Responses to Questions
1. Whether the apple is attached to a tree or falling, it exerts a gravitational force on the Earth equal to the force the Earth exerts on it, which is the weight of the apple (Newton’s third law).
2. The tides are caused by the difference in gravitational pull on two opposite sides of the Earth. The gravitational pull from the Sun on the side of the Earth closest to it depends on the distance from the Sun to the close side of the Earth. The pull from the Sun on the far side of the Earth depends on this distance plus the diameter of the Earth. The diameter of the Earth is a very small fraction of the total Earth–Sun distance, so these two forces, although large, are nearly equal. The diameter of the Earth is a larger fraction of the Earth–Moon distance, and so the difference in gravitational force from the Moon to the two opposite sides of the Earth will be greater.
3. The object will weigh more at the poles. The value of r² at the equator is greater, both from the Earth’s center and from the bulging mass on the opposite side of the Earth. Also, the object has centripetal acceleration at the equator. The two effects do not oppose each other.
4. Since the Earth’s mass is greater than the Moon’s, the point at which the net gravitational pull on the spaceship is zero is closer to the Moon. A spaceship traveling from the Earth towards the Moon must therefore use fuel to overcome the net pull backwards for over half the distance of the trip. However, when the spaceship is returning to the Earth, it reaches the zero point at less than half the trip
distance, and so spends more of the trip “helped” by the net gravitational pull in the direction of travel.
5. The gravitational force from the Sun provides the centripetal force to keep the Moon and the Earth going around the Sun. Since the Moon and Earth are at the same average distance from the Sun, they travel together, and the Moon is not pulled away from the Earth.
6. As the Moon revolves around the Earth, its position relative to the distant background stars changes.
This phenomenon is known as “parallax.” As a demonstration, hold your finger at arm’s length and look at it with one eye at a time. Notice that it “lines up” with different objects on the far wall depending on which eye is open. If you bring your finger closer to your face, the shift in its position against the background increases. Similarly, the Moon’s position against the background stars will shift as we view it in different places in its orbit. The distance to the Moon can be calculated by the amount of shift.
7. At the very center of the Earth, all of the gravitational forces would cancel, and the net force on the object would be zero.
8. A satellite in a geosynchronous orbit stays over the same spot on the Earth at all times. The satellite travels in an orbit about the Earth’s axis of rotation. The needed centripetal force is supplied by the component of the gravitational force perpendicular to the axis of rotation. A satellite directly over the North Pole would lie on the axis of rotation of the Earth. The gravitational force on the satellite in this case would be parallel to the axis of rotation, with no component to supply the centripetal force needed to keep the satellite in orbit.
9. According to Newton’s third law, the force the Earth exerts on the Moon has the same magnitude as the force the Moon exerts on the Earth. The Moon has a larger acceleration, since it has a smaller mass (Newton’s second law, F = ma).
10. The satellite needs a certain speed with respect to the center of the Earth to achieve orbit. The Earth rotates towards the east so it would require less speed (with respect to the Earth’s surface) to launch a satellite towards the east (a). Before launch, the satellite is moving with the surface of the Earth so already has a “boost” in the right direction.
11. If the antenna becomes detached from a satellite in orbit, the antenna will continue in orbit around the Earth with the satellite. If the antenna were given a component of velocity toward the Earth (even a very small one), it would eventually spiral in and hit the Earth.
12. Ore normally has a greater density than the surrounding rock. A large ore deposit will have a larger mass than an equal amount of rock. The greater the mass of ore, the greater the acceleration due to gravity will be in its vicinity. Careful measurements of this slight increase in g can therefore be used to estimate the mass of ore present.
13. Yes. At noon, the gravitational force on a person due to the Sun and the gravitational force due to the Earth are in the opposite directions. At midnight, the two forces point in the same direction.
Therefore, your apparent weight at midnight is greater than your apparent weight at noon.
