Calculate escape velocity from Earth surface — derive formula

Question

Derive the formula for escape velocity from the Earth’s surface. Calculate the escape velocity given $g = 9.8$ m/s² and $R_E = 6.4 \times 10^6$ m.

Solution — Step by Step

Escape velocity is the minimum speed a body must be given at the surface of a planet to escape its gravitational field — i.e., to travel to infinity and have zero kinetic energy left upon arriving at infinity.

We assume no air resistance and no other gravitational bodies. The launching is done by a single initial velocity; no rocket thrust is applied after launch.

At the surface:

Kinetic energy = $\frac{1}{2}mv_e^2$
Gravitational potential energy = $-\frac{GMm}{R_E}$

At infinity (final state):

Kinetic energy = 0 (just barely escapes, minimum energy case)
Gravitational potential energy = 0 (reference point)

By conservation of mechanical energy:

\frac{1}{2}mv_e^2 - \frac{GMm}{R_E} = 0 + 0

\frac{1}{2}mv_e^2 = \frac{GMm}{R_E}

Cancel $m$ from both sides (escape velocity is independent of the mass of the object):

\frac{1}{2}v_e^2 = \frac{GM}{R_E}

v_e = \sqrt{\frac{2GM}{R_E}}

Now use $g = GM/R_E^2$ , so $GM = gR_E^2$ :

\boxed{v_e = \sqrt{2gR_E}}

v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6}

v_e = \sqrt{125.44 \times 10^6}

v_e = \sqrt{1.2544 \times 10^8}

v_e \approx 1.12 \times 10^4 \text{ m/s}

\boxed{v_e \approx 11.2 \text{ km/s}}

This is about 40,000 km/h — significantly faster than any commercial aircraft.

Why This Works

The derivation uses energy conservation. We want the object to reach infinity with exactly zero kinetic energy — the minimum condition for escape. Any more initial speed would leave the object with kinetic energy at infinity; any less would mean it falls back.

The crucial insight is that gravitational potential energy at the surface is negative ( $-GMm/R_E$ ) — we’ve set zero at infinity. The object must gain positive kinetic energy equal to the magnitude of its initial (negative) potential energy.

Notice the result: $v_e = \sqrt{2gR_E}$ . Escape velocity depends only on $g$ at the surface and the planet’s radius. A planet with the same mass but smaller radius has a higher escape velocity (stronger surface gravity).

Alternative Method

You can also derive using the “work done against gravity” approach:

W = \int_{R_E}^{\infty} \frac{GMm}{r^2} dr = \left[-\frac{GMm}{r}\right]_{R_E}^{\infty} = \frac{GMm}{R_E}

Setting this equal to initial KE: $\frac{1}{2}mv_e^2 = \frac{GMm}{R_E}$ , same result.

Common Mistake

Students sometimes use $v_e = \sqrt{2gR_E}$ and substitute $g$ in cm/s² or km/s² without converting units. Always check: $g$ in m/s², $R_E$ in metres. The answer comes out in m/s, then divide by 1000 for km/s. If you use $g = 9.8$ and $R_E = 6400$ (in km without converting), you get a nonsensical unit — a sign that unit checking is essential.

Memorise $v_e \approx 11.2$ km/s for Earth. For the Moon: $g_{Moon} \approx 1.67$ m/s², $R_{Moon} \approx 1.74 \times 10^6$ m, giving $v_e \approx 2.4$ km/s. This is why the Moon has no atmosphere — gas molecules easily exceed 2.4 km/s at lunar temperatures and escape.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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