Gravitational fieldsAQA A-Level Physics: Topic test
20 questions, 54 marks
AQA A-Level Physics
Gravitational fields topic test
Total 54 marks
Name
Class
Date
- 1Saturn's moon Titan has a mass of 1.35 × 10²³ kg and a radius of 2.57 × 10⁶ m. Treat Titan as a uniform sphere. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)What is the gravitational field strength at the surface of Titan?[1 mark]
- A9.0 × 10¹² N kg⁻¹
- B2.0 × 10¹⁰ N kg⁻¹
- C3.5 × 10⁶ N kg⁻¹
- D1.4 N kg⁻¹
(b)What is the gravitational field strength at a height above the surface of Titan equal to the radius of Titan?[1 mark]- A0.34 N kg⁻¹
- B0.68 N kg⁻¹
- C0.15 N kg⁻¹
- D5.5 N kg⁻¹
(c)A probe of mass 300 kg lands on the surface of Titan. Show that the gravitational force between Titan and the probe is about 410 N.[2 marks]Total for question 1: 4 marks
- 2Two stars, A and B, form a binary system. Star A has a mass of 4.0 × 10³⁰ kg and star B has a mass of 1.0 × 10³⁰ kg. Their centres are 3.0 × 10¹¹ m apart. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)What is the magnitude of the gravitational force between the stars?[1 mark]
- A8.9 × 10³⁸ N
- B2.7 × 10⁵⁰ N
- C3.0 × 10²⁷ N
- D7.4 × 10²⁶ N
(b)The mass of star B were to double and the separation of the stars were also to double. By what factor would the force between them change?[1 mark]- A× 2
- B× ½
- C× ¼
- D× 1
(c)Explain why the force on star A due to star B is equal in magnitude to the force on star B due to star A, but the accelerations of the two stars are different. Calculate the acceleration of star B due to star A.[2 marks]Total for question 2: 4 marks
- 3Mercury has a mass of 3.30 × 10²³ kg and a radius of 2.44 × 10⁶ m. Treat Mercury as a uniform sphere with no atmosphere. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)Calculate the gravitational potential at the surface of Mercury, and explain the significance of the sign of your answer.[3 marks](b)A probe of mass 500 kg is raised from the surface of Mercury to a height of 2.44 × 10⁶ m above the surface. Calculate the work done on the probe.[4 marks]
Total for question 3: 7 marks
- 4Europa orbits Jupiter in a circular path of radius 6.71 × 10⁸ m with a time period of 3.55 days. Europa has a mass of 4.8 × 10²² kg. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)Show that for a satellite in a circular orbit is proportional to , and use the data to calculate the mass of Jupiter.[6 marks](b)Calculate the gravitational potential at the orbit of Europa and the total energy of Europa in its orbit. Explain the significance of the sign of the total energy.[6 marks]
Total for question 4: 12 marks
- 5A uniform spherical planet has a radius of 5.0 × 10⁶ m. The gravitational field strength at its surface is 12 N kg⁻¹. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)What is the mass of the planet?[1 mark]
- A4.5 × 10²⁴ kg
- B9.0 × 10¹⁷ kg
- C1.8 × 10²⁵ kg
- D2.0 × 10⁴ kg
(b)What is the gravitational potential at the surface of the planet?[1 mark]- A+6.0 × 10⁷ J kg⁻¹
- B−6.0 × 10⁷ J kg⁻¹
- C−2.4 × 10⁻⁶ J kg⁻¹
- D0 J kg⁻¹
(c)A 40 kg load is lifted vertically through 1.0 × 10³ m near the surface of the planet. Use the gravitational field strength to calculate the change in gravitational potential and the work done on the load.[2 marks]Total for question 5: 4 marks
- 6Engineers plan to place a communications satellite in an orbit that is synchronous with Mars, so that it stays above the same point on the Martian equator. Mars has a mass of 6.42 × 10²³ kg and takes 8.86 × 10⁴ s to rotate once. The radius of the satellite's orbit is 2.04 × 10⁷ m. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)Which row gives the conditions that the orbit must satisfy?[1 mark]
- Aequatorial plane; opposite to the rotation of Mars; time period equal to the rotation period of Mars
- Bpolar plane; same direction as the rotation of Mars; time period equal to the rotation period of Mars
- Cequatorial plane; same direction as the rotation of Mars; time period equal to the rotation period of Mars
- Dequatorial plane; same direction as the rotation of Mars; time period half the rotation period of Mars
(b)Another satellite orbits Mars with a time period eight times as long as that of a second satellite. By what factor is the radius of its orbit greater than that of the second satellite?[1 mark]- A× 2
- B× 8
- C× 64
- D× 4
(c)Calculate the orbital speed of the satellite.[2 marks]Total for question 6: 4 marks
- 7Venus has a mass of 4.87 × 10²⁴ kg and a radius of 6.05 × 10⁶ m. Treat Venus as a uniform sphere with no atmosphere. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)Calculate the escape velocity from the surface of Venus.[3 marks](b)A probe of mass 600 kg is in a circular orbit at a height of 3.0 × 10⁶ m above the surface of Venus. Calculate the minimum extra energy that must be given to the probe so that it can escape from Venus.[4 marks]
Total for question 7: 7 marks
- 8Ganymede has a mass of 1.48 × 10²³ kg and a radius of 2.63 × 10⁶ m. A probe is launched vertically from its surface at 1.5 × 10³ m s⁻¹. Ignore any atmosphere and the rotation of Ganymede. Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².(a)Calculate the gravitational field strength at the surface of Ganymede and the escape speed. Explain why the probe, launched at 1.5 × 10³ m s⁻¹, cannot escape.[6 marks](b)Calculate the maximum height above the surface reached by the probe. Compare your answer with the height predicted by assuming a uniform field equal to the surface value, and explain the difference.[6 marks]
Total for question 8: 12 marks
End of questions
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).