radius at perigee is,
Here,
is the minimum altitude,
Substitute 500 km for
is the radius of the Earth.
and
for
,
The radius at apogee is,
Here,
is the maximum altitude,
Substitute 2600 km for
is the radius of the Earth.
and
for
,
Comment
Step 2 of 5
The eccentricity of the orbit is,
Comment
Step 3 of 5
Substitute
for
and
for
.
Therefore the eccentricity of the orbit is
.
Comment
Step 4 of 5
(b)
The expression for the angular momentum h is,
Here, e is the eccentricity and µ is the gravitational parameter of the Earth.
The velocity at perigee
is equal to the angular momentum h divided by the radius at perigee
.
Substitute
for
Substitute
for
.
,
for
,
Therefore, the velocity at perigee is
for e.
.
Comment
Step 5 of 5
The velocity at apogee
is equal to the angular momentum h divided by the radius at apogee
.
Substitute
for
Substitute
for
.
,
for
,
Therefore, the velocity at apogee is
for
,
for e.
.
(c)
Write the expression forte the semi-major axis of the orbit in terms of radius at perigee and radius
at apogee.
Substitute
for
and
for
.
The expression for the period of the orbit is,
Here, µ is the gravitational parameter of the Earth.
Substitute
for
and
for
Therefore, the time period of the orbit is
.
.
Comment
Was this solution helpful?
17
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 24p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 24P
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Problem
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A satellite is placed into an earth orbit at perigee at an altitude of 500 km with a speed of 10
km/s. Calculate the flight path angle γ and the altitude of the satellite at a true anomaly of 120°.
My Textbook Solutions
Step-by-step solution
Step 1 of 5
The expression for the is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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2nd Edition
6th Edition
View all solutions
Here,
is the minimum altitude and
is the radius of the Earth.
Comment
Step 2 of 5
Substitute 500 km for
and
for
in the equation
.
The expression for the angular momentum is,
Here,
is the velocity at perigee and.
Substitute
for
and
for
.
Comment
Step 3 of 5
The expression for radius at perigee is given by,
Here, h is the angular momentum,
is the gravitational parameter and e is the eccentricity.
Rewrite the equation for e.
Substitute
for
,
for
and
for
.
Comment
Step 4 of 5
The relation between flight path angle, eccentricity e and is,
Here, ? is the true anomaly and e is the eccentricity.
Take the arctangent of both sides to solve for ?.
Substitute
for
and
for
Therefore, the path angle is
.
.
Comment
Step 5 of 5
The mathematical expression for the radius is,
Here, h is the angular momentum, ? is the true anomaly,
is the gravitational parameter and e
is the eccentricity.
Substitute
for
,
for
,
for
and
for
.
The height of the satellite above the Earth surface is,
Here, r is the radius and
Substitute
is the radius of the earth.
for r and
for
.
Therefore, the altitude of the satellite is
.
Comment
Was this solution helpful?
6
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 25p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 25P
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A satellite is launched into the earth’s orbit at an altitude of 1000 km with a speed of 10 km/s and
a flight path angle of 15°. Calculate the true anomaly of the launch point and the period of the
orbit.
My Textbook Solutions
Step-by-step solution
Step 1 of 6
The expression for the radius at perigee is,
Here, z is the altitude,
Orbital
Mechanics...
Orbital
Mechanics...
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s of...
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2nd Edition
6th Edition
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is the radius of the Earth.
Substitute 6378 km for
, and
for z.
The expression for the azimuthal velocity
Substitute 10 km/s for v, and
in terms of speed v, and the flight path angle, ? is,
for
Comment
Step 2 of 6
The angular momentum, h, is the product of radius r and the azimuthal velocity
Substitute
for r, and
.
for
Write the expression for the specific energy is,
Here,
is the earth’s gravitational parameter and v is the speed.
Substitute 10 km/s for v,
for r and
for
.
Comment
Step 3 of 6
The expression for the specific energy of the orbit, e in terms of semi-major axis of the orbit a
and earth’s gravitational parameter
, is,
Rewrite the equation for a.
Substitute
for
, and
for
.
Comment
Step 4 of 6
The eccentricity of the orbit, e, can be determined from the semi-major axis, a, gravitational
parameter, µ, and the angular momentum, h.
Rewrite the equation for e.
Substitute
for h,
for a, and
for
.
Comment
Step 5 of 6
The expression for the distance to the focus is,
Here, a is the semi-major axis, e is the eccentricity, and ? is the true anomaly.
Rewrite the equation for the true anomaly, ?.
Substitute
for a,
for e,
for r.
Therefore, the true anomaly launch point is
.
Comment
Step 6 of 6
Kepler’s third law states the square of the period of the revolutions is directly proportional to cube
of the semi major axis. For a satellite orbiting an asteroid having orbital radius a, and time period
, Kepler’s third law is given as:
Here,
is universal gravitational constant, and
is the mass of the asteroid.
Rewrite the equation for T.
