Q.1 Define Gravitation. How is it different from gravity?
Q.2 State Newton’s law of gravitation. Express it in vector form
Q.3 What is “G” ? Mention its value, units and dimensions.
Q.4 Mention 5 important characteristics of gravitational force
Q.5 What is acceleration due to gravity? On what factors does
it depend?
(b) What is its value at the earth’s surface?
Q.6 Assuming the earth to be a uniform sphere of radius
6400Kms and density 5.5gm/cc, find the value of g on the
surface of earth
Q.7 Two metal spheres of mass M each are kept at a distance R.
The gravitational force b/w them is found to be F. What will be
the force b/w them if
a.
b.
c.
Mass of each is doubled while distance
b/w them is made four times
Mass of one of them is doubles while
distance is kept R itself
Both of them are kept in water at a
distance R
Q.8 The mass of a planet is 1020 kg while its radius is 5000kms.
What will be the gravitational force experienced by a
spacecraft of mass 200kg standing on its surface.
(b) What is the acceleration near the surface of this planet
Q.9 The acceleration due to gravity near the earth’s surface is
found to be 9.81m/s2. Radius of the earth is 6400 kms. Using
this data find the mass of earth.
Q.10 Using the mass of earth calculated in the previous
question, find the density of earth. In real life, is the density of
earth uniform?
Q.11 Derive the value of acceleration due to gravity at a height
h above the earth’s surface
Q.12 Derive the value of acceleration due to gravity at a depth d
below the earth’s surface
Q.13 How does the shape of earth affect the value of “g” ?
Q.14 How does the rotation of earth affect the value of “g” ?
Derive the value of acceleration due to gravity at a place having
latitude λ.
Q.15 With what angular velocity should the earth rotate so that
a person at the equator does not experience any gravity?
Q.16 At what height above the earth’s surface does the value of
“g” become one tenth of its value at the surface.
Q.17 At what height above the earth’s surface does the value of
“g” become 64% of its value on the surface.
Q.18 Find the percentage change in the weight of a body if it is
taken to a height equal to the earth’s radius
Q.19 At what depth below the earth’s surface does the value of
“g” become 50% of its value on the surface.
Q.20 Find the value of acceleration due to gravity at the top of
Mount Everest ( height 10 km )
Q.21 Determine the speed with which the earth should rotate
so that a person on the equator would weigh 1/5th as much as
present.
Q.22 Define Gravitational field. Mention its SI unit.
Q.23 What is gravitational potential? Mention its units and
dimensions. Derive the value of gravitational potential at a
distance r from a planet of mass m.
Q.24 Give 4 differences b/w gravitational and inertial mass.
Q.25 What is a satellite? Define geostationary satellite. What
are the conditions required for it?
Q.26 A satellite of mass “m” is revolving about a planet of mass
“M” in an orbit of radius “r” . Derive the expression for orbital
velocity, time period and height of satellite.
Q.27 State Kepler’s laws of planetary motion. Derive the
second and the third.
Q.28 The time period of Jupiter is 11.6 years. How far is
Jupiter from the sun? Distance b/w earth and sun is known to
be 1.5x10 11m.
Q.29 what is escape velocity? Find the value of escape velocity
from earth.
(b) Find the value of escape velocity from a planet whose mass
if double while radius is 4 times that of earth
Q.30 Two particles of mass 5kg and 10kg are kept 100m apart.
Due to their mutual force of attraction they start approaching
each other. Find the velocity of each of them when the
separation b/w them has reduced to 25m.
(b) At what location will they collide?
Newton's Law of Gravitation
Idea: Newton's Universal Law of Gravitation states that any two objects exert a gravitational
force of attraction on each other. The direction of the force is along the line joing the objects
(See Fig.(7.3)). The magnitude of the force is proportional to the product of the gravitational
masses of the objects, and inversely proportional to the square of the distance between them. For
the two objects in Figure 7.3:
Figure 7.3: Gravitational Force Between Two Masses

m1 exerts a force


m2 exerts a force
on m1 .
By Newton's third law:
on m2 .
=
.
The magnitude of the gravitational force is:
F12 = G

Energy Relationships for Satellites
In the early 1600s, Johannes Kepler proposed three laws of planetary motion.
Kepler was able to summarize the carefully collected data of his mentor - Tycho
Brahe - with three statements that described the motion of planets in a suncentered solar system. Kepler's efforts to explain the underlying reasons for such
motions are no longer accepted; nonetheless, the actual laws themselves are still
considered an accurate description of the motion of any planet and any satellite.
Kepler's three laws of planetary motion can be described as follows:



The path of the planets about the sun is elliptical in shape, with the center of the
sun being located at one focus. (The Law of Ellipses)
An imaginary line drawn from the center of the sun to the center of the planet
will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
The ratio of the squares of the periods of any two planets is equal to the ratio of
the cubes of their average distances from the sun. (The Law of Harmonies)
The Law of Ellipses
Kepler's first law - sometimes referred to as the law of ellipses - explains that
planets are orbiting the sun in a path described as an ellipse. An ellipse can easily
be constructed using a pencil, two tacks, a string, a sheet
of paper and a piece of cardboard. Tack the sheet of paper
to the cardboard using the two tacks. Then tie the string
into a loop and wrap the loop around the two tacks. Take
your pencil and pull the string until the pencil and two
tacks make a triangle (see diagram at the right). Then
begin to trace out a path with the pencil, keeping the string
wrapped tightly around the tacks. The resulting shape will
be an ellipse. An ellipse is a special curve in which the sum of the distances from
every point on the curve to two other points is a constant. The two other points
(represented here by the tack locations) are known as the foci of the ellipse. The
closer together that these points are, the more closely that the ellipse resembles the
shape of a circle. In fact, a circle is the special case of an ellipse in which the two
foci are at the same location. Kepler's first law is rather simple - all planets orbit the
sun in a path that resembles an ellipse, with the sun being located at one of the foci
of that ellipse.
