Version 001 – test #2 Gravitation – tubman – (19112)
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AP B 1993 MC 6
001 10.0 points
If Spacecraft X has twice the mass of Spacecraft Y , then what is true about X and Y ?
I) On Earth, X experiences twice the gravitational force that Y experiences;
II) On the Moon, X has twice the weight of
Y;
III) When both are in the same circular orbit,
X has twice the centripetal acceleration
of Y .
Let : Te = 1 year ,
re = 150 million km ,
ra = 483 million km .
1
and
From Kepler’s laws,
Ta2
Te2
=
re3
ra3
3/2
ra
Ta =
Te
re
3/2
483 million km
(1 year)
=
150 million km
= 5.77808 year .
1. I only
2. I, II, and III
3. I and II only correct
003 (part 2 of 2) 10.0 points
What is the orbital velocity of the asteroid?
Assume there are 365 days in one year.
4. II and III only
Correct answer: 16654.7 m/s.
5. III only
Explanation:
I) gravitational force ∝ mass.
II) weight ∝ mass.
III) The centripetal acceleration is determined by
v2
ac =
,
r
so X and Y should have the same centripetal
acceleration when they are in the same circular orbit.
Explanation:
2 π ra
Ta
2 π (4.83 × 1011 m) 1 y
=
5.77808 year
365 d
1d 1h
×
24 h 3600 s
= 16654.7 m/s .
va =
Asteroid in Orbit
002 (part 1 of 2) 10.0 points
The period of the earth around the sun is
1 year and its distance is 150 million km from
the sun. An asteroid in a circular orbit around
the sun is at a distance 483 million km from
the sun.
What is the period of the asteroid’s orbit?
Circular Moon Orbit
004 (part 1 of 3) 10.0 points
A satellite is in a circular orbit just above the
surface of the Moon.
What is the satellite’s acceleration?
The value of gravitational constant is
6.67259 × 10−11 Ncdotm2 /kg2 and the mass
of the moon is 7.36 × 1022 kg and its radius is
2740.2 km .
Correct answer: 5.77808 year.
Correct answer: 0.654045 m/s2 .
Explanation:
Explanation:
Version 001 – test #2 Gravitation – tubman – (19112)
Let : G = 6.67259 × 10−11 Ncdotm2 /kg2 ,
Mmoon = 7.36 × 1022 kg , and
rmoon = 2740.2 km = 2.7402 × 106 m .
Fc = m ac = FG , so
G m Mmoon
r2
G Mmoon
ac =
r2
= (6.67259 × 10−11 Ncdotm2 /kg2 )
7.36 × 1022 kg
×
(2.7402 × 106 m)2
m ac =
= 0.654045 m/s2 .
2
007 10.0 points
Which are the correct statements regarding
Kepler’s laws?
A. They were obtained first by Tycho Brahe.
B. They were obtained by Kepler from
Brahe’s observations combined with Newton’s second law.
C. They were deduced by Kepler from
Brahe’s observations.
D. They were derived by Newton from his
gravitational and second laws.
E. They were derived by Newton using his
gravitational and second law together
with Brahe’s data.
1. C and D correct
2. B and D
005 (part 2 of 3) 10.0 points
What is the satellite’s speed?
Correct answer: 1338.74 m/s.
3. A, B, and C
4. A and E
5. C and E
Explanation:
v2
ac =
r
√
v = ac r
q
= (0.654045 m/s2 )(2.7402 × 106 m)
= 1338.74 m/s .
006 (part 3 of 3) 10.0 points
What is the period of the satellite’s orbit?
Correct answer: 3.57244 h.
Explanation:
2πR
v
1h
2 π (2.7402 × 106 m)
×
=
1338.74 m/s
3600 s
T =
= 3.57244 h .
Kepler
6. A and D
7. B and E
Explanation:
Energy in Lifting
008 10.0 points
Energy is required to move a 987 kg mass
from the Earth’s surface to an altitude 2.11
times the Earth’s radius RE .
What amount of energy is required to accomplish this move? The acceleration of gravity is 9.8 m/s2 .
Correct answer: 4.18027 × 1010 J.
Explanation:
Let :
m = 987 kg ,
h = 2.11 RE , and
g = 9.8 m/s2 .
