Gravitational fields and escape
How does a rock leave Mars, the Moon or the asteroid Vesta? Newton's law of gravitation and the idea of gravitational potential give the escape velocity, and explain which bodies send us meteorites.
- Use g = GM/r² to calculate surface field strength.
- Derive and use the escape velocity v = √(2GM/r).
- Explain why lunar and Martian meteorites exist but meteorites from Venus or Earth have not been identified.
Links to: A-level physics, gravitational fields: Newton's law of gravitation, field strength, gravitational potential and escape velocity.
Field strength at the surface
Newton's law of gravitation gives the force between two masses as F = GMm/r², with G = 6.67 × 10⁻¹¹ N m² kg⁻². The field strength at the surface of a spherical body of mass M and radius r is g = GM/r².
For Vesta, M ≈ 2.6 × 10²⁰ kg and r ≈ 2.6 × 10⁵ m, so g ≈ 0.25 N kg⁻¹, about 1/40 of Earth's. A small body's field is weak because g depends on M, which scales with r³, divided by r², so for similar densities g is roughly proportional to radius.
Escape velocity
The gravitational potential at distance r is V = −GM/r: the work done per kilogram to bring a mass from infinity. To escape, an object's kinetic energy must at least equal the energy needed to reach infinity: ½mv² = GMm/r, giving v_esc = √(2GM/r), independent of the object's mass.
Escape velocities are about 11.2 km s⁻¹ for Earth, 5.0 km s⁻¹ for Mars, 2.4 km s⁻¹ for the Moon and only about 0.36 km s⁻¹ for Vesta. Impacts eject some material above these speeds. That is why we have meteorites from the Moon, Mars and Vesta, while launching rock off Earth or Venus, with their high escape velocities and thick atmospheres, is far harder.
Surviving the launch
Accelerating rock to 5 km s⁻¹ without melting it seemed impossible until the process called spallation was understood. Near the surface, the shock wave from an impact reflects off the free surface and partly cancels, so a thin layer of rock is thrown off at high speed while experiencing relatively modest pressure. Martian meteorites show shock effects, such as plagioclase converted to glass (maskelynite), consistent with pressures of a few tens of gigapascals, high but survivable.
Escape velocity from Vesta
Vesta has a mass of 2.59 × 10²⁰ kg and a mean radius of 2.63 × 10⁵ m. Calculate its escape velocity. (G = 6.67 × 10⁻¹¹ N m² kg⁻²)
- v_esc = √(2GM/r).
- 2GM = 2 × 6.67 × 10⁻¹¹ × 2.59 × 10²⁰ = 3.455 × 10¹⁰ m³ s⁻².
- 2GM/r = 3.455 × 10¹⁰ ÷ 2.63 × 10⁵ = 1.314 × 10⁵ m² s⁻².
- v_esc = √(1.314 × 10⁵) ≈ 362 m s⁻¹.
Answer: About 360 m s⁻¹, about the speed of sound in air, which is why a large impact on Vesta could launch so much crust into space.
Calculate the escape velocity of the Moon (M = 7.35 × 10²² kg, r = 1.74 × 10⁶ m) and of Mars (M = 6.42 × 10²³ kg, r = 3.39 × 10⁶ m). Then plot escape velocity against radius for Vesta, the Moon, Mars and Earth (M = 5.97 × 10²⁴ kg, r = 6.37 × 10⁶ m) and comment on the trend.
Quiz
5 questions. Choose an answer to see whether you are right and why. Finish the quiz to complete the lesson.
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Question 1Which expression gives escape velocity from the surface of a body of mass M and radius r?
Setting ½mv² equal to GMm/r gives v = √(2GM/r). √(GM/r) is the speed of a low circular orbit.
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Question 2Why does escape velocity not depend on the mass of the escaping object?
½mv² = GMm/r: the mass m appears on both sides and cancels.
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Question 3What is the gravitational potential at the surface of a body?
Potential is the work done per unit mass bringing a mass from infinity, V = −GM/r; it is negative because the field is attractive.
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Question 4What process allows rock to be ejected from Mars at escape velocity without melting?
Near the free surface, reflected shock waves reduce the peak pressure while still accelerating a thin layer to high speed.
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Question 5For two bodies of the same density, how does surface field strength depend on radius?
g = GM/r² and M ∝ r³ for constant density, so g ∝ r.