Physics questions, chapter 3

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Gravitation

43 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first. G = 6.67 × 10⁻¹¹ N m² kg⁻², mass of the Earth = 5.97 × 10²⁴ kg, radius of the Earth = 6.37 × 10⁶ m.

Section A: objective questions

13 marks

Choose the best answer for each question.

Question 1

Which statement is Newton's Universal Law of Gravitation?

The gravitational force is directly proportional to the distance between two bodies
The gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centres
The gravitational force is inversely proportional to the product of the masses
The gravitational force is directly proportional to the square of the distance between the bodies

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F = Gm₁m₂/r²: proportional to m₁m₂ and inversely proportional to r² (textbook p. 80).

For the teacher: Remembering, low difficulty, standard 3.1.1

Textbook p. 80

Question 2

The distance between two bodies is doubled while their masses stay the same. The gravitational force becomes

twice as large
half as large
four times as large
one quarter as large

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F ∝ 1/r². Doubling r divides F by 2² = 4.

For the teacher: Applying, low difficulty, standard 3.1.2

Textbook p. 82

Question 3

Which quantity does the gravitational acceleration, g, at a point near a planet NOT depend on?

Mass of the planet
Distance from the centre of the planet
Mass of the object that falls
Radius of the planet

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g = GM/r². The mass of the object cancels out.

For the teacher: Understanding, low difficulty, standard 3.1.3

Textbook p. 83

Question 4

A piece of space junk of mass 24 kg is 7.00 × 10⁶ m from the centre of the Earth. What is the gravitational force between the junk and the Earth?

195 N
235 N
1.4 × 10³ N
8.5 × 10⁻⁶ N

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F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(24) ÷ (7.00 × 10⁶)² = 195 N.

For the teacher: Applying, low difficulty, standard 3.1.2

Textbook p. 95

Question 5

A satellite is 4.20 × 10⁷ m from the centre of the Earth. What is the gravitational acceleration at this position?

9.81 m s⁻²
0.226 m s⁻²
2.45 m s⁻²
0.0 m s⁻²

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g = GM/r² = (6.67 × 10⁻¹¹)(5.97 × 10²⁴) ÷ (4.20 × 10⁷)² = 0.226 m s⁻². It is small but not zero.

For the teacher: Applying, low difficulty, standard 3.1.3

Textbook p. 113

Question 6

What is the orbital period of a geostationary satellite?

1.5 hours
12 hours
27.3 days
24 hours

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A geostationary satellite has the same period as the Earth's rotation, 24 hours.

For the teacher: Remembering, low difficulty, standard 3.3.2

Textbook p. 105

Question 7

Which force provides the centripetal force that keeps a satellite in orbit around the Earth?

Gravitational force of the Earth
Thrust from the rocket engine
Friction with the atmosphere
Tension in a string

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Gravitational force acts towards the centre of the Earth and acts as the centripetal force. The satellite needs no engine in orbit.

For the teacher: Understanding, low difficulty, standard 3.1.5

Textbook p. 91

Question 8

A satellite of mass 400 kg orbits at radius 8.2 × 10⁶ m with linear speed 6.96 × 10³ m s⁻¹. What is the centripetal force on it?

1.2 × 10³ N
3.4 × 10⁵ N
2.4 × 10³ N
1.9 × 10¹⁰ N

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F = mv²/r = 400 × (6.96 × 10³)² ÷ 8.2 × 10⁶ = 2.36 × 10³ N ≈ 2.4 × 10³ N.

For the teacher: Applying, medium difficulty, standard 3.1.5

Textbook p. 95

Question 9

The Earth orbits the Sun at r = 1.50 × 10¹¹ m in 1 year. Saturn's orbital radius is 1.43 × 10¹² m. What is Saturn's orbital period?

9.5 years
29 years
91 years
910 years

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T₂ = 1 × (1.43 × 10¹² ÷ 1.50 × 10¹¹)^(3/2) = 9.53^1.5 = 29.4 years.