The Law of Equal Areas
Kepler's second law - sometimes referred to as the law of equal areas - describes
the speed at which any given planet will move while orbiting the sun. The speed at
which any planet moves through space is constantly changing. A planet moves
fastest when it is closest to the sun and slowest when it is furthest from the sun.
Yet, if an imaginary line were drawn from the center of the planet to the center of
the sun, that line would sweep out the same area in equal periods of time. For
instance, if an imaginary line were drawn from the earth to the sun, then the area
swept out by the line in every 31-day month would be the same. This is depicted in
the diagram below. As can be observed in the diagram, the areas formed when the
earth is closest to the sun can be approximated as a wide but short triangle;
whereas the areas formed when the earth is farthest from the sun can be
approximated as a narrow but long triangle. These areas are the same size. Since
the base of these triangles are shortest when the earth is farthest from the sun, the
earth would have to be moving more slowly in order for this imaginary area to be
the same size as when the earth is closest to the sun.
The Law of Harmonies
Kepler's third law - sometimes referred to as the law of harmonies - compares the
orbital period and radius of orbit of a planet to those of other planets. Unlike
Kepler's first and second laws that describe the motion characteristics of a single
planet, the third law makes a comparison between the motion characteristics of
different planets. The comparison being made is that the ratio of the squares of the
periods to the cubes of their average distances from the sun is the same for every
one of the planets. As an illustration, consider the orbital period and average
distance from sun (orbital radius) for Earth and mars as given in the table below.
Planet
Period
(s)
3.156 x 107 s
5.93 x 107 s
Earth
Mars
2
Average
Distance (m)
1.4957 x 1011
2.278 x 1011
T2/R3
(s2/m3)
2.977 x 10-19
2.975 x 10-19
Observe that the T /R ratio is the same for Earth as it is for mars. In fact, if the
2
3
same T /R ratio is computed for the other planets, it can be found that this ratio is
nearly the same value for all the planets (see table below). Amazingly, every planet
2
3
has the same T /R ratio.
Planet
3
Period
Average
T2/R3
(yr)
Distance (au)
(yr2/au3)
Mercury
0.241
0.39
0.98
Venus
.615
0.72
1.01
Earth
1.00
1.00
1.00
Mars
1.88
1.52
1.01
Jupiter
11.8
5.20
0.99
Saturn
29.5
9.54
1.00
Uranus
84.0
19.18
1.00
Neptune
165
30.06
1.00
Pluto
248
39.44
1.00
(NOTE: The average distance value is given in astronomical units where 1 a.u. is
11
equal to the distance from the earth to the sun - 1.4957 x 10 m. The orbital period
is given in units of earth-years where 1 earth year is the time required for the earth
7
to orbit the sun - 3.156 x 10 seconds. )
Kepler's third law provides an accurate description of the period and distance for a
2
3
planet's orbits about the sun. Additionally, the same law that describes the T /R
2
3
ratio for the planets' orbits about the sun also accurately describes the T /R ratio for
any satellite (whether a moon or a man-made satellite) about any planet. There is
2
3
something much deeper to be found in this T /R ratio - something that must relate
to basic fundamental principles of motion. In the next part of Lesson 4, these
principles will be investigated as we draw a connection between the circular motion
principles discussed in Lesson 1 and the motion of a satellite.
How did Newton Extend His Notion of Gravity to Explain Planetary Motion?
Newton's comparison of the acceleration of the moon to the acceleration of objects
on earth allowed him to establish that the moon is held in a circular orbit by the
force of gravity - a force that is inversely dependent upon the distance between the
two objects' centers. Establishing gravity as the cause of the moon's orbit does not
necessarily establish that gravity is the cause of the planet's orbits. How then did
Newton provide credible evidence that the force of gravity is meets the centripetal
force requirement for the elliptical motion of planets?
Recall from earlier in Lesson 3 that Johannes Kepler proposed three laws of
planetary motion. His Law of Harmonies suggested that the ratio of the period of
2
3
orbit squared (T ) to the mean radius of orbit cubed (R ) is the same value k for all
the planets that orbit the sun. Known data for the orbiting planets suggested the
following average ratio:
k = 2.97 x 10
-19
s /m = (T )/(R )
2
3
2
3
Newton was able to combine the law of universal gravitation with circular motion
principles to show that if the force of gravity provides the centripetal force for the
-19
2
3
planets' nearly circular orbits, then a value of 2.97 x 10 s /m could be predicted
2
3
for the T /R ratio. Here is the reasoning employed by Newton:
Consider a planet with mass Mplanet to orbit in nearly circular motion about the sun of
mass MSun. The net centripetal force acting upon this orbiting planet is given by the
relationship
2
Fnet = (Mplanet * v ) / R
This net centripetal force is the result of the gravitational force that attracts the
planet towards the sun, and can be represented as
Fgrav = (G* Mplanet * MSun ) / R
2
Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force
are equal. Thus,
2
(Mplanet * v ) / R = (G* Mplanet * MSun ) / R
2
Since the velocity of an object in nearly circular orbit can be approximated as v =
(2*pi*R) / T,
v = (4 * pi * R ) / T
2
2
2
2
Substitution of the expression for v into the equation above yields,
2
(Mplanet * 4 * pi * R ) / (R • T ) = (G* Mplanet * MSun ) / R
2
2
2
2
By cross-multiplication and simplification, the equation can be transformed into
T / R = (Mplanet * 4 * pi ) / (G* Mplanet * MSun )
2
3
2
The mass of the planet can then be canceled from the numerator and the
denominator of the equation's right-side, yielding
T / R = (4 * pi ) / (G * MSun )
2
3
2
The right side of the above equation will be the same value for every planet
2
3
regardless of the planet's mass. Subsequently, it is reasonable that the T /R ratio
would be the same value for all planets if the force that holds the planets in their
orbits is the force of gravity. Newton's universal law of gravitation predicts results
that were consistent with known planetary data and provided a theoretical
explanation for Kepler's Law of Harmonies.