14. Your apparent weight will be greatest in case (b), when the elevator is accelerating upward. The scale reading (your apparent weight) indicates your force on the scale, which, by Newton’s third law, is the same as the normal force of the scale on you. If the elevator is accelerating upward, then the net force must be upward, so the normal force (up) must be greater than your actual weight (down).
When in an elevator accelerating upward, you “feel heavy.”
Your apparent weight will be least in case (c), when the elevator is in free fall. In this situation your apparent weight is zero since you and the elevator are both accelerating downward at the same rate and the normal force is zero.
Your apparent weight will be the same as when you are on the ground in case (d), when the elevator is moving upward at a constant speed. If the velocity is constant, acceleration is zero and N = mg.
(Note that it doesn’t matter if the elevator is moving up or down or even at rest, as long as the velocity is constant.)
15. If the Earth’s mass were double what it is, the radius of the Moon’s orbit would have to double (if the Moon’s speed remained constant), or the Moon’s speed in orbit would have to increase by a factor of the square root of 2 (if the radius remained constant). If both the radius and orbital speed were free to change, then the product rv² would have to double.
16. If the Earth were a perfect, nonrotating sphere, then the gravitational force on each droplet of water in the Mississippi would be the same at the headwaters and at the outlet, and the river wouldn’t flow.
Since the Earth is rotating, the droplets of water experience a centripetal force provided by a part of the component of the gravitational force perpendicular to the Earth’s axis of rotation. The centripetal force is smaller for the headwaters, which are closer to the North pole, than for the outlet, which is closer to the equator. Since the centripetal force is equal to mg – N (apparent weight) for each droplet, N is smaller at the outlet, and the river will flow. This effect is large enough to overcome smaller effects on the flow of water due to the bulge of the Earth near the equator.
17. The satellite remains in orbit because it has a velocity. The instantaneous velocity of the satellite is tangent to the orbit. The gravitational force provides the centripetal force needed to keep the satellite in orbit, acting like the tension in a string when twirling a rock on a string. A force is not needed to keep the satellite “up”; a force is needed to bend the velocity vector around in a circle.
18. Between steps, the runner is not touching the ground. Therefore there is no normal force up on the runner and so she has no apparent weight. She is momentarily in free fall since the only force is the force of gravity pulling her back toward the ground.
19. If you were in a satellite orbiting the Earth, you would have no apparent weight (no normal force).
Walking, which depends on the normal force, would not be possible. Drinking would be possible, but only from a tube or pouch, from which liquid could be sucked. Scissors would not sit on a table (no apparent weight = no normal force).
20. The centripetal acceleration of Mars in its orbit around the Sun is smaller than that of the Earth. For both planets, the centripetal force is provided by gravity, so the centripetal acceleration is inversely proportional to the square of the distance from the planet to the Sun:
2
2
p s p
m v Gm m
r r so
2 2
v Gms
r r
Since Mars is at a greater distance from the Sun than Earth, it has a smaller centripetal acceleration.
Note that the mass of the planet does not appear in the equation for the centripetal acceleration.
21. For Pluto’s moon, we can equate the gravitational force from Pluto on the moon to the centripetal force needed to keep the moon in orbit:
2
2 p m
m Gm m
m v
r r
This allows us to solve for the mass of Pluto (mp) if we know G, the radius of the moon’s orbit, and the velocity of the moon, which can be determined from the period and orbital radius. Note that the mass of the moon cancels out.
22. The Earth is closer to the Sun in January. The gravitational force between the Earth and the Sun is a centripetal force. When the distance decreases, the speed increases. (Imagine whirling a rock around your head in a horizontal circle. If you pull the string through your hand to shorten the distance between your hand and the rock, the rock speeds up.)
2
2
E S E
m v Gm m
r r so GmS
v r
Since the speed is greater in January, the distance must be less. This agrees with Kepler’s second law.
23. The Earth’s orbit is an ellipse, not a circle. Therefore, the force of gravity on the Earth from the Sun is not perfectly perpendicular to the Earth’s velocity at all points. A component of the force will be parallel to the velocity vector and will cause the planet to speed up or slow down.
