2.3.1 In Class or Homework Exercise
1. The force of gravity between two identical bowling balls is 1.71 10 8 N. If the
bowling balls are 0.50 m apart, what is the mass of each bowling ball?
Since the bowling balls are identical, they will have the same mass.
Fg 1.71 10 8 N
r
0.50m
m ?
Fg
1.71 10
8
m2
m
Gm1m2
r2
(6.67 10 11 )mm
(0.50) 2
64.1
8.0kg
2. A hypothetical planet has a radius 1.6 times that of the earth, but has the
same mass. What is the acceleration due to gravity near its surface?
Although this problem could be solved using the known values for the earth’s
radius and mass, it is easier to set it up in terms of multiples of these
quantities.
R p 1.6 Re
Mp
gp
gp
Me
?
GM p
Rp 2
G(M e )
(1.6 Re ) 2
1 GM e
2.56 Re 2
1
ge
2.56
1
(9.80)
2.56
3.8m / s 2
3. Another hypothetical planet has a radius 20.0 times that of earth and a mass
100. times that of earth. What is g near the surface?
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Rp
20.0 Re
Mp
100.M e
gp
?
gp
GM p
Rp 2
G (100.M e )
(20.0 Re ) 2
100. GM e
400. Re 2
1
ge
4
1
(9.80)
4
2.45m / s 2
4. What is the effective value of g at a height of 1000.0 km above the earth's
surface? That is, what is the acceleration due to gravity of objects allowed to
fall freely at this altitude? Sketch a velocity-time graph of the object as it falls
toward the earth.
h 1000.0km 1.0000 106 m
r Re h
Re 6.38 106 m
6.38 106 1.0000 106
24
M e 5.98 10 kg
7.38 106 m
g ?
g
GM
R2
(6.67 10 11 )(5.98 1024 )
(7.38 106 ) 2
7.32m / s 2
As the object falls toward the earth, the acceleration should be increasing;
therefore, the slope of the velocity-time graph should be increasing with time.
If we take down as positive, the graph should look something like the
following:
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5. Determine the net force on the moon due to the gravitational attraction of both
the earth and the sun, assuming that the moon is between the sun and the
earth. The distance between the moon and the earth is 3.85 105 km , and the
distance between the moon and the sun is 1.50 108 km . All distances are
center to center.
mm 7.36 10 22 kg
me
5.98 10 24 kg
ms
1.99 1030 kg
rme
3.85 105 km 3.85 108 m
rms 1.50 108 km 1.50 1011 m

Fnet ?
If we draw a diagram of the situation (not to scale), we see that the earth and
the sun are pulling in opposite directions on the moon:
Drawing a free body diagram for the moon, we have
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Fme
Gmm me
rme 2
(6.67 10 11 )(7.36 1022 )(5.98 1024 )
(3.85 108 ) 2
1.98 1020 N
Fms
Gmm ms
rms 2
(6.67 10 11 )(7.36 10 22 )(1.99 1030 )
(1.50 1011 ) 2
4.34 1020 N
Using the direction toward the sun as positive,

Fnet

Fme
Fme

Fms
Fms
1.98 10 20
4.34 10 20
2.36 10 20 N
The positive sign tells us that the net force is toward the sun.
6. Do the previous question again, this time assuming that the earth and the sun
are pulling at right angles to one another.
If we draw a diagram of the situation (not to scale),
Drawing a free body diagram for the moon, we have
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Fme
Gmm me
rme 2
(6.67 10 11 )(7.36 1022 )(5.98 1024 )
(3.85 108 ) 2
1.98 1020 N
Fms
Gmm ms
rms 2
(6.67 10 11 )(7.36 10 22 )(1.99 1030 )
(1.50 1011 ) 2
4.34 1020 N

Since Fnet

Fme

Fms , a vector addition diagram gives us
tan
Fnet
Fme2
Fms2
20 2
(1.98 10 )
20 2
(4.34 10 )
20
4.77 10 N