Version 001 – test #2 Gravitation – tubman – (19112)
m ME
ME
and g = G 2 , so the
R
RE
change in the potential energy of the massEarth system is
(856 kg) (7 × 1022 kg)
1.7 × 106 m
= 2.35189 × 109 J .
U = −G
∆U = Uf − Ui
1
1
−
= G m ME
Ri Rf
1
1
−
= G m ME
RE
3.11 RE
m G ME
1
=
1−
RE
3.11
2.11
= m g RE
3.11
= (987 kg) (9.8 m/s2 ) (6.37 × 106 m)
2.11
3.11
= 4.18027 × 1010 J .
Moons Gravitational Field
009 10.0 points
Given: G = 6.67259 × 10−11 N m2 /kg2
A 856 kg meteor comes in from outer
space and impacts on the a moon’s surface
(this moon is not necessarily the Earth’s
moon). The mass of this moon is M =
7 × 1022 kg, and the radius of this moon is
R = 1.7 × 106 m.
How much work is done by the Moon’s
gravitational field?
Correct answer: 2.35189 × 109 J.
Explanation:
Using the formula relating the work W done
by the Moon’s gravitational field on the meteor with the change ∆U in the potential
energy of the meteor-Moon system,
W = −∆U
= − (Ufinal − Uinitial )
GmM
=− −
−0 ,
R
×
Estimate a Black Hole 01
010 10.0 points
A black hole is an object so heavy that neither
matter nor even light can escape the influence
of its gravitational field. Since no light can
escape from it, it appears black. Suppose a
mass approximately the size of the Earth’s
mass 4.7 × 1024 kg is packed into a small
uniform sphere of radius r.
Use:
The speed of light c = 2.99792 × 108 m/s .
The universal gravitational constant G =
6.67259 × 10−11 N m2 /kg2 .
Hint: The escape speed must be the speed
of light.
Based on Newtonian mechanics, determine
the limiting radius r0 when this mass (approximately the size of the Earth’s mass) becomes
a black hole.
Correct answer: 0.0069788 m.
Explanation:
Basic Concepts: Energy conservation
E=−
GmM
R
= (6.67259 × 10−11 N m2 /kg2 )
GmM
+K.
r
At minimum escape velocity, E = 0 (the projectile has just enough initial kinetic energy
to overcome the gravitational potential).
Solution: Technically speaking, in a region where gravity is extremely intense, Newton’s mechanics cannot be used. Rather, one
needs to apply the “general theory of relativity” developed by Albert Einstein. Knowing
this is the case, we still would like to see
what Newtonian mechanics tells us. Setting
vesc = c, the limiting radius is given by
1
1
GmM
2
,
m vesc
= m c2 =
2
2
r0
where M is the mass of the Moon, we obtain
W =
3
or
r0 =
2GM
.
c2
Version 001 – test #2 Gravitation – tubman – (19112)
It turns out that the theory of relativity gives
the same expression for this limiting radius,
referred to as the “Schwarzschild radius”. In
Newtonian mechanics, however, the description of what happens to objects attracted to
this sphere is often highly inaccurate. But in
any case,
2GM
c2
= 2 (6.67259 × 10−11 N m2 /kg2 )
(4.7 × 1024 kg)
×
(2.99792 × 108 m/s)2
r0 =
= 0.0069788 m .
Hewitt CP9 10 E48
011 10.0 points
If the Earth shrank in size, with all other
factors remaining the same, how would the
escape velocity from its surface change?
1. smaller
2. greater correct
3. Cannot be determined
4. the same
Explanation:
The gravitational field at the surface of the
Earth would increase if the Earth shrank;
escape velocity would be correspondingly
greater.
AP B 1998 MC 40
012 10.0 points
What is the kinetic energy of a satellite of
mass m that orbits the Earth of mass M in a
circular orbit of radius R?
1 GM m
2 R2
GM m
2. K =
R2
1. K =
4
1 GM m
correct
2
R
Explanation:
The gravitational force on the satellite provides the centripetal force needed to keep it
in circular orbit:
5. K =
v2
GM m
= FG = Fc = m
R2
R
GM m
2
mv =
, so
R
K=
1 GM m
1
m v2 =
.