For the teacher: Applying, medium difficulty, standard 3.2.3

Textbook p. 114

Question 10

Escape velocity from a planet depends on

the mass of the rocket and the mass of the planet
the mass of the rocket only
the mass and the radius of the planet
the orbital period of the rocket

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v = √(2GM/R) contains only the mass and radius of the planet; the mass of the object does not appear.

For the teacher: Understanding, low difficulty, standard 3.3.3

Textbook p. 107

Question 11

Diagram 1 shows three bodies in a line. The gravitational force between A and B is P. What is the gravitational force between B and C?

Bodies A 100 kg, B 100 kg and C 200 kg in a line, 5 m apart each — Diagram 1

P ÷ 2
P
4P
2P

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B and C are the same 5 m apart as A and B, but C has twice the mass of A. F ∝ m₁m₂, so the force is 2P.

For the teacher: Analysing, high difficulty, standard 3.1.2

Textbook p. 115

Question 12

Satellite Q orbits the Earth at four times the orbital radius of satellite P. Compared with P, the linear speed of Q is

half
double
one quarter
the same

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v = √(GM/r), so v ∝ 1/√r. Four times the radius gives 1/√4 = 1/2 of the speed.

For the teacher: Analysing, high difficulty, standard 3.3.1

Textbook p. 104

Question 13

According to Kepler's second law, a planet in an elliptical orbit moves fastest when it is

at the furthest point from the Sun
at the centre of the ellipse
at the nearest point to the Sun
on the minor axis

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The line to the Sun sweeps equal areas in equal times, so the planet must move faster when it is closer to the Sun.

For the teacher: Understanding, low difficulty, standard 3.2.1

Textbook p. 97

Section B: structured questions

30 marks

Answer all questions. Write your answers in the spaces provided.

Question 14

Mars (mass m) moves in a circular orbit of radius r around the Sun (mass M) with orbital period T and linear speed v.

[3 marks] (a)Write the formula for the gravitational force on Mars, the centripetal force on Mars and its linear speed.

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Gravitational force F = GMm/r²1
Centripetal force F = mv²/r1
v = 2πr/T1

Tick each point you got.

[2 marks] (b)Derive an expression for the mass of the Sun in terms of r and T.

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Equate: mv²/r = GMm/r², so v² = GM/r1
Substitute v = 2πr/T: M = 4π²r³/GT²1

Tick each point you got.

[2 marks] (c)r = 2.28 × 10¹¹ m and T = 687 days. Calculate the mass of the Sun.

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T = 687 × 86 400 = 5.94 × 10⁷ s1
M = 4π²(2.28 × 10¹¹)³ ÷ [(6.67 × 10⁻¹¹)(5.94 × 10⁷)²] = 1.99 × 10³⁰ kg1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 3.1.6

Textbook p. 112

Question 15

Diagram 2 shows a planet in an elliptical orbit around the Sun. X is the point nearest the Sun and Z is the furthest.

Elliptical orbit with the Sun at a focus and planet positions X, Y and Z — Diagram 2

[1 mark] (a)State Kepler's second law.

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A line joining a planet to the Sun sweeps out equal areas in equal times1

Tick each point you got.

[1 mark] (b)Compare the linear speed of the planet at X, Y and Z.

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Speed at X is greatest, then Y, and Z is the slowest (X > Y > Z)1

Tick each point you got.

[1 mark] (c)Explain your answer to (b).

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For equal times the area swept is equal, so the planet must cover a longer arc where it is closer to the Sun1

Tick each point you got.

For the teacher: Analysing, low difficulty, standard 3.2.1

Textbook p. 97

Question 16

A weather satellite orbits the Earth at a height of 560 km. Mass of the satellite is 1 200 kg.

[2 marks] (a)Calculate the gravitational acceleration at the position of the satellite.

