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Escape Speed

Key idea: Derive escape speed using energy: v_esc = √(2GM/r), apply it at different altitudes, and avoid common traps (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse escape speed using conservation of energy.

1. Definitions (Must Know)

A. Escape speed, v_esc

Escape speed (escape velocity), v_esc, at a distance r from the centre of a planet is the minimum speed needed to reach infinity with zero speed at infinity, assuming:

  • no air resistance,
  • no further propulsion,
  • only the planet’s gravity acts.

For a planet of mass M:

v_esc = square root of (2GM/r)

At the surface (r = R):

v_esc = square root of (2GM/R) = square root of 2gR

2. Key Ideas (What Earns Marks)

  • Escape speed is found using energy: (1/2)mv_esc² = Δ U
  • With U = -GMm/r and U(∞) = 0: (1/2)mv_esc² = GMm/r
  • v_esc depends on M and r, not on the rocket’s mass.
  • Higher starting altitude (larger r) gives smaller escape speed.
Common mix-up

Orbital speed for a circular orbit is v_orb = square root of (GM/r). Escape speed is larger by a factor of square root of 2: v_esc = square root of 2 v_orb.

Orbital speed vs escape speed (Earth, scaled)

Both speeds decrease with distance, and escape speed is always √2 times orbital speed at the same radius.

Scroll across the graph to read all labels.

Both speeds decrease with distance, and escape speed is always √2 times orbital speed at the same radius.Both speeds decrease with distance, and escape speed is always √2 times orbital speed at the same radius.
For the same r, v_esc = square root of 2 v_orb. Both decrease as 1/square root of r.
Open full-size graph
View figure data
Values for Orbital speed vs escape speed (Earth, scaled)
Distance from Earth's centre, r (10⁷ m)v_orb = √(GM/r)v_esc = √(2GM/r)
0.6377.911.17
16.328.93
24.476.32
4.223.084.35
Exam pitfall: misreading the escape condition

Escape speed means reaching infinity with zero final speed under gravity alone. It does not mean zero speed at launch, and it assumes no air resistance after launch.

3. Detailed Explanations

A. Derivation using energy

Escape condition: the object reaches infinity with zero speed, so total energy at infinity is:

E_∞ = 0

At radius r, total energy is:

Eᵣ = (1/2)mv² - GMm/r

Set Eᵣ = E_∞ = 0 for the minimum speed:

0 = (1/2)mv_esc²-GMm/r; (1/2)mv_esc² = GMm/r; v_esc = square root of (2GM/r)

B. Why square root of 2gR appears

At Earth’s surface, g = GM/R², so:

v_esc = square root of (2GM/R) = square root of 2gR

This is not the “uniform field to infinity” model. It is just an algebraic substitution using the exact surface value of g.

4. Common Mistakes

  • Using r = h instead of r = R + h.
  • Confusing escape speed with orbital speed.
  • Forgetting that v_esc is a speed (direction not required).
  • Adding mgΔ h all the way to infinity (uniform-field thinking) instead of using U = -GMm/r.

5. Exam Tips

  • Start from U = -GMm/r with U(∞) = 0 and write “at escape, v(∞) = 0”.
  • Keep r as centre-to-centre distance until the final substitution.
  • Use 2–3 s.f. unless the question specifies otherwise.

6. Worked Examples

Modelled example 1

Escape speed from Earth’s surface

Core

Problem

Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Find the minimum surface launch speed for arrival at infinity with zero final speed, neglecting atmosphere and further propulsion.
Study the worked solution

  1. Set the energy boundary

    Method

    For minimum escape, set total energy at launch equal to zero.

    Reason

    At infinity both U and the limiting final kinetic energy are zero.

    Working

    (1/2)mv_esc²-GMm/R = 0

  2. Solve for speed

    Method

    Use v_esc = square root of (2GM/R).

    Reason

    The launched object’s mass cancels from the energy equation.

    Working

    v_esc = square root of (2GM/R)

  3. Evaluate

    Method

    Obtain 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹.

    Reason

    The formula gives a speed magnitude.

    Working

    v_esc = square root of ((2(6.67 × 10⁻¹¹)(5.97 × 10²⁴))/(6.37 × 10⁶)) ≈ 1.12 × 10⁴ m s⁻¹

Guided practice 2

Escape speed from an altitude

About 5 min

Problem

Find the escape speed at altitude 300 km above Earth. Use R_E = 6.37 × 10⁶ m and GM_E = 3.99 × 10¹⁴ m³ s⁻².

Try this before viewing the solution

Unit: km s⁻¹

Hints

Hint 1: replace altitude with radius
Convert 300 km to 3.00 × 10⁵ m and add Earth’s radius.
View solution step by step

  1. Find the radial distance

    Method

    Obtain r = 6.67 × 10⁶ m.