Substitute
for
, and
for a
Therefore, the time period of the orbit is
.
Comment
Was this solution helpful?
4
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 26p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 26P
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Problem
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A satellite has perigee and apogee altitudes of 500 and 21,000 km. Calculate the orbit period,
eccentricity, and the maximum speed.
My Textbook Solutions
Step-by-step solution
Step 1 of 5
The expression for the radius at perigee is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
View all solutions
Here, z is the sum of the altitude at perigee,
Substitute
for
, and
is the radius of the Earth.
for
.
The radius at apogee, ra, is the sum of the altitude at apogee ( za) and the radius of the Earth
.
Substitute
for
, and
for
.
Write the expression forte the semi-major axis of the orbit in terms of radius at perigee and radius
at apogee.
Substitute
for
and
for
.
Comment
Step 2 of 5
The expression for the orbital period (T) is,
Here, µ is the gravitational parameter, a is the semi-major axis.
Comment
Step 3 of 5
Substitute
for
, and
Therefore, the orbital period is
for
.
.
Comment
Step 4 of 5
The expression for the eccentricity is,
Substitute
for
, and
for
.
Therefore, the eccentricity of the elliptical path is
.
Comment
Step 5 of 5
The expression for the angular momentum h is,
Here, e is the eccentricity and µ is the gravitational parameter of the Earth.
The velocity at perigee is maximum velocity, vmax.
Substitute
for h.
Substitute
for
,
Therefore, the maximum speed is
for e, and
for
.
.
Comment
Was this solution helpful?
10
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 27p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 27P
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A satellite is launched parallel to the earth’s surface with a speed of 7.6 km/s at an altitude of 500
km. Calculate the period.
My Textbook Solutions
Step-by-step solution
Step 1 of 4
The expression for the radius at perigee is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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2nd Edition
6th Edition
View all solutions
Here, z is the sum of the altitude at perigee,
is the radius of the Earth.
Substitute
.
for
, and
for
The angular momentum (h) is the product of the radius at perigee
and velocity at perigee
.
Substitute
for
, and
for
.
Comment
Step 2 of 4
The expression for the eccentricity is,
Here,
is the radius at perigee, µ is the gravitational parameter of the Earth, and h is the
angular momentum.
Substitute
for
,
for
, and
for
.
The expression for radius at apogee (ra) in terms of radius at perigee (rp) and the eccentricity (e)
is,
Substitute
for
,
for e.
Comment
Step 3 of 4
The altitude at apogee is the difference between the radius at apogee and the radius of the
Earth.
Substitute
for
,
for
.
The semi-major axis, a, is the average of the radius at perigee, rp, and the radius at apogee, ra.
Substitute
for
, and
for
.
Comment
Step 4 of 4
The orbital period is related to the semi-major axis, a, and the gravitational parameter of the
Earth, µ as:
Substitute
for
, and
for
.
Convert the orbital period, T, into hours.
Therefore, the orbital period is
.
Comment
Was this solution helpful?
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 28p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 28P
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Problem
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A satellite in orbit around the earth has a speed of 8 km/s at a given point of its orbit. If the period
is 2 h, what is the altitude at that point?
My Textbook Solutions
Step-by-step solution
Step 1 of 3
The expression for semi-major axis is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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2nd Edition
6th Edition
View all solutions
Here, µ is the gravitational parameter, T is the orbital period.
The orbital time period is
. Convert the units of the orbital period from h to s.
Substitute
for
, and
for T in equation
.
Comment
Step 2 of 3
The specific energy of an elliptical orbit, e, is related to the gravitational parameter of the Earth,
µ, and the semi-major axis of the orbit, a as:
Substitute
for
,
for
.
The specific energy of the orbit is the same for the entire orbit. This means that the position can
be determined from specific energy, e, and the velocity at that point, vp.
Rewrite the equation for position, rp.
Substitute
for
,
for
,
for
.
Comment
Step 3 of 3
The radius at perigee (rp) is the sum of the altitude at perigee (zp) and the radius of the Earth
.
Rearrange for altitude at perigee.
Substitute
for
,
for
Therefore, the altitude at perigee is
.
.
Comment
Was this solution helpful?
13
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 29p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 29P
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A satellite in polar orbit around the earth comes within 200 km of the North Pole at its point of
closest approach. If the satellite passes over the pole once every 100 min, calculate the
eccentricity of its orbit.
My Textbook Solutions
Step-by-step solution
Step 1 of 4
The expression for the radius at perigee is,
for
, and
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
View all solutions
Here, z is the sum of the altitude at perigee,
Substitute
Orbital
Mechanics...
is the radius of the Earth.
for
.
The orbital period, T, is the time the satellite takes to make one full orbit and reach the same
position again. If the satellite is over the pole every 100 minutes, then the orbital period is 100
minutes.
Convert the units of time from min to s.
Comment
Step 2 of 4
The expression for semi-major axis is,
Here, µ is the gravitational parameter, T is the orbital period.
Substitute
for
, and
for T.
