Active Physics Full Solutions to Textbook Exercises 12 Gravitation | p.

12 Gravitation 3. C
The gravitational force is
Checkpoint

Checkpoint 1 (p.353)
1. (a) F where R is the Earth’s radius.
The gravitational force exerted on an apple 4. C
by the Earth is the same as that exerted on Combining Newton’s law of universal gravitation
the Earth by the apple. They are an action– and the equation for centripetal force, we have
reaction pair.
(b) T
The period of the circular orbit does not 5. C
depend on the mass of the object rotating in
the orbit.
(c) F
In the formula , the radius R represents
the distance between the object and the 6. (a) The gravitational force exerted on the apple
Earth’s centre, not the height of the object by the Earth is
from the Earth’s surface.
2. Gravitational force between the Earth and the Sun

(downwards)
(b) The gravitational force exerted on the Earth
by the apple is 0.981 N (upwards) because it
forms an action–reaction pair with the
gravitational force exerted on the Earth by
Checkpoint 2 (p.358)
the apple.
1. Gravitational field strength on the Moon’s surface
(c) The acceleration of the apple is

The acceleration of the Earth is

Acceleration due to gravity on the Moon’s surface Owing to the enormous mass of the Earth, its
= 1.62 m s−2 acceleration is too small to be observed.
2. Weight of A = (0.1)(9.81) = 0.981 N 7. (a) The orbital speed of Io is
Weight of B = (0.1)(2.45) = 0.245 N
Weight of C = (0.1)(1.09) = 0.109 N

Exercise
(b)
Exercise 12.1 (p.353) So, the mass of Jupiter is
1.
distance between the gravitational force
masses between the masses
r F
2r F/4
3r F/9
8. (a)
4r F/16
2. B
The two forces form an action–reaction pair.

The ratio of their orbital speeds is 1.37:1.

Active Physics Full Solutions to Textbook Exercises 12 Gravitation | p.2

(b) Given that both P and Q experience the same 3. A

gravitational force.

The distance is d = r' − r 0 = 0.41r 0 above the

ground.
The mass ratio is 0.284 : 1 .
4. Since F ∝ g, we have, by proportion,
9. Let d be the distance between that point and the
Earth, and D be the Earth–Mars distance. Hence,
The weight of the astronaut on the Earth’s surface
is 785 N.
5. (a) The gravitational field strength at Venus’s
surface is
Substituting data, we have
0 = ( 5.328 × 1024) d2 − (9.361 × 1035) d
+ (3.670 × 1046 )
∴ d = 5.90 × 1010 m or 1.17 × 1011 m
Since the Earth–Mars distance D = 7.84 × 1010 m
and d < D, the answer is 5.90 × 1010 m.
10. (a) The gravitational force acting on the satellite (b)
provides the required centripetal force for it
to undergo uniform circular motion: The radius

The distance above the ground

= 6.699 × 106 − 6.37 × 106 = 3.29 × 105 m
The gravitational field strength at a distance
Since M denotes the mass of the planet (a
3.29 × 105 m above the Earth’s surface is the
constant), we have T2 ∝ r3.
same as that at Venus’s surface.
(b) (i) Since T2 ∝ r3, by proportion, we have
6. (a) The mass of Neptune M is given by

Therefore, the orbital period of satellite β is

9.08 × 105 s (or 10.5 days).
(b) The volume of the Neptune is
(ii) From (a),

Therefore, the mass of planet X is The average density of Neptune is

7. (a) The gravitational field strength at the top of

Mount Everest is
Exercise 12.2 (p.359)
1. A
2. B
Active Physics Full Solutions to Textbook Exercises 12 Gravitation | p.3

(b) The weight of the man at the top of Mount

Everest is

8. (a)
6. C
By proportion,

(b) The acceleration of the satellite due to the

Earth’s gravity is 7. A
Only statement (3) is incorrect. G is the
gravitational constant and does not depend on the
position of the object.
towards the centre of the Earth 8. D

Chapter Exercise

Multiple-choice Questions (p.361)

1. C
9. B
Statement (1) is incorrect. Since F = GMm/r2, the
On the surface of the Earth,
satellites experience gravitational forces of
different magnitudes because they have different
mass.
On the surface of the planet,
Statements (2) and (3) are correct because
g = GM/r2 and T2 ∝ r3.
2. B Dividing gives
Since g = GM/r2 ∝ 1/r2, a graph of g against
1/r2 shows a straight line passing through the
origin. 10. D
3. B Period T is independent of angle θ due to the
Statement (1) is incorrect. At an altitude of spherical symmetry. Both satellites have the same
350 km, the space station still experiences a period.
significant gravitational force from the Earth.  Note that
Statement (2) is incorrect. The altitude of a
geostationary object must be 3.59×107 m ≈
3500 km.
11. B
Statement (3) is correct. By F = mrω2, if the orbital
Option A is incorrect. The gravitational force
radius r of the space station increases, its angular
exerted on the astronaut by the Earth provides the
speed ω decreases.
centripetal force he needs to perform circular
4. C motion, and hence its effect on his motion is
significant.
Option C is incorrect. The feeling of apparent
5. B weightlessness arises because there is no reaction
force acting on the astronaut by the floor.
Option D is incorrect. The centripetal force is the
net force required by the astronaut to move in a
circular orbit, and it is provided by the
gravitational force acting on him.
Active Physics Full Solutions to Textbook Exercises 12 Gravitation | p.4

