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Escape speed through energy stores and transfers

Key idea: H2 Physics lessons on inverse-square gravitation, field strength, gravitational potential, escape speed and satellite orbits.

  • GCE A-Level H2 Physics 2027

Build the idea

Learn the idea

Big question: What minimum launch speed lets an object escape without further propulsion?

At the minimum escape speed, the object reaches infinity with zero kinetic energy. Equating initial kinetic energy plus gravitational potential energy to zero gives vesc = √(2GM/r). The escaping mass cancels, and the result ignores atmosphere, rotation and later propulsion.

Define the limiting journey

Escape speed is the minimum launch speed from a stated radius that allows an object to reach infinity without further propulsion. In the limiting case, its speed tends to zero at infinity; any smaller launch leaves negative total mechanical energy and a bound path.

The model assumes a spherical, non-rotating source, no atmosphere and no later engine work. It is therefore a clean gravitational-energy threshold, not a complete rocket-launch calculation.

Check your understanding: Does an escaping object stop at a finite height in the minimum-speed model?

No. Its speed approaches zero only as its distance approaches infinity.

Derive the speed from total energy

At launch, total energy is ½mv² − GMm/r. At infinity the limiting final kinetic and potential energies are both zero. Setting the totals equal gives vesc = √(2GM/r).

The escaping mass cancels, so all objects at one launch radius share the same ideal escape speed. Compared with circular-orbit speed √(GM/r), escape speed is √2 times larger at the same radius.

Check your understanding: Why does a more massive spacecraft not need a greater ideal escape speed?

Both its kinetic energy and gravitational potential energy scale with its mass, so mass cancels from the energy equation.

Gravitational potential and its radial gradientA graph of gravitational potential phi against outward radial distance r starts negative near a mass and rises towards zero. Its positive slope corresponds to an inward, negative radial field component through g sub r equals minus d phi by d r.φ = 0φrdφ/dr > 0gᵣ < 0: inwardφ = −GM/r approaches zero
Scroll diagram horizontally to read all labels.
With outward chosen positive, φ rises towards zero as r increases, so dφ/dr is positive and the radial field component gᵣ = −dφ/dr is negative (inward).

Key ideas to keep

  • Escape speed is not the speed needed for a circular orbit.
  • The launch point uses distance from the planet's centre, not height above its surface alone.
  • The derivation assumes only gravitational work after launch.

See the reasoning

Worked example

Find escape speed from a moon

Question: A moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Find surface escape speed using G = 6.67 × 10⁻¹¹ N m² kg⁻².

  1. Step 1: Write the limiting energy equation

    Why: Minimum escape ends with zero total energy at infinity.

    Working: ½mv² − GMm/R = 0.

  2. Step 2: Cancel mass and rearrange

    Why: The test body's mass affects both energy terms equally.

    Working: v = √(2GM/R).

  3. Step 3: Substitute

    Why: R is centre distance at the surface.

    Working: v = √[2(6.67×10⁻¹¹)(7.35×10²²)/(1.74×10⁶)] = 2.37×10³ m s⁻¹.

Answer: The surface escape speed is 2.37 km s⁻¹.

Check: It is √2 times the circular-orbit speed at the same radius, as the two formulae predict.

Another worked model

Question

Find the escape speed from r = 7.00 × 10⁶ m around Earth, using GM = 3.99 × 10¹⁴ m³ s⁻².

Check the worked solution

Set ½mv² − GMm/r = 0 for minimum escape. Then vesc = √(2GM/r) = √[2(3.99 × 10¹⁴)/(7.00 × 10⁶)] = 1.07 × 10⁴ m s⁻¹.

Use a hint if needed

Practise with support

Try this

Mars has GM = 4.28 × 10¹³ m³ s⁻² and radius 3.39 × 10⁶ m. Find its surface escape speed.

Hint: Use zero total energy at infinity, not constant surface g over an infinite distance.

Check your answer

vesc = √(2GM/R) = √[2(4.28 × 10¹³)/(3.39 × 10⁶)] = 5.03 × 10³ m s⁻¹.

Now work without the hint

Practise independently

Your turn

A planet has surface field strength 15.0 N kg⁻¹ and radius 4.00 × 10⁶ m. Derive an expression using g and R, then find its escape speed.

Check your answer

At the surface GM = gR², so vesc = √(2GM/R) = √(2gR). Thus vesc = √[2(15.0)(4.00 × 10⁶)] = 1.10 × 10⁴ m s⁻¹.

Avoid these traps

Common mistakes

Common mistake

Escape speed is the speed needed to keep an engine thrust equal to weight.

What is wrong with this reasoning?

Show better thinking

Escape speed is an initial-speed energy condition with no further propulsion: total energy is just zero for arrival at infinity with zero speed.

Common mistake

A more massive spacecraft needs a greater escape speed.

What is wrong with this reasoning?

Show better thinking

Both kinetic and gravitational potential energy contain spacecraft mass, so it cancels from vesc = √(2GM/r). The required energy still increases with spacecraft mass.

Write for the examiner

Exam guidance

Write the initial and final total energies explicitly; this makes the zero-at-infinity assumption visible.

Exam-style practice [7 marks]

A planet has radius R and surface gravitational field strength g. Derive its surface escape speed in terms of g and R. Then find the escape speed from altitude R above the surface as a fraction of the surface value.

Plan before you answer

  • Replace GM using g = GM/R².
  • Use centre distance 2R at altitude R.
  • Compare the two square-root expressions.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

At the surface, vesc = √(2GM/R). Since g = GM/R², GM = gR² and vesc = √(2gR). At altitude R, centre distance is 2R, so v = √[2GM/(2R)] = √(GM/R). Dividing by the surface value gives 1/√2. This ignores atmosphere, rotation and further propulsion.

Come back in three days

Check what stayed with you

Recall question 1

What is the total energy in the minimum escape case?

Check the answer

Zero, using potential zero at infinity.

Recall question 2

How does escape speed vary with launch radius?

Check the answer

It is proportional to r⁻¹/² for the same source mass.

Recall question 3

Compare escape and circular speeds at one radius.

Check the answer

Escape speed is √2 times circular-orbit speed.

Keep moving

Try this next

Continue to the next lesson in this topic.

Circular and geostationary orbits

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions. Models use centre-to-centre distance, zero potential at infinity, point masses or spherical bodies viewed externally, and ideal circular orbits unless a question states otherwise.

  • GCE A-Level H2 PhysicsTopic 8(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027