OCR · A-Level · Physics
Course version: H556 · Version from Sept 2015
Gravitational potential and energy: (e) escape velocity.
Escape velocity is the minimum speed a projectile needs at the surface of a mass M to escape its gravitational field and reach infinity with zero kinetic energy.
Checked against the OCR H556 specification (opens in a new tab) on .
- Point 5 of 5
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Escape velocity is the minimum speed a projectile needs at the surface of a mass M to escape its gravitational field and reach infinity with zero kinetic energy. At the surface, radius R, the total energy is ½mv² − GMm/R. Setting this equal to zero at infinity gives ½mv² = GMm/R, so v = √(2GM/R). The mass m cancels, so escape velocity depends only on the mass and radius of the body, not on the projectile. For example, for a planet of mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m, v = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.4 × 10⁶) ≈ 1.1 × 10⁴ m s⁻¹. Escape velocity is a scalar speed, and the direction of launch is ignored in this idealised treatment.
What earns the marks
- States that escape velocity is the minimum speed for a projectile to reach infinity with zero kinetic energy.
- Derives v = √(2GM/R) by equating total energy at the surface to zero at infinity.
- Explains that the projectile mass m cancels, so escape velocity is independent of the projectile.
- Calculates escape velocity using G = 6.67 × 10⁻¹¹ N m² kg⁻² and consistent SI units.
- Recognises that escape velocity depends on the mass and radius of the body being escaped from.
Common mistakes and how to put them right
Including the projectile mass m in the final expression.
Put it right: m cancels, leaving v = √(2GM/R).
Using the orbital speed formula v = √(GM/r) instead of the escape formula.
Put it right: Escape requires √2 times the circular orbital speed at the same radius.
Substituting the radius in kilometres without converting to metres.
Put it right: Convert all quantities to SI units before calculating.
Exam tips
- Start from the energy condition ½mv² − GMm/R = 0 and cancel m before substituting numbers.
- Convert radii from km to m and use G = 6.67 × 10⁻¹¹ N m² kg⁻² in SI units.
- Give the answer as a speed with the correct unit, and check the order of magnitude against typical values such as 1.1 × 10⁴ m s⁻¹ for Earth.
Try the lesson
The start of the Gravitational potential and energy: (e) escape velocity. lesson, made from checked teaching notes for this specification point.
Listen
Escape velocity is the minimum speed a projectile needs at the surface of a mass M to escape its gravitational field and reach infinity with zero kinetic energy. At the surface, radius R, the total energy is ½mv² − GMm/R. Setting this equal to zero at infinity gives ½mv² = GMm/R, so v = √(2GM/R). The mass m cancels, so escape velocity depends only on the mass and radius of the body, not on the projectile. For example, for a planet of mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m, v = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.4 × 10⁶) ≈ 1.1 × 10⁴ m s⁻¹. Escape velocity is a scalar speed, and the direction of launch is ignored in this idealised treatment. A strong answer earns marks for these points. States that escape velocity is the minimum speed for a projectile to reach infinity with zero kinetic energy. Derives v = √(2GM/R) by equating total energy at the surface to zero at infinity. Explains that the projectile mass m cancels, so escape velocity is independent of the projectile. Calculates escape velocity using G = 6.67 × 10⁻¹¹ N m² kg⁻² and consistent SI units. A common slip: Including the projectile mass m in the final expression. Instead, m cancels, leaving v = √(2GM/R). Exam tip: Start from the energy condition ½mv² − GMm/R = 0 and cancel m before substituting numbers.
See it
Key words
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Escape velocity
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Energy condition for escape
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Independence from projectile mass
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Escape velocity formula
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See every definition
- Escape velocity
- The minimum speed a projectile needs at the surface of a mass M to escape its gravitational field and reach infinity with zero kinetic energy.
- Energy condition for escape
- At the surface, radius R, the total energy is ½mv² − GMm/R, and setting this equal to zero at infinity gives ½mv² = GMm/R.
- Independence from projectile mass
- The mass m cancels, so escape velocity depends only on the mass and radius of the body, not on the projectile.
- Escape velocity formula
- The expression v = √(2GM/R) obtained by equating total energy at the surface to zero at infinity.
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