10.2.5 Escape Speed

Escape Speed

To escape a gravitational field, a mass must travel at the escape velocity
- This is dependent on the mass and radius of the object creating the gravitational field, such as a planet, a moon or a black hole
Escape velocity is defined as:

The minimum speed that will allow an object to escape a gravitational field with no further energy input

It is the same for all masses in the same gravitational field ie. the escape velocity of a rocket is the same as a tennis ball on Earth

An object reaches escape velocity when all its kinetic energy has been transferred to gravitational potential energy
This is calculated by equating the equations:

\frac{1}{2} \times m \times v^{2} = \frac{G \times M \times m}{r}

Where:
- m = mass of the object in the gravitational field (kg)
- v = escape velocity of the object (m s^-1)
- G = Newton's Gravitational Constant
- M = mass of the object to be escaped from (ie. a planet) (kg)
- r = distance from the centre of mass M (m)

Since mass m is the same on both sides of the equations, it can cancel on both sides of the equation:

\frac{1}{2} \times v^{2} = \frac{G \times M}{r}

Multiplying both sides by 2 and taking the square root gives the equation for escape velocity, v:

v = \sqrt{\frac{2 \times G \times M}{r}}

This equation is not given on the datasheet. Be sure to memorise how to derive it

Escape Velocity Diagram, downloadable AS & A Level Physics revision notes

For an object to leave the Earth's gravitational field, it will have to travel at a speed greater than the Earth's escape velocity, v

Rockets launched from the Earth's surface do not need to achieve escape velocity to reach their orbit around the Earth
This is because:
- They are continuously given energy through fuel and thrust to help them move
- Less energy is needed to achieve orbit than to escape from Earth's gravitational field

The escape velocity is not the velocity needed to escape the planet but to escape the planet's gravitational field altogether
- This could be quite a large distance away from the planet

Worked Example

Calculate the escape velocity at the surface of the Moon given that its density is 3340 kg m^-3 and has a mass of 7.35 × 10²² kg.Newton's Gravitational Constant = 6.67 × 10^-11 N m²kg^-2

Step 1: Rearrange the density equation for radius:

$ρ = \frac{M}{V}$ and $V = \frac{4}{3} π r^{3}$

$ρ = \frac{M}{\frac{4}{3} π r^{3}} = \frac{3 M}{4 π r^{3}}$

$r = \sqrt[3]{\frac{3 M}{4 π ρ}}$

Step 2: Calculate the radius by substituting in the values:

M = 7.35 × 10²² kg
ρ = 3 340 kg m⁻³

^{$r = \sqrt[3]{\frac{3 \times (7.35 \times 10^{22})}{4 π \times 3340}} = 1.7384 \times 10^{6} m$}

Step 3: Substitute r into the escape velocity equations:

^{$V = \sqrt{\frac{2 G M}{r}} = \sqrt{\frac{2 \times (6.67 \times 10^{- 11}) \times (7.35 \times 10^{22})}{1.7384 \times 10^{6}}}$}

V = 2.37 km s⁻¹

Exam Tip

When writing the definition of escape velocity, avoid terms such as 'gravity' or the 'gravitational pull / attraction' of the planet. It is best to refer to its gravitational field.

DP IB Physics: HL

Revision Notes

10.2.5 Escape Speed

Escape Speed

Worked Example

Exam Tip

Author: Katie

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