Investigate!
Scientists know much more about the planets than they did in Kepler's days. Use
The Planets widget bleow to explore what is known of the various planets.
The Planets
Select a planet from the pull-down menu to retrieve information about it.
Mercury
Retrieve Information
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Check Your Understanding
1. Our understanding of the elliptical motion of planets about the Sun spanned
several years and included contributions from many scientists.
a. Which scientist is credited with the collection of the data necessary to support the
planet's elliptical motion?
b. Which scientist is credited with the long and difficult task of analyzing the data?
c. Which scientist is credited with the accurate explanation of the data?
See Answer
Tycho Brahe gathered the data. Johannes Kepler analyzed the data. Isaac Newton
explained the data - and that's what the next part of Lesson 4 is all about.
2. Galileo is often credited with the early discovery of four of Jupiter's many moons.
The moons orbiting Jupiter follow the same laws of motion as the planets orbiting
the sun. One of the moons is called Io - its distance from Jupiter's center is 4.2 units
and it orbits Jupiter in 1.8 Earth-days. Another moon is called Ganymede; it is 10.7
units from Jupiter's center. Make a prediction of the period of Ganymede using
Kepler's law of harmonies.
See Answer
Answer: T = 7.32 days
Given:
Io: Rio = 4.2 and Tio = 1.8
Ganymede: Rg = 10.7 Tg=???
Use Kepler's 3rd law to solve.
(Tio)^2/(Rio) = 0.04373;
3
so (Tg)^2 / (Rg) = 0.04373
3
Proper algebra would yield (Tg)^2 = 0.04373 • (Rg)
(Tg) = 53.57 so Tg = SQRT(53.57) = 7.32 days
2
3
3. Suppose a small planet is discovered that is 14 times as far from the sun as the
11
Earth's distance is from the sun (1.5 x 10 m). Use Kepler's law of harmonies to
2
3
-19
2
3
predict the orbital period of such a planet. GIVEN: T /R = 2.97 x 10 s /m
See Answer
Answer: Tplanet = 52.4 yr
Use Kepler's third law:
(Te)^2/(Re)^3 = (Tp)^2/(Rp)
3
Rearranging to solve for Tp:
2
3
(Tp)^2=[ (Te) / (Re) ] • (Rp)
3
or (Tp) = (Te) • [(Rp) / (Re)] where (Rp) / (Re) = 14
2
2
2
3
so (Tp) = (Te) • [14] where Te=1 yr
2
2
3
2
(Tp) =(1 yr) *[14]^3 = 2744 yr
2
Tp = SQRT(2744 yr )
2
4. The average orbital distance of Mars is 1.52 times the average orbital distance of
the Earth. Knowing that the Earth orbits the sun in approximately 365 days, use
Kepler's law of harmonies to predict the time for Mars to orbit the sun.
See Answer
Given: Rmars = 1.52 • Rearth and Tearth = 365 days
Use Kepler's third law to relate the ratio of the period squared to the ratio of radius
cubed
2
2
3
(Tmars) / (Tearth) • (Rmars) / (Rearth)
2
2
(Tmars) = (Tearth) • (Rmars) / (Rearth)
2
3
3
(Tmars) = (365 days) * (1.52)
2
3
3
(Note the Rmars / Rearth ratio is 1.52)
Tmars = 684 days
Orbital radius and orbital period data for the four biggest moons of Jupiter are listed
27
in the table below. The mass of the planet Jupiter is 1.9 x 10 kg. Base your
answers to the next five questions on this information.
Jupiter's Moon
Io
Europa
Ganymede
Callisto
2
Period (s)
1.53 x 105
3.07 x 105
6.18 x 105
1.44 x 106
Radius (m)
4.2 x 108
6.7 x 108
1.1 x 109
1.9 x 109
5. Determine the T /R ratio (last column) for Jupiter's moons.
3
See Answer
T2/R3
a.
b.
c.
d.
2
3
-16
2/
3
2
3
-16
2/
3
2
3
-16
2/
3
2
3
-16
2/
3
a. (T ) / (R ) = 3.16 x 10 s m
b. (T ) / (R ) = 3.13 x 10 s m
c. (T ) / (R ) = 2.87 x 10 s m
d. (T ) / (R ) = 3.03 x 10 s m
6. What pattern do you observe in the last column of data? Which law of Kepler's
does this seem to support?
See Answer
The (T ) / (R ) ratios are approximately the same for each of Jupiter's moons. This is
what would be predicted by Kepler's third law!