Comment
Step 3 of 4
The semi-major axis of the orbit, a, is the average of radius at perigee (rp), and the radius at
apogee (ra).
Rewrite the equation for
Substitute
.
for a,
for
.
Comment
Step 4 of 4
The expression for the eccentricity is,
Substitute
for
, and
for
.
Therefore, the eccentricity of the elliptical path is
.
Comment
Was this solution helpful?
13
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 30p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 30P
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Problem
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The following position data for an earth orbiter are given:
Altitude = 1000 km at a true anomaly of 40°.
Altitude = 2000 km at a true anomaly of 150°.
My Textbook Solutions
Calculate (a) The eccentricity, (b) the perigee altitude (kilometers), and (c) the semimajor axis
(kilometers).
Step-by-step solution
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s of...
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Step 1 of 4
(a)
The radius, r, at any true anomaly, (?) is equal to the altitude at that true anomaly (z), and the
radius of the Earth (earth).
Substitute
for
Substitute
for
,
for
,
,
for
for
,
.
for
.
The expression for the angular momentum h is,
Here, e is the eccentricity, r is the radius, ? is the true anomaly, and µ is the gravitational
parameter of the Earth.
Take square on both sides of the equation.
Comment
Step 2 of 4
The angular momentum is conserved for an elliptical orbit. So we can write as,
Substitute in the calculations of angular momentum.
Rewrite the equation for e.
Substitute
for
, and
Therefore, the eccentricity is
for
.
.
Comment
Step 3 of 4
(b)
The expression for the angular momentum h is,
Here, e is the eccentricity,
is the radius at perigee, ? is the true anomaly, and µ is the
gravitational parameter of the Earth.
Substitute
for
,
for
,
for
.
Comment
Step 4 of 4
The true anomaly, ?, is 0°. This allows the radius at perigee, rp, to be calculated from the angular
momentum, h, and the gravitational parameter of the Earth, µ, and the eccentricity, e.
Substitute
for
,
for e, and
for
.
The radius at perigee (rp) is the sum of the altitude at perigee (zp) and the radius of the Earth
.
Rearrange for altitude at perigee.
Substitute
for
,
for
Therefore, the altitude at perigee is
.
.
(c)
When the satellite is at apogee, ra, the true anomaly, ?, is 180°. The expression for radius at
apogee, ra is,
Substitute
for
,
for e, and
for
.
The semi-major axis, a, is the average of the radius at perigee, rp, and the radius at apogee, ra.
Substitute
for
, and
Therefore, the semi major axis is
for
.
.
Comment
Was this solution helpful?
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 31p
Orbital Mechanics for Engineering Students
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homework questions
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Chapter 2, Problem 31P
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Problem
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An earth satellite has a speed of 7.5 km/s and a flight path angle of 10° when its radius is 8000
km. Calculate (a) the true anomaly (degrees) and (b) the eccentricity of the orbit.
My Textbook Solutions
Step-by-step solution
Step 1 of 4
(a)
The expression for azimuthal velocity is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
View all solutions
Here, v is the total velocity; ? is the flight path angle.
Substitute
for v,
for
.
The angular momentum (h) is the product of the azimuthal velocity
Substitute
for
, and
for
and position (r).
.
The expression for energy of the orbit is,
Here, µ is gravitational parameter of the Earth, r is the position of the satellite, v is the velocity of
the satellite.
Substitute
for v,
for
, and
for r.
Comment
Step 2 of 4
The expression for semi-major axis is,
Substitute
for
, and
for
.
The expression for the eccentricity of the orbit is,
Substitute
for h,
The eccentricity is
for
, and
for a.
.
Comment
Step 3 of 4
The orbit equation is,
Rewrite for angle (?).
Substitute
for h,
for
, and
for e, and
for
r.
Therefore, the angle is
.
Comment
Step 4 of 4
(b)
The value of eccentricity is calculated in the part (a). The eccentricity is
.
Comment
Was this solution helpful?
7
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 32p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 32P
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If, for an earth satellite, the specific angular momentum is 70,000 km2/s and the specific energy
is − 10 km2/s2, calculate the apogee and perigee altitudes.
My Textbook Solutions
Step-by-step solution
Step 1 of 4
Use the gravitational parameter of the Earth and the angular momentum to find the eccentricity.
The expression for the specific energy is,
Here,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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2nd Edition
6th Edition
View all solutions
is the semi-major axis and µ is the gravitational parameter of the Earth.
Rewrite the equation for specific energy.
Consider the magnitude only.
Substitute
for
,
for
.
Comment
Step 2 of 4
The expression for the angular momentum h is,
Here, e is the eccentricity, r is the radius, ? is the true anomaly, and µ is the gravitational
parameter of the Earth.
Take square on both sides.
Rewrite the equation for e.
Substitute
for h,
for
, and
for a.
Comment
Step 3 of 4
The radius at apogee ra can be found from the semi-major axis, a, and the eccentricity, e.
Substitute
for a, and
for e.