    Reason

    The gravitational potential energy uses distance from Earth’s centre.

    Working

    r = R_E + h = 6.37 × 10⁶ + 3.00 × 10⁵ = 6.67 × 10⁶ m

  2. Calculate escape speed

    Method

    Obtain 1.09 × 10⁴ m s⁻¹, or 10.9 km s⁻¹.

    Reason

    The larger starting radius makes the required speed slightly lower than at the surface.

    Working

    v_esc = square root of ((2(3.99 × 10¹⁴))/(6.67 × 10⁶)) ≈ 1.09 × 10⁴ m s⁻¹

Common misconception 3

Comparing escape speeds (scaling)

Find and correct the mistake

Learner claim

An exoplanet has twice Earth’s mass and half Earth’s radius. A learner says its surface escape speed is four times Earth’s because M/R is four times larger. Diagnose the claim.

Try this before viewing the solution

View solution step by step

  1. Form the ratio inside the root

    Method

    The exoplanet has (M'/R')/(M/R) = 2/0.5 = 4.

    Reason

    Both the source mass and surface radius change.

    Working

    (M'/R')/(M/R) = 2/0.5 = 4

  2. Apply the square root

    Method

    The escape-speed ratio is 2.

    Reason

    v_esc ∝ square root of (M/R), so the factor of four is under a square root.

    Working

    v'/v = square root of 4 = 2

Examiner practice 4

Escape speed from g and R (using v_esc = square root of 2gR)

3 marks

Examination question

The Moon has radius 1.74 × 10⁶ m and surface gravitational field strength 1.62 N kg⁻¹. Derive the usable form from g = GM/R² and estimate the Moon’s escape speed. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Replace GM

    1 mark

    Method

    From g = GM/R², use GM = gR².

    Reason

    This expresses the source parameter using the supplied surface data.

    Working

    GM = gR²

  2. Simplify escape speed

    1 mark

    Method

    Obtain v_esc = square root of 2gR.

    Reason

    Substitute GM = gR² into square root of (2GM/R).

    Working

    v_esc = square root of (2gR²/R) = square root of 2gR

  3. Evaluate

    1 mark

    Method

    Obtain 2.38 × 10³ m s⁻¹, or 2.4 km s⁻¹.

    Reason

    The square-root expression has dimensions of speed.

    Working

    v_esc = square root of (2(1.62)(1.74 × 10⁶)) ≈ 2.38 × 10³ m s⁻¹

Challenge 5

How does v_esc change with altitude?

Minimal support

Independent transfer

At Earth’s surface r = R_E; at altitude h, r = R_E + h. Derive v_esc(R_E + h)/v_esc(R_E) and use it to justify how escape speed changes with altitude.

Try this before viewing the solution

Hints

Hint 1: write one square root for each radius
Use v_esc = square root of (2GM/r) in numerator and denominator before cancelling.
View solution step by step

  1. Form the speed ratio

    Method

    Divide the altitude expression by the surface expression.

    Reason

    The same planet means the factor 2GM cancels.

    Working

    (v_esc(R_E + h))/v_esc(R_E) = (square root of (2GM/(R_E + h)))/(square root of (2GM/R_E))

  2. Simplify and interpret

    Method

    Obtain square root of (R_E/(R_E + h)), which is less than one for h > 0.

    Reason

    A higher starting point is less deeply bound gravitationally.

    Working

    (v_esc(R_E + h))/v_esc(R_E) = square root of (R_E/(R_E + h)) < 1

7. Mind Stretchers

Mind stretcher 1: Show v_esc = square root of 2 v_orbExtension

Show that the escape speed from radius r is square root of 2 times the circular orbital speed at the same radius.

Show answer

Circular orbit: v_orb = square root of (GM/r).

Escape: v_esc = square root of (2GM/r) = square root of 2 square root of (GM/r) = square root of 2 v_orb.

Mind stretcher 2: Launch at 0.80v_escExtension

An object is launched upwards from Earth’s surface with speed 0.80v_esc. Explain why it will not escape, and describe what happens far from Earth.

Show answer

Escape requires total energy E = 0. With v = 0.80v_esc, the initial kinetic energy is only (0.80)² = 0.64 of what is needed to reach E = 0, so the total energy is negative (bound).

The object rises, slows down as kinetic energy is converted to gravitational potential energy, reaches a maximum distance where its speed becomes zero, then falls back.

Mind stretcher 3: Optional (Enrichment)Extension

A. Black holes (beyond syllabus)

If the escape speed from a radius exceeds the speed of light, even light cannot escape. This is one way to motivate the idea of a black hole, but it is not required for A Level calculations.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027