2
3
7. Use the graphing capabilities of your TI calculator to plot T vs. R (T should be
plotted along the vertical axis) and to determine the equation of the line. Write the
equation in slope-intercept form below.
2
See Answer
T = (3.03 * 10 ) * R - 4.62 * 10
2
-16
3
+9
3
2
Given the uncertainty in the y-intercept value, it can be approximated as 0.
2
-16
Thus, T = (3.03 * 10 ) * R
3
See graph below.
2
3
8. How does the T /R ratio for Jupiter (as shown in the last column of the data
2
3
table) compare to the T /R ratio found in #7 (i.e., the slope of the line)?
See Answer
The values are almost the same - approximately 3 x 10 .
-16
9. How does the T /R ratio for Jupiter (as shown in the last column of the data
2
3
-11
table) compare to the T /R ratio found using the following equation? (G=6.67x10
2
2
27
N*m /kg and MJupiter = 1.9 x 10 kg)
2
3
T / R = (4 * pi ) / (G * MJupiter )
2
3
2
See Answer
The values in the data table are approx. 3 x 10 . The value of 4*pi/(G*MJupiter) is
-16
approx. 3.1 x 10 .
-16
Graph for Question #6
Escape velocity
From Wikipedia, the free encyclopedia
Jump to: navigation, search
For other senses of this term, see Escape velocity (disambiguation).
This article needs additional citations for verification. Please help improve this article by
adding citations to reliable sources. Unsourced material may be challenged and removed.
(January 2012)
In physics, escape velocity is the speed at which the sum of an object's kinetic energy and its
gravitational potential energy is equal to zero.[nb 1] It is the speed needed to "break free" from the
gravitational attraction of a massive body, without further propulsion, i.e., without spending
more fuel.
For a spherically symmetric massive body, the escape velocity at a given distance is calculated
by the formula[1]
where G is the universal gravitational constant (G = 6.67×10−11 m3 kg−1 s−2), M the mass of the
planet, star or other massive body, and r the distance from the center of gravity.[nb 2]
In this equation atmospheric friction (air drag) is not taken into account. A rocket moving out of
a gravity well does not actually need to attain escape velocity to do so, but could achieve the
same result at any speed with a suitable mode of propulsion and sufficient fuel. Escape velocity
only applies to ballistic trajectories.
The term escape velocity is actually a misnomer, and it is often more accurately referred to as
escape speed since the necessary speed is a scalar quantity which is independent of direction
(assuming a non-rotating planet and ignoring atmospheric friction or relativistic effects).
Contents
[hide]










1 Overview
2 Orbit
3 List of escape velocities
4 Calculating an escape velocity
5 Deriving escape velocity using calculus
6 Multiple sources
7 See also
8 Notes
9 References
10 External links
Overview[edit]
Luna 1, launched in 1959, was the first man-made object to attain escape velocity from Earth (see below
table).[2]
A barycentric velocity is a velocity of one body relative to the center of mass of a system of
bodies. A relative velocity is the velocity of one body with respect to another. Relative escape
velocity is defined only in systems with two bodies. For systems of two bodies the term "escape
velocity" is ambiguous, but it is usually intended to mean the barycentric escape velocity of the
less massive body. In gravitational fields "escape velocity" refers to the escape velocity of zero
mass test particles relative to the barycenter of the masses generating the field.
The existence of escape velocity is a consequence of conservation of energy. For an object with a
given total energy, which is moving subject to conservative forces (such as a static gravity field)
it is only possible for the object to reach combinations of places and speeds which have that total
energy; and places which have a higher potential energy than this cannot be reached at all.
For a given gravitational potential energy at a given position, the escape velocity is the minimum
speed an object without propulsion needs to be able to "escape" from the gravity (i.e. so that
gravity will never manage to pull it back). For the sake of simplicity, unless stated otherwise, we
will assume that an object is attempting to escape from a uniform spherical planet by moving
directly away from it (along a radial line away from the center of the planet) and that the only
significant force acting on the moving object is the planet's gravity.
Escape velocity is actually a speed (not a velocity) because it does not specify a direction: no
matter what the direction of travel is, the object can escape the gravitational field (provided its
path does not intersect the planet). The simplest way of deriving the formula for escape velocity
is to use conservation of energy. Imagine that a spaceship of mass m is at a distance r from the
center of mass of the planet, whose mass is M. Its initial speed is equal to its escape velocity, .
At its final state, it will be an infinite distance away from the planet, and its speed will be
negligibly small and assumed to be 0. Kinetic energy K and gravitational potential energy Ug are
the only types of energy that we will deal with, so by the conservation of energy,
Kƒ = 0 because final velocity is zero, and Ugƒ = 0 because its final distance is infinity, so
Defined a little more formally, "escape velocity" is the initial speed required to go from an initial
point in a gravitational potential field to infinity with a residual velocity of zero, with all speeds
and velocities measured with respect to the field. Additionally, the escape velocity at a point in
space is equal to the speed that an object would have if it started at rest from an infinite distance
and was pulled by gravity to that point. In common usage, the initial point is on the surface of a
planet or moon. On the surface of the Earth, the escape velocity is about 11.2 kilometers per
second (~6.96 mi/s), which is approximately 33 times the speed of sound (Mach 33) and several
times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in
"space", it is slightly less than 7.1 km/s.