The altitudes at apogee za is,
Substitute
for
, and
for
.
Therefore, the altitude at apogee za is
.
Comment
Step 4 of 4
The radius at perigee rp, can be found from the semi-major axis, a, and the eccentricity, e.
Substitute
for
and
The altitudes at perigee,
Substitute
for e.
is,
for
, and
Therefore, the altitude at perigee
for
is
.
.
Comment
Was this solution helpful?
10
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 33p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 33P
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A rocket launched from the surface of the earth has a speed of 7 km/s when the powered flight
ends at an altitude of 1000 km. The flight path angle at this time is 10°. Determine the
eccentricity and the period of the orbit.
My Textbook Solutions
Step-by-step solution
Step 1 of 5
Use the angular momentum, semi-major axis, and gravitational parameter of the Earth to find the
eccentricity of the orbit.
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
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2nd Edition
6th Edition
View all solutions
The radius of the satellite is the sum of the altitude of the satellite and the radius of the Earth.
Here, z is the altitude and
Substitute
is the radius of the Earth.
for z, and
for
.
The azimuthal velocity in terms of velocity of the satellite (v), and the flight path angle (?), is,
Substitute
for v, and
for
.
The relation between angular momentum (h), the distance of the satellite from the focus (r), and
azimuthal velocity, (
) is,
Substitute
for
, and
for r.
Comment
Step 2 of 5
Write the expression for the specific energy is,
Here,
is the earth’s gravitational parameter, v is the speed, and r is the distance between the
satellite and the focus of the orbit.
Substitute
for v,
for
, and
for r.
The specific energy can also be used to calculate the semi-major axis of the orbit.
Solve this equation for a.
Substitute
for
,
for
.
Comment
Step 3 of 5
The semi-major axis of the orbit, a, is also dependent on the angular momentum, h, the
gravitational parameter of the Earth, µ, and the eccentricity, e.
Solve the semi-major axis equation to get the eccentricity.
Substitute
for
,
for a, and
Therefore, the eccentricity of the orbit is
for
.
.
Comment
Step 4 of 5
The period of the orbit (T) can be found from the semi-major axis of the orbit (a) and the
gravitational parameter of the Earth (µ).
Substitute
for
, and
for a.
Comment
Step 5 of 5
Convert the units of the period from seconds to minutes.
Therefore, the time period of the orbit after rounding off to three significant figures is
.
Comment
Was this solution helpful?
4
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 34p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 34P
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Problem
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If the perigee velocity is c times the apogee velocity, calculate the eccentricity of the orbit in
terms of c.
My Textbook Solutions
Step-by-step solution
Step 1 of 1
The velocity at perigee, vp, is some constant, c, time the velocity at apogee, va.
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The velocity at perigee, vp, in terms of angular momentum, h, and the radius at perigee, rp is,
The velocity at apogee, va, in terms of angular momentum, h, and the radius at apogee, ra is,
Now re-write the initial condition in terms of angular momentum, h, and radii at perigee and
apogee, rp and ra.
Solve for the ratio of the radius at apogee, ra, to radius at perigee, rp.
The ratio of radius at perigee, rp, to radius at apogee, ra, can also be expressed in terms of the
eccentricity, e.
Now substitute the constant, c for
.
Solve the expression to get the eccentricity, e, in terms of the constant, c.
Therefore, the eccentricity is
.
Comments (1)
Was this solution helpful?
3
1
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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Chapter 2, Problem 35P
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At what true anomaly does the speed on a parabolic trajectory equal α times the speed at the
periapsis, where α ≤ 1?
My Textbook Solutions
Step-by-step solution
Step 1 of 3
The distance to the focus, r, is dependent upon the angular momentum, h, the gravitational
parameter of the Earth, μ, the eccentricity of the orbit, e, and the true anomaly, θ.
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In case of parabolic trajectory, the eccentricity is equal to 1. The orbit equation can be given as
follows:
The speed at any point on the parabolic trajectory can be given as follows:
Here,
is the gravitational parameter and r is the distance to the focus.
Comment
Step 2 of 3
The position on the parabolic trajectory can be given as follows:
The speed at the periapsis on a parabolic trajectory is given as follows:
At periapsis, the position on the parabolic trajectory can be given as follows:
At periapsis, the true anomaly is 0 degrees.
Substitute 0 degrees for
in the above expression.
Comment
Step 3 of 3
The speed at the parabolic trajectory is
Substitute
for v and
Substitute
times the speed at the periapsis.
for v’ in the above expression.
for r’ and
Thus, the true anomaly is
for r in the above expression.
.
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 36p
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Chapter 2, Problem 36P
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What velocity, relative to the earth, is required to escape the solar system on a parabolic path
from earth’s orbit?
My Textbook Solutions
Step-by-step solution
Step 1 of 2
The Earth makes one complete orbit in one year, so ? can be calculated from the radians in a
revolution divided by the seconds in a year.
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The two expressions of Earth’s orbital velocity, v, can be used to solve for the radius of Earth’s
orbit, r.
