OCR A Physics · Module 5: Newtonian world and astrophysics
Gravitational fields: practice questions with answers
6 exam-style questions on gravitational fields. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Using the right equation2 marks
Mars has a mass of 6.42 × 10²³ kg and a radius of 3.39 × 10⁶ m. Calculate the gravitational field strength at its surface.
g = N kg⁻¹
Show the answer and working
Answer g = 3.73 N kg⁻¹
g = GM/r² = (6.67 × 10⁻¹¹ × 6.42 × 10²³) ÷ (3.39 × 10⁶)² = 3.73 N kg⁻¹.
- Got 1.26 × 10⁷ N kg⁻¹? You did not square the radius. g = GM ÷ r² = 3.73 N kg⁻¹.
- Got 3.73 × 10⁶ N kg⁻¹? You used the radius in km. Use r = 3.39 × 10⁶ m: g = GM ÷ r² = 3.73 N kg⁻¹.
- Unit prefixes3 marks
A satellite orbits 400 km above the surface of the Earth. The Earth has radius 6.37 × 10⁶ m and mass 5.97 × 10²⁴ kg. Calculate its orbital speed, in km s⁻¹.
v = km s⁻¹
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Answer v = 7.67 km s⁻¹
r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m. v = √(GM/r) = 7.67e+03 m s⁻¹ = 7.67 km s⁻¹.
- Got 31.6 km s⁻¹? You used the height as the orbital radius. The radius is measured from the Earth's centre: r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m, giving 7.67 km s⁻¹.
- Got 58800 km s⁻¹? You did not take the square root. GMm/r² = mv²/r gives v = √(GM/r) = 7.67 km s⁻¹.
- Using the right equation3 marks
The Moon orbits the Earth at a radius of 3.84 × 10⁸ m with a period of 27.3 days. Use Kepler's third law to calculate the orbital radius of a geostationary satellite (period 1.00 day), in km.
r = km
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Answer r = 42400 km
T² ∝ r³, so r₂ = r₁ × (T₂/T₁)^(2/3) = 3.84 × 10⁸ × (1 ÷ 27.3)^(2/3) = 4.24 × 10⁷ m = 4.24e+04 km.
- Got 14100 km? You scaled the radius directly with the period. Kepler's third law is T² ∝ r³, so r = 3.84 × 10⁸ × (1 ÷ 27.3)^(2/3) = 4.24e+04 km.
- Got 2690 km? You used the exponent the wrong way round. r³ ∝ T², so r ∝ T^(2/3): r = 4.24e+04 km.
- Using the right equation3 marks
Calculate the gain in gravitational potential energy of a 1200 kg satellite raised from the Earth's surface (radius 6.37 × 10⁶ m) to an orbit of radius 1.274 × 10⁷ m. The Earth has mass 5.97 × 10²⁴ kg. Give your answer in GJ.
ΔE = GJ
Show the answer and working
Answer ΔE = 37.5 GJ
ΔE = GMm(1/r₁ - 1/r₂) = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 1200 × (1/6.37 × 10⁶ - 1/1.274 × 10⁷) = 37.5 GJ.
- Got 75 GJ? You used the potential at the surface only. ΔE = GMm(1/r₁ - 1/r₂) = GMm × (1/6.37 × 10⁶ - 1/1.274 × 10⁷) = 37.5 GJ.
- Got 18.8 GJ? You halved your answer. The change in energy is GMm(1/r₁ - 1/r₂) = 37.5 GJ.
- Using the right equation1 mark
Why is gravitational potential always negative (with zero at infinity)?
- A Because gravity is always repulsive.
- B Because the mass of the Earth is negative.
- C Because work must be done against the field to move a mass to infinity.
Show the answer and working
Answer C: Because work must be done against the field to move a mass to infinity.
Gravity is attractive, so energy must be supplied to take a mass out to infinity, where the potential is zero.
- Why not A? Gravity is attractive, which is why the potential is negative.
- Why not B? Masses are positive; the sign comes from choosing zero potential at infinity.
- Using the right equation1 mark
Which statement about a geostationary satellite is correct?
- A It orbits above the equator with a period of 24 hours.
- B It orbits over the poles with a period of 24 hours.
- C It orbits at the same height as the International Space Station.
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Answer A: It orbits above the equator with a period of 24 hours.
To stay above a fixed point it must orbit in the equatorial plane with a 24 hour period, in the same direction as Earth rotates.
- Why not B? Over the poles it would not remain above a fixed point on Earth.
- Why not C? A geostationary orbit is much higher than the ISS, as the period of 24 hours needs a larger radius.