24. Standing at rest, you feel an upward force on your feet. In free fall, you don’t feel that force. You would, however, be aware of the acceleration during free fall, possibly due to your inner ear.
25. If we treat gG
as the acceleration due to gravity, it is the result of a force from one mass acting on another mass and causing it to accelerate. This implies action at a distance, since the two masses do not have to be in contact. If we view gG
as a gravitational field, then we say that the presence of a mass changes the characteristics of the space around it by setting up a field, and the field then interacts with other masses that enter the space in which the field exists. Since the field is in contact with the mass, this conceptualization does not imply action at a distance.
Solutions to Problems
1. The spacecraft is at 3.00 Earth radii from the center of the Earth, or three times as far from the Earth’s center as when at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance, the force of gravity on the spacecraft will be one-ninth of its weight at the Earth’s surface.
2
1 Earth's 9
surface
1480 kg 9.80 m s
1610 N
G 9 F mg
This could also have been found using Eq. 6-1, Newton’s law of universal gravitation.
2. The force of gravity on an object at the surface of a planet is given by Newton’s law of universal gravitation, Eq. 6-1, using the mass and radius of the planet. If that is the only force on an object, then the acceleration of a freely falling object is acceleration due to gravity.
Moon 2 Moon Moon G
M m
F G mg
r o
22
11 2 2 2
Moon
Moon 2 6 2
Moon
7.35 10 kg
6.67 10 N m kg 1.62 m s
1.74 10 m g GM
r
u u
u
3. The acceleration due to gravity at any location on or above the surface of a planet is given by
2
planet planet ,
g G M r where r is the distance from the center of the planet to the location in question.
2
Planet Earth Earth 2
planet 2 2 2 2 2 Earth 2
Earth Earth
1 1 9.80 m s
1.9 m s
2.3 2.3 2.3
2.3
M M M
g G G G g
r R R
4. The acceleration due to gravity at any location at or above the surface of a planet is given by
2
planet Planet ,
g G M r where r is the distance from the center of the planet to the location in question.
2 2
Planet Earth Earth
planet 2 2 2 Earth
Earth Earth
1.80 1.80 1.80 1.80 9.80 m s 17.6 m s
M M M
g G G G g
r R R
Đ ã
ă á
â ạ
5. The acceleration due to gravity is determined by the mass of the Earth and the radius of the Earth.
0 new 0 0 2
0 2 new 2 2 2 9 0
0 new 0 0
2 2
3 9
GM GM G M GM
g g g
r r r r
So g is multiplied by a factor of 2 9 .
6. The acceleration due to gravity at any location at or above the surface of a planet is given by
2
planet Planet ,
g G M r where r is the distance from the center of the planet to the location in question.
For this problem, MPlanet MEarth 5.97 10 kg.u 24
(a) r REarth6400 m 6.38 10 m 6400 mu 6
24
11 2 2 2
Earth
2 6 2
5.98 10 kg
6.67 10 N m kg 9.78 m s
6.38 10 m 6400 m g GM
r
u
u < u
(b) r REarth6400 km 6.38 10 m 6.4 10 m 12.78 10 m 3 sig figu 6 u 6 u 6
24
11 2 2 2
Earth
2 6 2
5.98 10 kg
6.67 10 N m kg 2.44 m s
12.78 10 m g GM
r
u
u < u
7. The distance from the Earth’s center is r REarth300 km 6.38 10 m 3 10 mu 6 u 5
6.68 10 m 2 sig figu 6 . Calculate the acceleration due to gravity at that location.
24
11 2 2 2
Earth Earth
2
2 2 6
2
2
5.97 10 kg
6.67 10 N m kg 8.924 m s
6.68 10 m 1" "
8.924 m s 0.91 's
9.80 m s
M M
g G G
r r
g g
u
u u
Đ ã
ă á
â ạ
<
This is only about a 9% reduction from the value of g at the surface of the Earth.