The escape velocity relative to the surface of a rotating body depends on direction in which the
escaping body travels. For example, as the Earth's rotational velocity is 465 m/s at the equator, a
rocket launched tangentially from the Earth's equator to the east requires an initial velocity of
about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the
Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth. The
surface velocity decreases with the cosine of the geographic latitude, so space launch facilities
are often located as close to the equator as feasible, e.g. the American Cape Canaveral (latitude
28°28' N) and the French Guiana Space Centre (latitude 5°14' N).
The barycentric escape velocity is independent of the mass of the escaping object. It does not
matter if the mass is 1 kg or 1,000 kg, what differs is the amount of energy required. For an
object of mass the energy required to escape the Earth's gravitational field is GMm / r, a
function of the object's mass (where r is the radius of the Earth, G is the gravitational constant,
and M is the mass of the Earth, M=5.9736×1024 kg).
For a mass equal to a Saturn V rocket, the escape velocity relative to the launch pad is 253.5
am/s (8 nanometers per year) faster than the escape velocity relative to the mutual center of
mass. When the mass reaches the Andromeda galaxy the earth will have recoiled 500 m away
from the mutual center of mass.
Orbit[edit]
If an object attains escape velocity, but is not directed straight away from the planet, then it will
follow a curved path. Although this path does not form a closed shape, it is still considered an
orbit. Assuming that gravity is the only significant force in the system, this object's speed at any
point in the orbit will be equal to the escape velocity at that point (due to the conservation of
energy, its total energy must always be 0, which implies that it always has escape velocity; see
the derivation above). The shape of the orbit will be a parabola whose focus is located at the
center of mass of the planet. An actual escape requires a course with an orbit that does not
intersect with the planet, or its atmosphere, since this would cause the object to crash. When
moving away from the source, this path is called an escape orbit. Escape orbits are known as C3
= 0 orbits. C3 is the characteristic energy, = −GM/a, where a is the semi-major axis, which is
infinite for parabolic orbits.
When there are many gravitating bodies, such as in the solar system, a rocket that travels at
escape velocity from one body, say Earth, will not travel to an infinite distance because it needs
an even higher speed to escape the Sun's gravity. Near the Earth, the rocket's orbit will appear
parabolic, but will become an ellipse around the Sun, except when it is perturbed by the Earth,
whose orbit it must still intersect, and other bodies.
List of escape velocities[edit]
To leave planet Earth, an escape velocity of 11.2 km/s (approx. 40,320 km/h, or 25,000 mph) is required;
however, a speed of 42.1 km/s is required to escape the Sun's gravity (and exit the Solar System) from
the same position.
Location
with respect
to
at Mercury,
the Sun's
gravity:
67.7
at Venus,
the Sun's
gravity:
49.5
11.2[4]:200
at the
Earth/Moon,
the Sun's
gravity:
42.1
2.4
at the Moon,
the Earth's
gravity:
at Mars,
the Sun's
gravity:
34.1
at Jupiter,
the Sun's
gravity:
18.5
Location
with respect to
Ve (km/s)[3]
on the Sun,
the Sun's gravity:
617.5
on Mercury,
Mercury's gravity:
4.3[4]:230
on Venus,
Venus's gravity:
on Earth,
Earth's gravity:
on the Moon,
the Moon's
gravity:
on Mars,
Mars' gravity:
on Jupiter,
Jupiter's gravity:
on Ganymede,
Ganymede's
gravity:
10.3
5.0[4]:234
59.6[4]:236
2.7
Ve
(km/s)[3]
1.4
on Saturn,
Saturn's gravity:
35.6[4]:238
at Saturn,
the Sun's
gravity:
13.6
on Uranus,
Uranus' gravity:
21.3[4]:240
at Uranus,
the Sun's
gravity:
9.6
on Neptune,
Neptune's gravity:
23.8[4]:240
at Neptune,
the Sun's
gravity:
7.7
on Pluto,
Pluto's gravity:
at Solar System
galactic radius,
the Milky Way's
gravity:
on the event horizon, a black hole's
gravity:
1.2
492–594 [5] [6]
≥ 299,792
(Speed of light)
Because of the atmosphere it is not useful and hardly possible to give an object near the surface
of the Earth a speed of 11.2 km/s (40,320 km/h), as these speeds are too far in the hypersonic
regime for most practical propulsion systems and would cause most objects to burn up due to
aerodynamic heating or be torn apart by atmospheric drag. For an actual escape orbit a spacecraft
is first placed in low Earth orbit (160–2,000 km) and then accelerated to the escape velocity at
that altitude, which is a little less — about 10.9 km/s. The required change in speed, however, is
far less because from a low Earth orbit the spacecraft already has a speed of approximately
8 km/s (28,800 km/h).
Calculating an escape velocity[edit]
To expand upon the derivation given in the Overview,
where is the barycentric escape velocity, G is the gravitational constant, M is the mass of the
body being escaped from, r is the distance between the center of the body and the point at which
escape velocity is being calculated, g is the gravitational acceleration at that distance, and μ is the
standard gravitational parameter.[7]
The escape velocity at a given height is
times the speed in a circular orbit at the same height,
(compare this with the velocity equation in circular orbit). This corresponds to the fact that the
potential energy with respect to infinity of an object in such an orbit is minus two times its
kinetic energy, while to escape the sum of potential and kinetic energy needs to be at least zero.