When r increases, a decreases. So, the

Structured Questions (p.362) asteroid gains kinetic energy at a lower
rate at a higher altitude. (1A)
12. (a) The gravitational field strength of a planet is
defined as the gravitational force per unit (ii) It comes from the gravitational
mass experienced by a small object at potential energy lost by the
distance r from the centre of the planet. (2A) asteroid. (1A)

By Newton’s law of universal gravitation, we 14. (a) The mass (or the density) of the planet (1A)

have The radius of the planet (1A)

(b) (i) The radius of the granite rock is

r = 0.2 km = 200 m (1A)
Therefore, the gravitational field strength is
The volume of the granite rock is

(b) The asteroid is r = 4 × 106 + 2 × 107 =

2.4 × 107 m from the centre of the planet.
From the graph, the corresponding The difference between the mass is
gravitational field strength is g = 1.4 N
kg−1. (1M)
The gravitational force experienced by the
asteroid is
F = mg = (5 × 106)(1.4) = 7 × 106 N (1A) (ii) The difference in the gravitational field
strength comes from the difference of
(c) From the graph, the gravitational field
mass. (1A)
strength on the planet’s surface is 2 N
kg−1. (1A)

The mass of the planet is

(iii)

13. (a) Suppose the gravitational force exerted on

the asteroid by the Earth is the only force
acting on it.
By Newton’s second law, we have

Therefore, the acceleration of the asteroid is

independent of its mass. (1A)
(b) When the asteroid is at an altitude of
10 000 m, its acceleration is
(Dumbbell shape curve that is always
below the original one: 1A)
15. (a) The variation in the height is still small
compared with the Earth’s radius. (1A)
(b) (i) The height of the satellite is
r = 940 × 103 + 6400 × 103
(c) (i) The acceleration a of the asteroid has
= 7.34 × 106 m
an inverse-square relation with its
distance r from the Earth’s centre
(i.e. a = GM/r2). (1A)
Active Physics Full Solutions to Textbook Exercises 12 Gravitation | p.5

By the law of universal gravitation, the (ii) The orbital speed of the satellite is
gravitation force is

(b) (i)

(ii) The unbalanced force provides the

necessary centripetal force for the
satellite to perform uniform circular
motion. (1A)
(iii) The gravitational force provides the
Correct force arrow: 1A
necessary centripetal force for the
satellite to perform uniform circular (ii) Since the gravitational force has to
motion: provide the necessary centripetal force
for the satellite, the plane of the orbit of
the satellite must pass through the
The angular velocity of the satellite is
centre of the Earth. (1A)

17. (a) The gravitational force acting on the satellite

provides the centripetal force it requires:

(iv) The orbital time of the satellite is

Hence, the orbital radius of a geostationary
satellite is
or 1.74 hours.
(c) Work has to be done against the frictional
drag force. (1A)
Hence, the velocity of the satellite
decreases. (1A)
16. (a) (i) Since the gravitational force acting on
the satellite provides the necessary (b) Consider a satellite which is orbiting around
centripetal force for the satellite to the Earth at an orbital period T , an orbital
undergo uniform circular motion, we radius r and an altitude h.
have

Since r E is the radius of the Earth, the orbital

Considering the gravitational force
period of the satellite only depends on its
acting on an object on the Earth’s
altitude. (1A)
surface, we have
(c) The polar-orbiting satellites pass over the
poles many times a day, and hence their
orbital periods are shorter than those of the
geostationary satellites. (1A)
Combining the two above gives
By the result of (b), the shorter the period T,
the lower the altitude h. (1A)

Thus, the orbital radius of the satellite

is r = 4.235 × 107 ≈ 4.24 × 107 m. (1A)
Active Physics Full Solutions to Textbook Exercises 12 Gravitation | p.6

(d) (i) The period of their orbital motion is By proportion, we have

Therefore, the speed of the satellite orbiting near

the surface of planet X is 10v. (1A)

(ii) The number of times the satellites pass

over the North Pole in on day is

Shoot-the-stars Questions (p.364)

1. B
The Earth, which has a larger mass, exerts a
greater gravitational force on the spacecraft when
it is midway between the Earth and the Moon.
Therefore, the position where the gravitational
forces exerted by the Earth and the Moon cancel
each other should be closer to the Moon.
2. The acceleration due to gravity g is numerically
equal to the gravitational field strength, which is
given by

Bob assumes that the acceleration due to gravity g

is a constant of 9.81 m s−2 regardless of the change
in the altitude of the object. To be precise, this
value is only valid at the Earth’s surface (i.e. h =
0). (1A)
When the object is thrown upwards, its altitude h
becomes larger and hence the actual value of g
should be lower. Therefore, the calculated value of
v would be higher than the actual value. (1A)
3. The centripetal force required by the satellite to
under circular motion is provided by the
gravitational force exerted by the Earth:

The density of a planet is

Combining the above two equations gives

The speed of the satellite is given by