8. We are to calculate the force on Earth, so we need the distance of each planet from Earth.
6 10 6 11
Earth Earth
Venus Jupiter
6 12
Earth Saturn
150 108 10 km 4.2 10 m 778 150 10 km 6.28 10 m 1430 150 10 km 1.28 10 m
r r
r
u u u u
u u
Jupiter and Saturn will exert a rightward force, while Venus will exert a leftward force. Take the right direction as positive.
Earth Jupiter Earth Saturn Earth Venus
Earth- 2 2 2
planets Earth Earth Earth
Jupiter Saturn Venus
2
Earth 11 2 12 2 10 2
11 2 2
318 95.1 0.815
6.28 10 m 1.28 10 m 4.2 10 m
6.67 10 N m kg 5.97 10
M M M M M M
F G G G
r r r
GM
u u u
u u
Đ ã
ă á
ă á
â ạ
< 24kg 2 4.02 10 mu 22 2 9.56 10 Nu 17 | 9.6 10 Nu 17
The force of the Sun on the Earth is as follows.
24 30
11 2 2 22
Earth Sun
Earth- 2 11 2
Sun Earth
Sun
5.97 10 kg 1.99 10 kg
6.67 10 N m kg 3.52 10 N
1.50 10 m
M M
F G
r
u u
u u
< u
And so the ratio is Earth- Earth- 17 22 5
planets Sun
9.56 10 N 3.52 10 N 2.7 10 ,
F F u u u which is 27 millionths.
9. Calculate the force on the sphere in the lower left corner, using the free- body diagram shown. From the symmetry of the problem, the net forces in the x and y directions will be the same. Note T q45 .
2 2 2
right dia 2 2 2
1 1
cos 1
2 2 2
x 2
m m m
F F F G G G
d d d
T
Đ ã
ă á
â ạ
Thus
2 2
1 1
y x 2 2 F F Gm
d Đ ã
ă á
â ạ. The net force can be found by the
m
m m
m d d
T FGup
right
FG FGdiag
Pythagorean combination of the two component forces. Due to the symmetry of the arrangement, the net force will be along the diagonal of the square.
2 2
2 2 2
2 2
2
11 2 2 8 o
2
1 1
2 2 1 2 2
2 2 2
8.5 kg 1
6.67 10 N m kg 2 1.4 10 N at 45
0.80 m 2
x y x x
m m
F F F F F G G
d d
u u
Đ ã Đ ã
ă á
ă á â ạ
â ạ
Đ ã
ă á
â ạ
<
The force points towards the center of the square.
10. Assume that the two objects can be treated as point masses, with m1 m and m2 4.00 kgm. The gravitational force between the two masses is given by the following.
11 2 2 2 10
1 2
2 2 2
4.00 4.00
6.67 10 N m kg 2.5 10 N
0.25 m
m m
m m m m
F G G
r r
u u
This can be rearranged into a quadratic form of m2 4.00m0.234 0. Use the quadratic formula to solve for m, resulting in two values which are the two masses.
1 3.94 kg , 2 0.06 kg
m m
11. The force on m due to 2m points in the ˆidirection. The force on m due to 4m points in the ˆj direction. The force on m due to 3m points in the direction given by 1 0
0
tan y .
T x Add the force vectors together to find the net force.