The velocity corresponding to the circular orbit is sometimes called the first cosmic velocity,
whereas in this context the escape velocity is referred to as the second cosmic velocity[8]
For a body with a spherically-symmetric distribution of mass, the barycentric escape velocity
from the surface (in m/s) is approximately 2.364×10−5 m1.5kg−0.5s−1 times the radius r (in meters)
times the square root of the average density ρ (in kg/m³), or:
Deriving escape velocity using calculus[edit]
Let G be the gravitational constant and let M be the mass of the earth (or other gravitating body)
and m be the mass of the escaping body or projectile. At a distance r from the centre of
gravitation the body feels an attractive force[9]
The work needed to move the body over a small distance dr against this force is therefore given
by
The total work needed to move the body from the surface r0 of the gravitating body to infinity is
then
This is the minimal required kinetic energy to be able to reach infinity, so the escape velocity v0
satisfies
which results in
Multiple sources[edit]
The escape velocity from a position in a field with multiple sources at rest with respect to each
other is derived from the total potential energy per kg at that position, relative to infinity. The
potential energies for all sources can simply be added. For the escape velocity it can be shown
that this gives an escape velocity which is equal to the square root of the sum of the squares of
the individual escape velocities due to each source.
For example, at the Earth's surface the escape velocity for the combination Earth and Sun would
be
.
See also[edit]
Spaceflight portal
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Orbital speed
Characteristic energy (C3)
Delta-v budget – speed needed to perform manoeuvres.
Gravitational slingshot – a technique for changing trajectory.
Gravity well
Two-body problem
Black hole – an object with an escape velocity greater than the speed of light
Oberth effect – burning fuel deep in a gravity field gives higher change in velocity.
Newton's cannonball
List of artificial objects escaping from the Solar System
List of artificial objects in heliocentric orbit
Notes[edit]
1. Jump up ^ The gravitational potential energy is negative since gravity is an attractive force and
the potential energy has been defined for this purpose to be zero at infinite distance from the
centre of gravity
1.
2.
A thief stole a box with valuable article of weight ‘W’ and jumped
down a wall of height h. Before he reach the ground he
experienced a load of
(a) zero
(b) W / 2
(c) W
(d) 2 W
The acceleration due to gravity g and mean density of the earth 
are related by which of the following relation? Where g is
gravitational constant and R is radius of the earth
(a)  =
(b)  =
(c)  =
3.
4.
5.
6.
(d)  =
When the planet comes nearer the sun moves
(a) fast
(b) slow
(c) constant at every point
(d) none of the above
Kepler’s second law regarding constancy of arial velocity of a
planet is a consequence of the law of conservation of
(a) energy
(b) angular momentum
(c) linear momentum
(d) none of these
The period of geostationary artificial satellite is
(a) 24 hours
(b) 6 hours
(c) 12 hours
(d) 48 hours
A geostationary satellite is orbiting the earth at a height of 6R
above the surface of the earth, R being the radius of the earth.
The time period of another satellite at a height of 2.5 R from the
surface of earth is
(a) 6
(b) 6 hr
7.
8.
hr
(c) 5
hr
(d) 10 hr
The distance of Neptune and Saturn from the sun are nearly 1013
m and 1012 m respectively. Assuming that they move in circular
orbits, their periodic times would be in the ratio of
(a) 10
(b) 100
(c) 10
(d) 1000
A satellite is orbiting close to the surface of the earth, then its
speed is
(a)
(b) Rg
(c)
9.
(d)
If the gravitational force between two objects were proportional
to 1/R (and not as 1/R2) where R is separation between them,
then a particle in circular orbit under such a force would have its
orbital speed v proportional to
(a)
(b) R0
(c) R1
10.
(d)
Imagine a light planet revolving around a very massive star in a
circular orbit of radius R with a period of revolution T. If the
gravitational force of attraction between the planet and the star is
proportional to
(a) T2  R2
then
(b) T2 
11.
12.
(c) T2 
(d) T2  R3
The period of a satellite in a circular orbit of radius R is T. The
period of another satellite in circular orbit of radius 4R is
(a) T/4
(b) 8T
(c) 2T
(d) T/8
A planet moves around the sun. At a point A, it is closest from the
Sun at a distance d1 and has a speed v1. At another point B, when
it is farthest from the sun at a distance d2, its speed will be
(a)
(b)
(c)
13.
14.
(d)
The period of geostationary artificial satellite of earth is
(a) 6 hours
(b) 12 hours
(c) 24 hours
(d) 365 days
If ‘r’ represents the radius of the orbit of a satellite of mass ‘m’
moving round a planet of mass ‘M’, the velocity of the satellite is
given by
(a) v2 =
(b) v2 =
(c) v =
15.
16.
17.
18.
(d) v =
A missile is launched with a velocity less than the escape velocity.
The sum of its kinetic and potential energy is
(a) Positive
(b) Negative
(c) Zero
(d) may be positive or negative
The escape velocity of projection from the earth is approximately
(R = 6400 km)
(a) 7 km/sec
(b) 112 km/sec
(c) 12.2 km/sec
(d) 1.1 km/sec
If the earth is 1/4th of its present distance from the sun, the
duration of the year would be
(a) 1/4 of the present year
(b) 1/6 of the present year
(c) 1/8 of the present year
(d) 1/16 of the present year
The relation between escape velocity and orbit velocity is
(a) ve =
(b) ve =
vorb
(c) ve = 2vorb
19.
20.
21.
22.
(d) ve =
vorb
There is no atmosphere on the moon because
(a) it is closer ot the earth
(b) it revolves round the earth
(c) it gets light from the sun
(d) the escape velocity of gas molecules is less than their root
mean square velocity here
If the radius of the earth were to shrink by 1% its mass
remaining the same, the acceleration due to gravity on the earth’s
surface would
(a) decrease by 2%
(b) remain unchanged
(c) increase by 2%
(d) will increase by 9.8%
Fg and Fe represents gravitational and electrostatic forces
respectively, between the two electrons situated at a
distance of 10 m. The ratio Fg/Fe is of the order of
(a) 1043
(b) 1036
(c) 1043
(d) 1036
The value of ‘g’ at a particular point is 9.8 m/sec2 suppose the
23.