2
2
R
Apollo 11 in Orbit 01
013 (part 1 of 2) 10.0 points
When it orbited the Moon, the Apollo 11
spacecraft’s mass was 12500 kg, and its
mean distance from the Moon’s center was
1.93304 × 106 m.
Find the orbital speed of the spacecraft.
Assume its orbit to be circular and the
Moon to be a uniform sphere of mass
7.36 × 1022 kg. The gravitational constant
is 6.67259 × 10−11 N · m2 /kg2 .
Correct answer: 1593.92 m/s.
Explanation:
The gravitational force supplies the centripetal force:
GM m
v2
=
m
d2
r d
GM
v=
q d
=
6.67259 × 10−11 N · m2 /kg2
s
7.36 × 1022 kg
×
1.93304 × 106 m
= 1593.92 m/s .
3. K = 0
4. K =
1 GM m
4
R
014 (part 2 of 2) 10.0 points
What is the minimum energy required for the
Version 001 – test #2 Gravitation – tubman – (19112)
craft to leave the orbit and escape the Moon’s
gravitational field?
Correct answer: 1.58786 × 1010 J.
Explanation:
The total energy is
1
GmM
E = K + U = m v2 −
2
d
1
GmM
GmM
GM
= m
−
=−
2
d
d
2d
−11
2
2
= −6.67259 × 10
N · m /kg
(12500 kg)(7.36 × 1022 kg)
×
2(1.93304 × 106 m)
= −1.58786 × 1010 J .
The minimum energy required for the craft to
leave the orbit and escape the Moon’s gravitational field is
~ = 1.58786 × 1010 J .
Emin = kEk
New Satellite Orbit 02
015 (part 1 of 3) 10.0 points
What is the kinetic energy of a satellite of
mass m which is in a circular orbit of radius
3 Re about the earth?
G Me m
1. K =
3 Re
G Me m
2. K =
Re2
m v2
3. K =
3 Re
4. K = m g Re
5. K =
G Me m
correct
6 Re
6. K = 3 m g Re
7. K =
m v2
6 Re
8. K = 3 G Me m
9. K = −
G Me m
3 Re
5
G Me m
6 Re2
Explanation:
The acceleration of the satellite in circular
v2
orbit of radius 3 Re is ac =
, so the force
3 Re
on the satellite is
G Me m
m v2
=
F = m ac =
3 Re
(3 Re)2
G Me m
m v2 =
e Re
and the kinetic energy is
1
G Me m
K = m v2 =
.
2
6 Re
10. K = −
016 (part 2 of 3) 10.0 points
What is the total energy of the satellite?
1
m v2
2
G Me m
=−
correct
6 Re
G Me m
=
Re
G Me m
=
3 Re
G Me m
=−
6 Re2
Gm
=
Re
G Me m
= 3 m g Re +
3 Re
1. E = m g Re +
2. E
3. E
4. E
5. E
6. E
7. E
8. E = 3 G Me m
G Me m
3 Re
Explanation:
The potential energy of the satellite is
G Me m
U =−
3 Re
so the total energy is
G Me m G Me m
E =K +U =
−
6 Re
3 Re
G Me m
.
=−
6 Re
9. E = −
Version 001 – test #2 Gravitation – tubman – (19112)
017 (part 3 of 3) 10.0 points
How much work must an external force
do on the satellite to move it from a circular orbit of radius 2 Re to 3 Re , if its mass
is 2000 kg?
The universal gravitational
constant 6.67 × 10−11 N · m2 /kg2 , the mass
of the Earth 5.98 × 1024 kg and its radius
6.37 × 106 m.
Correct answer: 1.04361 × 1010 J.
Explanation:
Let :
G = 6.67 × 10−11 N · m2 /kg2 ,
Me = 5.98 × 1024 kg , and
Re = 6.37 × 106 m .
The work done by an external force to move
the satellite from the closer orbit to the further orbit will be the work against gravity (a
positive number which yields the change in
potential energy) plus the change in kinetic
energy (a negative number since the kinetic
energy is smaller in the orbit with the greatest
radius):
G Me m
G Me m
W = Ef − Ei = −
− −
6 Re
4 Re
G Me m
=
12 Re
1
=
(6.67 × 10−11 N · m2 /kg2 )
12
(5.98 × 1024 kg)(2000 kg)
×
6.37 × 106 m
= 1.04361 × 1010 J .
6