2 2 2 2 2 2
0 0 0 0 0 0
2 2 2
0 0
2 2 2 2 2 2 2 2 2 2
0 0 0 0 0 0 0 0 0 0
2 0 0
3 / 2 3 / 2
2 2 2 2 2 2
0 0 0 0 0 0
2 ˆ 4 ˆ 3 cos ˆ 3 sin ˆ
3
2 ˆ 4 ˆ 3 ˆ ˆ
2 3 ˆ 4 3 ˆ
m m m m m m m m
G G G G
x y x y x y
m m
m m m x y
G G G G
x y x y x y x y x y
x y
Gm x x y y x y
T T
êĐ ã Đ ã º
ă á ă á
ô ằ
ă á ă á
ôâ ạ â ạ ằ
ơ ẳ
F i j i j
i j i j
i j
G
12. With the assumption that the density of Europa is the same as Earth’s, the radius of Europa can be calculated.
1/ 3
Europa Earth Europa
Europa Earth 4 3 4 3 Europa Earth
3 Europa 3 Earth Earth
M M M
r r
r r M
U U
S S
o o Đ ã
ă á
â ạ
1/ 3 2 / 3 1/ 3 1/ 3
Europa Europa Europa Earth Earth Europa Europa
Europa 2 1/ 3 2 2 2 1/ 3 Earth
Europa Earth Earth Earth Earth
Europa Earth
Earth 22 1/ 3 2
24
4.9 10 kg 9.80 m s
5.98 10 kg
GM GM GM M GM M M
g g
r M r r M M
r M
u u
Đ ã
ă á
Đ Đ ã ã â ạ
ă ă á á
ă â ạ á
â ạ
Đ ã
ă á
â ạ
2 2
1.98 m s | 2.0 m s
13. To find the new weight of objects at the Earth’s surface, the new value of g at the Earth’s surface needs to be calculated. Since the spherical shape is being maintained, the Earth can be treated as a point mass. Find the density of the Earth using the actual values, and use that density to find g under the revised conditions.
1/ 3
E E E E
original 2 4 3 3 E
3
E E E
1/ 3 1/ 3 1/ 3
1/ 3 1/ 3
E E E
E
original 2 / 3 2 / 3 new 2 / 3 2 / 3
E
3 3
;
4 4
; 2 2 2
3 3 3 3
4 4 4 4
m m m m
g G r
r r r
m m m
g G m G g G G g
m
U S S SU
SU SU SU SU
o Đ ã o
ă á
â ạ
Đ ã Đ ã Đ ã Đ ã
ă á ă á ă á ă á
â ạ â ạ â ạ â ạ
Thus g is multiplied by 21/ 3, and so the weight would be multiplied by 21/ 3.
14. The expression for the acceleration due to gravity at the surface of a body is body 2body
body
M ,
g G
R where Rbody is the radius of the body. For Mars, gMars 0.38gEarth.
Mars Earth
2 2
Mars Earth
2 2
24 23
Mars
Mars Earth
Earth
0.38
3400 km
0.38 0.38 5.98 10 kg 6.5 10 kg
R 6380 km
M M
G G
R R
M M R
o
u u
Đ ã Đ ã
ă á
ă á â ạ
â ạ
15. For the net force to be zero means that the gravitational force on the spacecraft due to the Earth must be the same as that due to the Moon.
Write the gravitational forces on the spacecraft, equate them, and solve for the distance x. We measure from the center of the bodies.
Earth spacecraft Moon spacecraft
Earth- 2 Moon 2
spacecraft spacecraft
2 2 Earth spacecraft Moon spacecraft
2 2
Earth Moon Earth Moon
;
M m M m
F G F G
x d x
M m M m x d x x d x
G G
x d x M M M M
o o
24
Earth 8 8
22 24
Moon Earth
5.97 10 kg
3.84 10 m 3.46 10 m
7.35 10 kg 5.97 10 kg x d M
M M
u u u
u u
This is only about 22 Moon radii away from the Moon. Or, it is about 90% of the distance from the center of the Earth to the center of the Moon.
16. The speed of an object in an orbit of radius r around the Sun is given by v G MSun r, and is also given by v 2Sr T, where T is the period of the object in orbit. Equate the two expressions for the speed and solve for MSun, using data for the Earth.
d
spacecraft
Earth Moon
d-x x
2 11 3
2 3 Sun 30
Sun 2 11 2 2 7 2
4 1.50 10 m
2 4
2.01 10 kg
6.67 10 N m kg 3.15 10 sec
M r r
G M
r T GT
S S S
o u u
u < u
This is the same result obtained in Example 6-9 using Kepler’s third law.