24.
earth suddenly shrink uniformly to half its present size
without losing any mass. The value of ‘g at the same
point (assuming that the distance of the point from the
centre of the earth does not shrink) will become
(a) 9.8 m/sec2
(b) 4.9 m/sec2
(c) 19.6 m/sec2
(d) 2.45 m/sec2
The planet mercury is revolving in an elliptical orbit around the
sun as shown in figure. The kinetic energy of mercury will be
greater at
(a) A
(b) B
(c) C
(d) D
The orbit velocity of an artificial satellite in a circular orbit just
above the earth’s surface is v. For a satellite orbiting at an
altitude of half of the earth’s radius, the orbital velocity is
(a)
(b)
(c)
25.
(d)
If the change in the value of g at the height h above the surface
of the earth is the same as at a depth x below it, then
(both x and h being much smaller than the radius of the
earth)
(a) x = h
(b) x = 2 h
(c) x =
(d) x = h2
Answers to Gravitation, Paper 1 (Go to Top)
1. Ans.: (a)
2. Ans.: (c)
3.Ans.: (a)
4. Ans.: (b)
5. Ans.: (a)
6. Ans.: (a)
7. Ans.: (c)
8. Ans.: (c)
9. Ans.: (b)
10. Ans.: (b)
11. Ans.: (b)
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1.
2.
A satellite is revolving around the sun in a circular orbit with uniform velocity v. If the
gravitational force suddenly disappears, the velocity of the satellite will be
(a) zero
(b) v
(c) 2v
(d) infinity
Who among the following first gave the experimental velocity of G?
(a) Cavendish
(b) Copernicus
(c) Brook Taylor
(d) none of these
3.
4.
The mean radius of the earth is R, its angular speed on its own axis is  and the acceleration
due to gravity at earth’s surface is g. The cube of the radius of the orbit of a geo-stationary
satellite will be
(a) r2g / 
(b) R22 / g
(c) RG 2
(d) R2g / 2
The largest and the shortest distance of the earth from are r 1 and r2. It’s distance from the sun
when it is perpendicular to the major-axis of the orbit drawn from the sun.
(a)
(b)
(c)
5.
6.
(d)
Geo-stationary satellite
(a) revolves about the polar axis
(b) has a time period less than that of the earth’s satellite
(c) moves faster than a near earth satellite
(d) is stationary in the space
A spherical planet far out in space has a mass M 0 and diameter D0. A particle of mass m falling
freely near the surface of this planet will experience an acceleration due to gravity which is
equal to
(a)
(b)
(c)
7.
(d)
Two planets of radii r1 and r2 are made from the same material. The ratio of the acceleration
due to gravity g1/g2 at the surface of the two planets is
(a)
(b)
(c)
8.
(d)
If g is the acceleration due to gravity of the earth’s surface the gain in the potential energy of
an object of mass m raised from the surface of the earth to a height equal to the radius R of
the earth is
(a)
mgR
(b) 2mgR
(c) mgR
9.
(d)
mgR
An earth’s satellite of mass m revolves in a circular orbit at a height h from the surface g is
acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is
given by
(a)
(b) gR
(c)
10.
11.
12.
(d)
If the radius of the earth were to shrink by one percent, its mass remaining the same, the
acceleration due to gravity on the earth’s surface would
(a) decrease
(b) remains unchanged
(c) increase
(d) none of these
The escape velocity from the earth’s surface is 11 km/sec. A certain planet has a radius twice
that of the earth but its mean density is the same as that of the earth. The value of the escape
velocity from this planet would be
(a) 22 km/sec
(b) 11 km/sec
(c) 5.5 km/sec
(d) 16.5 km/sec
The escape velocity from earth is 11.2 km per sec. If a body is to be projected in a direction
making an angle 45 to the vertical, then the escape velocity is
(a) 11.2  2 km/sec
(b) 11.2 km/sec
(c) 11.2 
13.
km/sec
(d) 11.2 
km/sec
What would be the duration of the year if the distance between the earth and the sun gets
doubled?
(a) 1032 days
(b) 129 days
(c) 365 days
(d) 730 days
14.
15.
16.
17.
18.
19.
20.
21.
If escape velocity from the earth’s surface is 11.2 km/sec. then escape velocity from a planet of
mass same as that of earth but radius one fourth as that of earth is
(a) 11.2 km/sec
(b) 22.4 km/sec
(c) 5.65 km/sec
(d) 44.8 km/sec
A thin uniform, circular ring is rolling down an inclined plane of inclination 30 without slipping.
Its linear acceleration along the inclined plane will be
(a) g/2
(b) g/3
(c) g/4
(d) 2g/3
A artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential)
energy E0. Its potential energy is
(a) 2E0
(b) E0
(c) 1.5 E0
(d) E0
The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6
 1024 kg and G = 6.66  1011 Nm2/kg2. The speed of the moon is nearly
(a) 1 km/sec
(b) 4 km/sec
(c) 8 km/sec
(d) 11.2 km/sec
When body is raised to a height equal to radius of earth, the P.E. change is
(a) MgR
(b)
(c) 2 MgR
(d) none of these
A planet has twice the radius but the mean density is 1/4th as compared to earth. What is the
radio of the escape velocity from the earth to that from the planet?