Fnet
Fme
Fms
1.98 1020
4.34 1020
24.5
4.77 1020 N , 24.5 from the sun
7. Frank is really concerned about his weight. But Frank is lazy, and doesn't
really want to exercise in order to lose weight. How far above the surface of
the earth will Frank have to go so that his weight will be only half of what it is
on the surface of the earth? How will this affect Frank's mass?
Weight (or force of gravity) is determined by two factors – mass and the
acceleration due to gravity. Since he is not exercising, he is not going to lose
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any mass – his mass will not change. The acceleration due to gravity must be
half of what it is on earth if his weight is to be cut in half.
4.90m / s 2
g
5.98 1024 kg
Me
6.38 106 m
Re
h ?
We will first find the distance that he must be from the center of the earth.
GM
g
r2
(6.67 10 11 )(5.98 1024 )
4.9
r2
r 9.02 106 m
We can now find the height above the earth:
r Re h
9.02 106
h
6.38 106 h
2.64 106 m
8. A spaceship is travelling from the earth to the moon. At what distance from the
earth will the spaceship experience zero net force because the earth and the
moon pull with equal and opposite forces? The distance (center to center)
between the earth and the moon is 3.85 105 km .
Me
5.98 1024 kg
Mm
7.36 1022 kg
de
?
If the net force on the spaceship is zero, then the magnitude of the force of
gravity being exerted by the earth must be the same as the magnitude of the
force of gravity being exerted by the moon.
Fse Fsm
Gms M e
de 2
Gms M m
dm2
5.98 1024
de 2
7.36 1022
(3.85 108 d e ) 2
5.98 1024
de 2
7.36 1022
1.48 1017 7.70 108 d e
7.36 1022 d e 2
de 2
8.85 1041 4.60 1033 d e 5.98 1024 d e 2
0 5.91 1024 d e 2 4.60 1033 de 8.85 1041
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This must be solved using the quadratic formula, giving the following two
answers:
de
4.30 108 m
de
3.48 108 m
Since the first answer is greater than the total distance between the earth and
the moon, the second answer is the correct one for this problem.
9. A force of 40.0 N is required to pull a 10.0 kg wooden block at a constant
velocity across a smooth glass surface on earth. A physics class is planning a
class trip to Jupiter ( M J 1.90 1027 kg , RJ 6.98 107 m ), and would like to
figure out beforehand what force would be necessary to pull the same wooden
block across the same glass surface on Jupiter. Can you help them out?
It is first necessary to find the coefficient of friction on the earth; this will be the
same on Jupiter since the same two surfaces are involved.
Earth
Fp 40.0 N
Fg
m 10.0kg
?

ma

ma

Fp
0
Fp
Fp

F
mg
(10.0)(9.80)
98.0 N

Ff
Ff
FN
40.0
(98.0)
0.408
Jupiter
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0.408
MJ
1.90 1027 kg
RJ
6.98 107 m
Fp
?
On Jupiter, only the force of gravity will be different:
Fg
Gm1m2
r2
(6.67 10 11 )(1.90 10 27 )(10.0)
(6.98 107 ) 2
260. N

ma

ma

Fp
0
Fp
Fp

F

Ff
Ff
FN
(0.408)(260.)
106 N
10. Four 8.0 kg spheres are located at the corners of a square of sides 0.50 m.
Calculate the magnitude and direction of the gravitational force on one sphere
due to the other three.
Drawing a free body diagram for sphere 1, we have
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Gm1m2
r12 2
Fg12
Fg13
Gm1m3
r132
(6.67 10 11 )(8.0)(8.0)
(0.50) 2
(6.67 10 11 )(8.0)(8.0)
(0.50) 2
1.71 10 8 N
1.71 10 8 N
To find F14 we must find the distance between spheres 1 and 4:
(0.50)2 (0.50) 2
r14
0.71m
Fg14
Gm1m4
r14 2
(6.67 10 11 )(8.0)(8.0)
(0.71) 2
8.5 10 9 N

Since Fnet

F12
F14 x
F14 y
(8.5 10 9 )sin 45
Fnetx
Fnety
6.0 10
Fnet
Fnetx 2


F13 F14
9
6.0 10 9 N
1.71 10
8
2.31 10 8 N
Fnety 2
(2.31 10 8 ) 2 (2.31 10 8 ) 2
3.3 10 8 N
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
Fnet
3.3 10 8 N toward the centre of the square (along the diagonal)
11. What happens to the gravitational force between 2 masses when the distance
between the masses is doubled?
Gm1m2
Fg 2
Gm1m2
(2r ) 2
Fg1
2
r
1 Gm1m2
4 r2
1
Fg1
4
The force will be one quarter the original force.
12. What happens to the gravitational force between two objects if the distance
between the objects is tripled and one of the masses is doubled?
G (2m1 )m2
Fg 2
Gm1m2
(3r ) 2
Fg1
2
r
2 Gm1m2
9 r2
2
Fg1
9
The force will be two-ninths the original force.
13. What happens to the gravitational force between two objects if the distance
between the objects is halved and each of the masses is tripled?
G (3m1 )(3m2 )
Fg 2
Gm1m2
( 12 r ) 2
Fg1
2
r
9 Gm1m2
1
r2
4
36 Fg1
The force will be 36 times the original force.
14. If Earth were twice as massive but remained the same size, what would
happen to the value of G?
Since G is a universal constant, its value will not change.
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