Escape Speed (HL) | HL IB Physics Revision Notes 2025

Escape Speed

To escape a gravitational field, a mass must travel at, or above, the minimum escape speed
- 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 speed 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
- For example, the escape speed of a rocket is the same as a tennis ball on Earth

The escape speed of an object is the speed at which all its kinetic energy has been transferred to gravitational potential energy
This is calculated by equating the equations:

$\frac{1}{2} m {v_{e s c}}^{2} = \frac{G M m}{r}$

Where:
- m = mass of the object in the gravitational field (kg)
- $v_{e s c}$ = escape velocity of the object (m s⁻¹)
- G = Newton's Gravitational Constant
- M = mass of the object to be escaped from (i.e. a planet) (kg)
- r = distance from the centre of mass M (m)

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

$\frac{1}{2} {v_{e s c}}^{2} = \frac{G M}{r}$

Multiplying both sides by 2 and taking the square root gives the equation for escape velocity $v_{e s c}$ :

$v_{e s c} = \sqrt{\frac{2 G M}{r}}$

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 speed at the surface of the Moon.

Density of the Moon = 3340 kg m⁻³
Mass of the Moon = 7.35 × 10²² kg

Answer:

Step 1: List the known quantities

Gravitational constant, G = 6.67 × 10⁻¹¹ N m²kg⁻²
Density of the Moon, ρ = 3340 kg m⁻³
Mass of the Moon, M = 7.35 × 10²² kg

Step 2: Rearrange the density equation for radius r

Density: $ρ = \frac{M}{V}$ and volume of a sphere: $V = \frac{4}{3} π r^{3}$

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

$r = \sqrt[3]{\frac{3 M}{4 π ρ}}$

Step 3: Calculate the radius by substituting in the values

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

Step 4: Substitute r into the escape speed equation

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

Escape speed of the Moon: $v_{e s c}$ = 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. This equation is given on the data sheet, but make sure you know how it is derived.

Escape Speed (HL) (HL IB Physics)

Revision Note

Escape Speed

Worked example

Exam Tip

Author: Katie M