(a) 3 : 1
(b) 1 : 2
(c) 1 : 1
(d) 2 : 1
The masses of two planets are in the ratio 1 : 2. Their radii are in the ratio 1 : 2. The
acceleration due to gravity on the planets are in the ratio.
(a) 1 : 2
(b) 2 : 1
(c) 3 : 5
(d) 5 : 3
If the acceleration due to gravity of a planet is half the acceleration due to gravity of earth’s
surface and radius of planet is half the radius of the earth, the mass of planet in terms of mass
of earth is
(a)
(b)
(c)
22.
23.
(d)
The radii of the earth and the moon are in the ratio 10 : 1 while acceleration due to gravity on
th eearth’s surface and moon’s surface are in the ratio 6 : 1. The ratio of escape velocities from
earth’s surface to that of moon surface is
(a) 10 : 1
(b) 6 : 1
(c) 1.66 : 1
(d) 7.74 : 1
Acceleration due to gravity g in terms of mean density of Earth d (where R is radius of earth
and G – universal gravitational constant) is
(a) g = 4R2 d G
(b) g =
(c) g =
24.
25.
(d) g =
RdG
The dimensions of universal gravitational constant are
(a) M2 L2 T2
(b) M1 L3 T2
(c) M L1 T2
(d) M L2 T2
If R is radius of the earth and g the acceleration due to gravity on the earth’s surface, the
mean density of the earth is
(a)
(b)
(c)
1.
(d)
A planet is moving around the Sun in a circular orbit of circumference C. The work done on the
planet by the gravitational force F of the Sun is
FC
F/C
FC/2
4
Zero.
0,1,2,3
2.
A satellite is moving around the Earth in a circular orbit with a velocity V. If the
gravitational force of the Earth were to suddenly disappear, then the satellite would
5
move with a velocity V, tangentially to its circular orbit.
fall towards the surface of the Earth.
move radially outwards with a velocity V.
spirally move away from the Earth.
6,3,4,5
3.
Two planets of radii R1 and R2 have the same density. The ratio of their
accelerations due to gravity at the surface is
R2 /R1
6
(R2 /R1)2
R1 /R2
(R1 /R2)2
2,4,1,3
4.
A reference frame attached to the Earth
3
cannot be an inertial frame of reference because of the Earth's
rotation and revolution.
is an inertial frame by definition.
is not an inertial frame because the Earth moves with respect to
the Sun.
is an inertial frame because Newton's laws of motion are
applicable inside it.
6,3,5,4
5.
A planet's density is 2 times that of the Earth. But the acceleration due to gravity on
its surface is exactly the same as on the Earth's surface. The radius of the planet in
terms of the Earth's radius R is
2R
R/4
11
R/2
4R
Class-ix-Science-physics-Gravitation Solved
Questions :Numerical problems
Gravitation Solved Questions :Numerical problems
1. How does the force of gravitation between two objects change when the distance between them is reduced to
half?
Answer: when the distance between the objects is reduced to half the gravitational force increases by four
times the original force.
2. The gravitational force acts on all objects in proportion to their masses. Why, then, a heavy object does not
fall faster than a light object?
Answer: Acceleration due to gravity does not depend on mass of object . Hence, all bodies fall with the same
acceleration provided there is no air or other resistance
3. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon
with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Answer: According to Newton’s 3rd law of motion Every action has equal reaction in opposite direction.
Since, The earth surface attracts the moon with the same force with which the moon attracts the earth and
cancel them
4. If the moon attracts the earth, why does the earth not move towards the moon?
Answer: The earth is much larger than the moon so, the acceleration produced on the earth surface cannot be
noticed.
5. What is the importance of Universal Law of Gravitation?
Ans: There are many importance of Universal Law of Gravitation
1. The force of attraction that binds us to the earth,
2. The motion of planets moving around the sun,
3. the motion of moon around the earth
4. The occurring of tides due to sun and moon.
6 What is Gravitation?
Answer: Gravitation is the force of attraction between two objects in the universe.
i) Gravitation may be the attraction of objects by the earth. Eg :- If a body is dropped from a certain height, it
falls downwards due to earth’s gravity. If a body is thrown upwards, it reaches a certain height and then falls
downwards due to the earth’s gravity.
ii) Gravitation may be the attraction between objects in outer space.Eg :- Attraction between the earth and
moon. Attraction between the sun and planets
7. What is Centripetal force?
Answer: When a body moves in a circular path, it changes its direction at every point. The force which keeps
the body in the circular path acts towards the centre of the circle. This force is called centripetal force.
If there is no centripetal force, the body will move in a straight line tangent to the circular path.
8. State Universal law of gravitation?
Answer: The universal law of gravitation states that, ‘Every object in the universe attracts every other object
with a force which is directly proportional to product of the masses and inversely proportional to the square of
the distance between them.’
9. In what direction does the buoyant force on an object immersed in a liquid act?
Ans: The buoyant force acts on an object in the vertically upward direction through the center of gravity of the
displaced liquid.
10. A stone is released from the top of a tower of height 19.6 m. calculate its final velocity just before touching
the ground.
Ans: Given that,
u = 0, g = 9.8 ms–2, s = 19.6 m
Now, v2 - u2 = 2gs
or,
v2 - 0 = 2 x 9.8 x 19.6 = (19.6)2
or,
v = 19.6 ms–1 ( v is +ve due to downward direction)