Rearrange above equation and solve for
.
Substitute
and
for
.
Comment
Step 2 of 2
The velocity of Earth’s orbit, v, can be determined from the gravitational parameter of the Sun, µ,
and the radius of Earth’s orbit, r.
The velocity needed to escape an orbit can be rewritten in terms of the velocity of Earth’s orbit, v.
Substitute
for
The necessary change in velocity
geosynchronous orbit velocity
Substitute
for
is the difference between the escape velocity
, and
.
and
for
.
Thus, the speed required to escape from the solar system is
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 37p
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Chapter 2, Problem 37P
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A hyperbolic earth departure trajectory has a perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the hyperbolic excess speed (kilometers per second). (b) Find the radius
(kilometers) when the true anomaly is 100°. (c) Find vr and v⊥ (kilometers per second) when the
My Textbook Solutions
true anomaly is 100°.
Step-by-step solution
Step 1 of 6
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(a)
Use the specific energy of the trajectory and the perigee radius to find the hyperbolic excess
speed.
The perigee radius
, can be found from the perigee altitude
, and the radius of the Earth
.
Substitute
for
and
for
.
The expression for the specific energy of the trajectory in terms of perigee velocity is,
Here,
is the perigee velocity,
is the gravitational parameter of the Earth, and
is the
perigee radius.
Substitute
for
,
for
, and
for
.
Comment
Step 2 of 6
The specific energy of the trajectory (e) can be written in terms of the excess velocity
is,
Thus, the hyperbolic excess speed is,
Substitute
for
.
Therefore, the hyperbolic excess speed is
.
Comments (2)
Step 3 of 6
(b)
The angular momentum of the trajectory in terms of radius at perigee and the velocity at perigee
is,
Substitute
for
and 11 km/s for
.
The radius at perigee, rp, can also be expressed in terms of the angular momentum, h, the
gravitational parameter of the Earth, µ, and the eccentricity of the orbit, e.
Solve this relation for the eccentricity, e.
Substitute
for
,
for
, and
for
.
The distance of the trajectory from the center of the Earth is,
Here, h is the function of angular momentum,
is the gravitational parameter of the Earth,
is
the true anomaly, and e is the eccentricity.
Comment
Step 4 of 6
Substitute
for
,
for
Therefore, the radius when the true anomaly
, and
is
for e.
.
Comment
Step 5 of 6
(c)
The azimuthal velocity at a true anomaly of 100°, v?, is,
Substitute
for
and
for r.
Therefore, the perpendicular component of velocity is
.
Comment
Step 6 of 6
The radial velocity is,
Substitute
for
,
Therefore, the radial velocity is
for h,
for e, and
for
.
.
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 38p
Orbital Mechanics for Engineering Students
(3rd Edition)
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A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the
earth with a true anomaly of 150°. If the speed of the meteoroid at that time is 2.23 km/s,
calculate (a) the eccentricity of the trajectory; (b) the altitude at closest approach; and (c) the
My Textbook Solutions
speed at the closest approach.
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Step-by-step solution
Step 1 of 7
The orbit formula is given as,
Here,
the eccentricity of the orbit,
anomaly,
is the radius,
is the true
is the distance between the earth and the approaching particle.
The specific energy of the trajectory
Here,
is the semi-major axis and
is,
is the gravitational parameter
And specific energy of the trajectory is also given by,
Here,
is velocity of satellite.
Comment
Step 2 of 7
Perigee altitude is given as,
Here,
is the radius at perigee and
is radius of earth,
is perigee altitude.
Comment
Step 3 of 7
The radius at perigee is given by,
Here,
is the specific angular momentum.
Comment
Step 4 of 7
The radius at perigee is also given by,
(a)
Specific energy of the trajectory is given by,
Here,
is velocity of meteoroid ,
is the gravitational parameter,
is distance between the
earth and meteoroid.
Substitute
for
,
for
, and
for
in the above
expression,
Specific energy of the trajectory is also given by,
Rearrange the above expression for
Substitute
for
,
and
for
in the above expression.
Orbit formula is given as,
Rearrange this expression,
Substitute
for
,
for
, and
for
,
……. (1)
Simple quadratic equation is given by,
From the comparison of the equation with the equation (1).
,
,
Solve the quadratic equation using the following formula,
Comments (2)
Step 5 of 7
Substituting
for
,
for
,
for
First considering the positive sign for calculating
,
for
in the above expression.
,
This value is negative so it is to be discarded.
Now, considering the negative sign for calculating
,
This value is negative so it is to be discarded.
Hence, The acceptable value of eccentricity of the trajectory is
.
Comment
Step 6 of 7
(b)
The radius of the trajectory at perigee is
Substitute
for
and 1.086 for
.
Perigee altitude is given by,
Substitute 11465.434 for
and 6378 km for
in the above expression.
Hence, the value of altitude at closest approach is
.