17. Each mass M will exert a gravitational force on mass m. The vertical components of the two forces will sum to be 0, and so the net force on m is directed horizontally. That net force will be twice the horizontal component of either force.
2 2
Mm
F GMm
x R o
3 / 2
2 2 2 2 2 2 2 2
net 2 2 3 / 2
cos 2 2
Mm x
x Mm x
GMm GMm x GMmx
F x R x R x R x R
F F GMmx
x R
T
18. From the symmetry of the problem, we can examine diametrically opposite infinitesimal masses and see that only the horizontal
components of the force will be left. Any off-axis components of force will add to zero. The
infinitesimal horizontal force on m due to an infinitesimal mass dM is
2 2
dMm
dF Gm dM
x r .
The horizontal component of that force is given by the following.
dMmx 2 2cos 2 2 2 2 2 23 / 2
Gm Gm x Gmx
dF dM dM dM
x r T x r x r x r
The total force is then found by integration.
2 23 / 2 2 23 / 2 2 23 / 2
x x x
Gmx dM Gmx dM GMmx
dF dF F
x r o x r o x r
³ ³
From the diagram we see that it points inward towards the center of the ring.
19. The expression for g at the surface of the Earth is 2E
E
m . g G
r Let g 'g be the value at a distance of rE 'r from the center of Earth, which is 'r above the surface.
(a)
2
E E E E
2 2
2 2
E E 2 E E E
E E
1 1 2
1
m m m m r r
g G g g G G G g
r r r r r r r
r r
' '
o ' | o
' '
Đ ã Đ ã
ă á ă á
Đ ã â ạ â ạ
ă á
â ạ
TT R
R
x
2 2
x R
2 2
x R
TT r
r
x
2 2
x r
2 2
x r dM
dM
E
2 r
g g
r ' | '
(b) The minus sign indicated that the change in g is in the opposite direction as the change in r. So, if r increases, g decreases, and vice-versa.
(c) Using this result:
2 56 2 2
E
1.25 10 m
2 2 9.80 m s 0.384 m s 9.42 m s
6.38 10 m
g g r g
r
' u
' | o
u Direct calculation:
24
11 2 2 2
E
2 6 5 2
5.98 10 kg
6.67 10 N m kg 9.43m s
6.38 10 m 1.25 10 m g Gm
r
u
u < u u
The difference is only about 0.1%.
20. We can find the actual g by taking g due to the uniform Earth, subtracting away g due to the bubble as if it contained uniform Earth matter, and adding in g due to the oil-filled bubble. In the diagram, r = 1000 m (the diameter of the bubble, and the distance from the surface to the center of the bubble). The mass of matter in the bubble is found by taking the density of the matter times the volume of the bubble.
oil uniform bubble bubble
present Earth (Earth (oil)
matter)
oil uniform bubble bubble
present Earth (oil) (Earth
matter)
g g g g
g g g g g
o
'
bubble
bubble (Earth
oil matter) 4 3
bubble bubble oil Earth 3 bubble
2 2 2 2
(Earth matter
oil
matter)
GM GM
G G
M M r
r r r Đăăâ ãááạ r ĐăâU U ãáạ S The density of oil is given, but we must calculate the density of a uniform Earth.
24
3 3
E
Earth 4 3 4 6 3
matter 3 E 3
5.98 10 kg
5.50 10 kg m 6.38 10 m
m U r
S S
u u
u
4 3
oil Earth 3 bubble
2 matter
11 2 2
2 3 3 3 4 2 3
3 3 2
6.67 10 N m kg
8.0 10 kg m 5.50 10 kg m 5.0 10 m
1.00 10 m
g G r
r U U S
S
'
u u u u
u
Đ ã
ă á
â ạ
<
4 2 4 2
1.6414 10 m su | 1.6 10 m su Finally we calculate the percentage difference.
4 2
3 2
1.6414 10 m s
% 100 1.7 10 %
9.80 m s g
g
' u u u
The negative sign means that the value of g would decrease from the uniform Earth value.
rE
rEr r