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g = GM/(R + h)² = (6.67 × 10⁻¹¹)(5.97 × 10²⁴) ÷ (6.37 × 10⁶ + 5.6 × 10⁵)²1
= 8.29 m s⁻²1

Tick each point you got.

[2 marks] (b)Another satellite, Proba-1, orbits at a height of 700 km. Calculate its linear speed.

Show answerHide answer

v = √(GM/r), r = 6.37 × 10⁶ + 7.0 × 10⁵ = 7.07 × 10⁶ m1
v = 7.50 × 10³ m s⁻¹1

Tick each point you got.

[1 mark] (c)Why does the linear speed of a satellite not depend on its mass?

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The mass of the satellite appears on both sides (in gravitational force and centripetal force) and cancels1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 3.3.1

Textbook p. 110

Question 17

Escape velocity is calculated with v = √(2GM/R).

[1 mark] (a)What is meant by escape velocity?

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The minimum velocity needed by an object at the surface to overcome the gravitational force and escape into outer space1

Tick each point you got.

[2 marks] (b)Planet Saturn has mass 5.68 × 10²⁶ kg and radius 6.03 × 10⁷ m. Calculate the escape velocity of Saturn.

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v = √[2 × (6.67 × 10⁻¹¹)(5.68 × 10²⁶) ÷ 6.03 × 10⁷]1
= 3.54 × 10⁴ m s⁻¹1

Tick each point you got.

[1 mark] (c)Saturn's rings are made of small particles. Why do they not escape into outer space?

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Their speed is lower than the escape velocity, so gravity keeps them in orbit around Saturn1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 3.3.4

Textbook p. 114

Question 18

Table 1 shows three possible orbits X, Y and Z around the Earth. Satellite P is an Earth imaging satellite and satellite Q is a communication satellite.

Table of orbits: X ellipse 6.70 × 10³ m 1.41 hours; Y circle 3.59 × 10⁷ m 24.04 hours; Z circle 5.43 × 10⁷ m 41.33 hours — Table 1

[2 marks] (a)Which orbit is suitable for satellite Q? Explain.

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Orbit Y1
Its period is about 24 hours, so the satellite is geostationary and always above the same place for communication1

Tick each point you got.

[2 marks] (b)Which orbit is suitable for satellite P? Explain.

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Orbit X1
It is a low orbit with a short period, so the satellite passes over many different places and is close enough for clear images1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 3.3.2

Textbook p. 115

Question 19

The Moon orbits the Earth at radius 3.83 × 10⁸ m with a period of 27.3 days.

[1 mark] (a)State Kepler's third law.

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The square of the orbital period is directly proportional to the cube of the radius of the orbit (T² ∝ r³)1

Tick each point you got.

[2 marks] (b)Calculate the orbital radius of a geostationary satellite (T = 1 day).

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r³ = (3.83 × 10⁸)³ × (1 ÷ 27.3)²1
r = 4.22 × 10⁷ m1

Tick each point you got.

[1 mark] (c)What is the height of the satellite above the surface of the Earth?

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h = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.59 × 10⁷ m1

Tick each point you got.

For the teacher: Applying, high difficulty, standard 3.2.3

Textbook p. 102

Question 20

A space agency plans a communication satellite that stays above one point on Mars. Mars has mass 6.42 × 10²³ kg, radius 3.40 × 10⁶ m and one rotation takes 24.6 hours.

[1 mark] (a)State the orbital period the satellite should have.

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24.6 hours (same as Mars's rotation), moving in the same direction1

Tick each point you got.

[2 marks] (b)Calculate the orbital radius needed. Use r³ = GMT²/4π².

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T = 24.6 × 3600 = 8.86 × 10⁴ s; r³ = (6.67 × 10⁻¹¹)(6.42 × 10²³)(8.86 × 10⁴)² ÷ 4π²1
r = 2.04 × 10⁷ m (height ≈ 1.70 × 10⁷ m above the surface)1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 3.3.2

Textbook p. 116

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