Comment
Step 7 of 7
(c)
Radius of the perigee is given by,
Rearrange the above expression for
Substitute
for
.
,
for
, and 1.806 for e,
The velocity at perigee can be given as,
Here,
is velocity of the perigee.
Substitute
for
and
for rp.
Hence, the speed at closest approach is
.
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 39p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 39P
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If α is a number between 1 and
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, calculate the true anomaly at which the
speed on a hyperbolic trajectory is α times the hyperbolic excess speed.
My Textbook Solutions
Step-by-step solution
Step 1 of 2
The distance to the focus, r, is dependent upon the angular momentum, h, the gravitational
parameter of the Earth, μ, the eccentricity of the orbit, e, and the true anomaly, θ.
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The orbit equation can be given as follows:
The hyperbolic access speed is the speed with which a body on a hyperbolic path arrives at
infinity. The expression of hyperbolic excess speed can be given as follows:
Here,
is the gravitational parameter and a is the half distance between periapsis and
apoapsis.
The expression of escape speed can be given as follows:
Here,
is the gravitational parameter and r is the distance from the focus.
Comment
Step 2 of 2
For a hyperbolic trajectory,
Substitute
for
and
for
in the above expression.
According to the given condition,
Substitute
for
and
for
in the above expression.
Tae square on both sides.
Substitute
for r in the above expression.
Thus, the true anomaly is
.
Comment
Was this solution helpful?
0
0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 40p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 40P
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For a hyperbolic orbit, find the eccentricity in terms of the radius at periapsis rp and the
hyperbolic excess speed v∞.
My Textbook Solutions
Step-by-step solution
Step 1 of 2
Use the expression for the radius at periapsis and the expression for the hyperbolic excess
speed to find the expression for the eccentricity of hyperbola.
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The expression for the radius at periapsis is,
Here, a is the semi major axis of the hyperbola and e is the eccentricity.
Rewrite the equation for a.
The expression for the hyperbolic excess speed is,
Here,
is the gravitational parameter of the Earth.
Take square on both sides.
Rewrite the equation for a.
Comment
Step 2 of 2
Substitute
for
in the equation
.
Therefore, the eccentricity of hyperbola in terms of radius at periapsis and hyperbolic excess
speed is
.
Comment
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 41p
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Chapter 2, Problem 41P
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A space vehicle has a velocity of 10 km/s in the direction shown when it is 10,000 km from the
center of the earth. Calculate its true anomaly.
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Step-by-step solution
Step 1 of 3
The expression for specific energy is,
Here, v is the velocity,
is the gravitational parameter and
is the radius of trajectory.
The expression for specific energy can also be written as,
Here,
is the semi major axis.
The radius at periapsis is evaluated as,
Here,
is eccentricity.
The expression for radius at periapsis can also be written as,
Here,
is the specific relative angular momentum.
Comment
Step 2 of 3
Consider the following provided figure,
Here,
is the component of velocity perpendicular to position vector
component of velocity along the position vector
and
is the
.
Comment
Step 3 of 3
The velocity
of vehicle can be resolved into two components. The angle
is the angle
made by the velocity vector to the perpendicular on the trajectory as shown in the figure.
Component of velocity perpendicular to position vector
Substitute
for
and
for
is,
.
The radial component of velocity is,
Substitute
for
and
for
.
The specific angular momentum is,
Substitute
for v◊ and
for
.
for
, and
From the relation,
Substitute
for
,
for
.
From the relation,
Substitute
for
and
for
.
From the relations,
and,
Equate the both expressions of
Substitute
,
for
,
for
, and
for
.
From the equation (2.49),
Substitute
for
,
for
,
for
, and
for
.
Hence, the value of the true anomaly is
.
Comments (1)
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / advanced physics / advanced physics questions and answers / a spacecraft at a radius r has a speed v and a ight path angle …
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Q: A spacecraft at a radius r has a speed v and a ight path angle gamma. nd an expression for the eccentricity of its orbit.
A: See answer
Q: 2.42 A spacecraft at a radius r has a speed v and a ight path angle y. Find an expression for the eccentricity of its orbit.
Ansie = V1 + 0(0 – 2)cosy where = rv2/u;
A: See answer
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 43p
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Chapter 2, Problem 43P
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For an orbiting spacecraft, r = r1 when θ = θ1, and r = r2 when θ = θ2. What is the eccentricity?
My Textbook Solutions
Step-by-step solution
Step 1 of 2
The distance to the focus, r, is dependent upon the angular momentum, h, the gravitational
parameter of the Earth, μ, the eccentricity of the orbit, e, and the true anomaly, θ.
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Comment
Step 2 of 2
For an orbiting spacecraft, the distance to the focus
with true anomaly
can be written as
with true anomaly
can be written as
follows:
For an orbiting spacecraft, the distance to the focus
follows:
Divide equation
by equation
and solve for e.
