Escape Velocity Formula: vₑ = √(2GM/R), Derivation, Examples

The escape velocity formula is $v_e = \sqrt{\frac{2GM}{R}}$ , where G is the gravitational constant, M is the mass of the planet and R is its radius. Since $g = \frac{GM}{R^2}$ at the surface, the same result can be written $v_e = \sqrt{2gR}$ . For Earth it comes to about 11.2 km/s.

Escape velocity is the smallest speed you must give a body at the surface so that, with no further push and no air resistance, it never falls back. Gravity keeps slowing it, but more and more weakly as the distance grows, and at this launch speed the body only just reaches 'infinity' with zero speed left. The idea belongs to the gravitation chapter of Class 11 and AP Physics 1.

Quick facts

Formula
$v_e = \sqrt{\frac{2GM}{R}}$
With g
$v_e = \sqrt{2gR}$
Earth
about 11.2 km/s
Moon
about 2.4 km/s
Link to orbit
$v_e = \sqrt{2}\, v_o$ near the surface
SI unit
m/s, often quoted in km/s
Does not depend on
The mass of the escaping body or the launch direction

Escape velocity formula derivation from energy

Give a body of mass m a speed $v_e$ at the surface. Its kinetic energy is $\frac{1}{2}mv_e^2$ and its gravitational potential energy is $-\frac{GMm}{R}$ . Far away, both are zero if the body only just escapes.

Total energy at the surface: $\frac{1}{2}mv_e^2 - \frac{GMm}{R}$ .
Total energy at infinity, arriving with zero speed: 0.
Energy is conserved, so $\frac{1}{2}mv_e^2 = \frac{GMm}{R}$ .
Cancel m and solve: $v_e = \sqrt{\frac{2GM}{R}}$ .
Put $GM = gR^2$ : $v_e = \sqrt{2gR}$ .

The mass m cancels in step 4. A pebble and a spacecraft need the same escape speed from the same planet. This derivation is a standard board question, and examiners look for the energy statement in step 3.

Symbols and units

What goes into vₑ = √(2GM/R)
Symbol	Quantity	SI unit or value
$v_e$	Escape velocity	m/s
$G$	Universal gravitational constant	$6.674 \times 10^{-11}$ N m²/kg²
$M$	Mass of the planet, moon or star	kg; Earth 5.97 × 10²⁴ kg
$R$	Distance from the centre at launch, usually the radius	m; Earth 6.37 × 10⁶ m
$g$	Surface gravity	m/s²; Earth 9.8 m/s²

Inside the root, $\frac{GM}{R}$ has unit (N m²/kg²)(kg)/m = N m/kg = J/kg = m²/s², so the root gives m/s. A unit check like this confirms the formula before you plug in numbers. The value of G is discussed on the gravitational constant page.

Worked examples

Example 1: escape velocity of Earth formula in action

Use G, Earth's mass and radius to find Earth's escape velocity.

GM for Earth is about $3.98 \times 10^{14}\ \text{m}^3/\text{s}^2$ .

$v_e = \sqrt{\frac{2 \times 3.98 \times 10^{14}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 1.12 \times 10^4\ \text{m/s}$

That is 11.2 km/s. Using $\sqrt{2gR}$ with g = 9.8 m/s² gives the same 11.2 km/s.

Example 2: from the Moon

The Moon has mass 7.35 × 10²² kg and radius 1740 km. Find its escape velocity.

R = 1.74 × 10⁶ m. $\frac{2GM}{R} = \frac{2 \times 6.674 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6} \approx 5.64 \times 10^6\ \text{m}^2/\text{s}^2$ .

$v_e \approx 2.37 \times 10^3\ \text{m/s} \approx 2.4\ \text{km/s}$

Example 3: ratio method, exam style

A planet has 4 times Earth's mass and 2 times its radius. Find its escape velocity.

$v_e \propto \sqrt{\frac{M}{R}}$ , so $\frac{v_p}{v_E} = \sqrt{\frac{4}{2}} = \sqrt{2}$ .

$v_p = 1.414 \times 11.2 \approx 15.8\ \text{km/s}$

Ratio methods avoid big powers of ten and are quicker in objective tests.

Escape velocity and orbital velocity

A satellite skimming the surface needs the orbital velocity $v_o = \sqrt{\frac{GM}{R}} \approx 7.9$ km/s. Compare: $v_e = \sqrt{2}\, v_o$ . So to go from a low circular orbit to escape you must raise the speed by about 41 percent. Between $v_o$ and $v_e$ the path is an ellipse; at or above $v_e$ the body leaves.

Escape speeds compared
Body	Escape velocity
Moon	about 2.4 km/s
Mars	about 5.0 km/s
Earth	about 11.2 km/s

Why the Moon has no air

Gas molecules move at speeds that depend on temperature. On the sunlit Moon, a fair share of light molecules move fast enough to approach its low 2.4 km/s escape speed, and over long times they leak away. Earth's higher value holds on to nitrogen and oxygen. Setting $v_e = c$ gives the radius $R = \frac{2GM}{c^2}$ at which not even light escapes; for a body of the Sun's mass that is only about 3 km, the size of a black hole of that mass.

Escape velocity assumes a single push at launch and no air drag. A rocket that keeps firing its engines can leave Earth at a lower speed.

Common mistakes

Leaving R in kilometres. Convert to metres before you take the root.
Dropping the 2 and getting the orbital speed instead.
Thinking direction matters. Without air, any launch direction that clears the ground works if the speed is enough.
Thinking a heavier rocket needs a higher escape velocity. It needs more energy, but the same speed.

The surface gravity used in √(2gR) comes from the acceleration due to gravity formula.

Frequently asked questions

What is the formula of escape velocity?

vₑ = √(2GM / R), or vₑ = √(2gR) using surface gravity. G is 6.674 × 10⁻¹¹ N m²/kg², M the planet's mass and R its radius.

What is the escape velocity of Earth?

About 11.2 km/s, which you get by putting surface gravity and Earth's radius into √(2gR).

Does escape velocity depend on the mass of the object?

No. The object's mass cancels in the energy equation, so every object needs the same launch speed from a given planet.

What is the relation between escape velocity and orbital velocity?

Near the surface, vₑ = √2 × vₒ. Earth's orbital speed of 7.9 km/s times 1.414 gives 11.2 km/s.

Why is it called escape velocity if direction does not matter?

Strictly it is an escape speed. The name stuck, but the result depends only on the size of the launch speed, as long as the path does not hit the planet.