On further solving,
Thus, the eccentricity can be written as follows:
Comment
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Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 44p
Orbital Mechanics for Engineering Students
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Chapter 2, Problem 44P
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Problem
Continue to post
At a given instant, a spacecraft has the position and velocity vectors
20 questions remaining
and
relative to an earth-centered nonrotating frame. (a) What is the position vector
after the true anomaly increases by 90°? (b) What is the true anomaly of the initial point?
My Textbook Solutions
Step-by-step solution
Step 1 of 12
Start by calculating the magnitude of the velocity, v, by taking the square root of the dot product
for
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
View all solutions
of the velocity vector, v, with itself.
Substitute
Orbital
Mechanics...
.
Comment
Step 2 of 12
Next calculate the distance to the focus¸ ro, by taking the square root of the dot product of the
position vector, r, with itself.
Substitute
for r.
Comment
Step 3 of 12
The velocity in the radial direction can be calculated by taking the dot product of the velocity
vector, v, with the position vector, r, and dividing by the magnitude of the position vector, ro.
Substitute
for
and
for r, and
for
.
Comment
Step 4 of 12
The azimuthal velocity, v?, is the square root of the difference of the squares of the magnitude of
the velocity, v, and the radial velocity, vr.
Substitute
for
and
for
Comment
Step 5 of 12
The angular momentum, h, is the azimuthal velocity, v?, multiplied by the magnitude of the
position vector, ro.
Substitute
for
and
for
.
Comment
Step 6 of 12
The position, r, after a change in true anomaly,
can be calculated from the angular
momentum, h, the gravitational parameter of the earth, µ, and the change in true anomaly,
Substitute
for
,
for
,
for
, and
for
.
.
Comment
Step 7 of 12
Use the gravitational parameter of the Earth, µ, the position after the true anomaly change, r, the
angular momentum, h, and the change in true anomaly, ?? to calculate the value of the first
Lagrange function, f.
Substitute
for
,
for
,
for
, and
for
.
Use the initial radius, ro, and the radius after the change in true anomaly, r, and the angular
momentum to calculate the values of the second Lagrange function, g.
Substitute
for
,
for
,
for
, and
for
.
Comment
Step 8 of 12
Calculate the final position vector from the values of the two Lagrange functions, f and g, the
initial position vector, ro, and the initial velocity vector¸ vo.
Substitute
for
,
for
,
for
The position vector after the true anomaly increased
, and
for
is
.
Comment
Step 9 of 12
(b) To calculate the initial true anomaly, ?o, start with the equation relating the initial position, ro,
to the angular momentum, h, gravitational parameter of the Earth, µ, the eccentricity, e, and the
initial true anomaly, ?o.
There are two unknowns in this relationship: the eccentricity, e, and the initial true anomaly, ?o.
Solve the relationship for those two quantities in terms of the others.
Substitute
for
,
for
, and
for
Comment
Step 10 of 12
Next, use the equation relating the radial initial velocity, vro, to the gravitational parameter, µ, the
angular momentum, h, the eccentricity, e, and the initial true anomaly, ?o.
Again, this equation has two unknowns: the eccentricity, e, and the initial true anomaly, ?o. Solve
the relationship for those two quantities.
Substitute
for
,
for
, and
for
Comment
Step 11 of 12
To calculate the eccentricity, add the square of
and
.
To solve for e, factor the square of e out of each term and use the Pythagorean identity with sin
and cos.
Now take the square root of both sides to get a value for the eccentricity, e.
Comment
Step 12 of 12
Finally, use the eccentricity in the initial radial velocity equation to solve for the initial true
anomaly, ?o.
Substitute
for
for
,
for
, and
for
, and
.
Thus, the true anomaly of initial point is
.
Comment
Was this solution helpful?
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 45p
Orbital Mechanics for Engineering Students
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Answers from our experts for your tough
homework questions
(3rd Edition)
Show all steps:
8 Bookmarks
Chapter 2, Problem 45P
Enter question
ON
Problem
Continue to post
20 questions remaining
Relative to an earth-centered, nonrotating frame the position and velocity vectors of a spacecraft
are
and
, respectively. (a)
Find the distance and speed of the spacecraft after the true anomaly changes by 82°. (b) Verify
My Textbook Solutions
that the specific angular momentum h and total energy ε are conserved.
Step-by-step solution
Step 1 of 8
The orbit equation when true anomaly changes by
Here,
is radius of orbit after some time,
gravitational parameter and
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
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is,
is the specific angular momentum,
is the
is the initial radius.
Specific relative angular momentum is,
And
Here,
is initial velocity.
The velocity of the spacecraft is,
Here,
and
is the time derivative of Lagrange coefficients.
The Lagrange coefficients are,
Comment
Step 2 of 8
(a)
The magnitude of
can be calculated as,
Substitute,
for
,
The magnitude of initial velocity vector is,
Substitute,
for
,
The initial velocity in the radial direction
Substitute,
is,
for
for
,
for
and
,
From orbit equation,
Substitute,
for
for
,
for
,
for
,
for
,
,
Thus the value of radius is
.
Comment
Step 3 of 8
From the relation,
Substitute,
for
for
,
for
,
for
,
for
and
,
Comment
Step 4 of 8
From the relation,
Substitute,
for
,
for
,
for
and
for
,
Comment
Step 5 of 8
From the relation,
Comment
Step 6 of 8
Substitute,
for
for
,
for
,
for
and
,
Magnitude of speed is,
Comment
Step 7 of 8
Therefore, the speed of spacecraft is
.
(b)
Let the specific energy before true anomaly change is
Substitute,
for
,
for
and
Let specific energy after true anomaly change is
Substitute,
for
,
, then from relation,
for
,
, then from relation,
for
and
for
,
Since the specific energy is the same with the initial position and velocity and with the final
position and velocity, hence energy is conserved.
Comment
Step 8 of 8
Final radial velocity is,
Substitute,
for
for
,
for
and
,
Azimuth velocity,
Substitute,
for
and
for
,
The final specific angular momentum is,
Substitute,
for
and
,
Since initial and final specific angular momentum are same, therefore, the specific angular
momentum,
, is conserved.
Comments (1)
Was this solution helpful?
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 46p
Orbital Mechanics for Engineering Students
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Answers from our experts for your tough
homework questions
(3rd Edition)
Show all steps:
Bookmark
Chapter 2, Problem 46P
Enter question
ON
Problem
Continue to post
20 questions remaining
Relative to an earth-centered, nonrotating frame the position and velocity vectors of a spacecraft
are
and
. (a) Find the position vector 10 min later. (b)
Calculate the change in true anomaly over the 10-min time span.
My Textbook Solutions
Step-by-step solution
Step 1 of 6
The initial radial velocity is,
Here,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
View all solutions
is the initial velocity vector,
is the initial position vector of spacecraft and
is the
magnitude of initial position of spacecraft.
Comment
Step 2 of 6
The initial azimuthal velocity is,
The specific angular momentum is,
Lagrange’s functions are,
And,
Here,
is the change in time and
is the gravitational parameter of earth.
The position vector of spacecraft is,
Change in true anomaly is,
Here,
is specific angular momentum.
Comment
Step 3 of 6
(a)
The magnitude of initial radius is,
Substitute,
for
in
.
The magnitude of velocity is,
Substitute
for
in
.
From the relation,
Substitute
for
,
for
, and
for
in
.
From the above calculation,
From the relation,
Substitute,
for
and
for
in
.
From the relation,
Substitute
for
and
for
in
.
Comment
Step 4 of 6
From the relation,
Substitute
For
for
,
for
,
for
,
for
and
,
And from the relation,
Substitute,
for
,
for
,
for
and
for
.
Comment
Step 5 of 6
From the relation of position,
Substitute,
for
in
,
for
,
for
and
for
.
Thus, the position vector of spacecraft is
.
Comment
Step 6 of 6
(b)
The magnitude of position vector is,
Substitute
for
in
.
From the relation,
Here,
is the change in true anomaly.
Substitute
for
in
,
for
,
for
and
for
.
Rounding off to three significant figures, the change in true anomaly after ten-minute is
.
Comment
Was this solution helpful?
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0
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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home / study / science / physics / calculus based physics / calculus based physics solutions manuals / orbital mechanics for engineering students / 3rd edition / chapter 2 / problem 47p
Orbital Mechanics for Engineering Students
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Answers from our experts for your tough
homework questions
(3rd Edition)
Show all steps:
6 Bookmarks
Chapter 2, Problem 47P
Enter question
ON
Problem
Continue to post
20 questions remaining
For the sun–earth system, find the distance of the L1, L2, and L3 Lagrange points from the center
of mass of the system.
My Textbook Solutions
Step-by-step solution
Step 1 of 5
Write the expression for the dimensionless mass ratio
is,
Orbital
Mechanics...
Orbital
Mechanics...
Fundamental
s of...
3rd Edition
2nd Edition
6th Edition
View all solutions
Here,
is the mass of the sun and
Substitute
for
is the mass of the Earth.
and
for
,
Write the expression for the elliptical orbit.
Substitute
for
.
Comment
Step 2 of 5
Plot the graph for the above function.
Comment
Step 3 of 5
It is observed that the graph of
has two crossings on the right. One point is at
.
The solutions are:
The distance from sun to Earth is
. Now, calculate the value of
as
. Now, calculate the value of
as
. Now, calculate the value of
as
follows,
Therefore, the value of
is
.
Comment
Step 4 of 5
The distance from sun to Earth is
follows,
Therefore, the value of
is
.
Comment
Step 5 of 5
The distance from sun to Earth is
follows,
Therefore, the value of
is
.
Comment
Was this solution helpful?
6
1
Recommended solutions for you in Chapter 2
Chapter 2, Problem 37P
Chapter 2, Problem 10P
A hyperbolic earth departure trajectory has a
perigee altitude of 250 km and a perigee speed of
11 km/s. (a) Calculate the...
Show that for any orbit.
See solution
See solution
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