Escape Velocity from a Planet

Escape velocity is the minimum speed at which an object can leave the surface of a planet and never come back — assuming no atmosphere and no further propulsion. The idea: the kinetic energy at launch must be at least as large as the gravitational potential energy that needs to be overcome.

The Setup

The gravitational potential energy of an object of mass $m$ at distance $r$ from a mass $M$ (taking PE = 0 at infinity) is:

$U(r) = -\dfrac{GMm}{r}$

At the surface ($r = R$), the PE is $-GMm/R$. To escape, the total energy must be ≥ 0:

$\tfrac{1}{2}m v_{\text{esc}}^{2} - \dfrac{GMm}{R} \ge 0$

Setting equality and solving:

$v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}}$

Using the surface gravity $g = GM/R^{2}$, we can rewrite this as:

$v_{\text{esc}} = \sqrt{2gR}$

Step-by-Step Solution

Given: Earth surface gravity $g = 9.81\;\text{m/s}^{2}$, Earth radius $R = 6.371 \times 10^{6}\;\text{m}$.

Find: Escape velocity from Earth's surface.

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Step 1 — Plug into the formula.

$v_{\text{esc}} = \sqrt{2gR} = \sqrt{2(9.81)(6.371 \times 10^{6})}$

Step 2 — Compute the product inside the square root.

$2 \times 9.81 \times 6.371 \times 10^{6}$

$= 19.62 \times 6.371 \times 10^{6}$

$= 124.99 \times 10^{6}$

$\approx 1.250 \times 10^{8}\;\text{m}^{2}/\text{s}^{2}$

Step 3 — Take the square root.

$v_{\text{esc}} = \sqrt{1.250 \times 10^{8}}$

$= \sqrt{1.250} \times 10^{4}$

$\approx 1.118 \times 10^{4}\;\text{m/s}$

$\approx 11{,}183\;\text{m/s}$

$\approx 11.18\;\text{km/s}$

Step 4 — Convert to more familiar units.

$v_{\text{esc}} \approx 11.18\;\text{km/s} \approx 40{,}248\;\text{km/h} \approx 25{,}010\;\text{mph}$

For comparison: a jet airliner cruises at about 900 km/h. A bullet might travel at 1000 m/s (3600 km/h). Even the fastest projectile humans have launched needs 11 times that to escape Earth.

Step 5 — Other planets.

Plug in different $g$ and $R$ values to compute escape velocities elsewhere:

• Moon ($g = 1.62$, $R = 1.737 \times 10^{6}$): $v_{\text{esc}} \approx 2.38\;\text{km/s}$
• Mars ($g = 3.71$, $R = 3.39 \times 10^{6}$): $v_{\text{esc}} \approx 5.03\;\text{km/s}$
• Jupiter ($g = 24.79$, $R = 6.991 \times 10^{7}$): $v_{\text{esc}} \approx 59.5\;\text{km/s}$
• Sun ($g = 274$, $R = 6.96 \times 10^{8}$): $v_{\text{esc}} \approx 617.5\;\text{km/s}$

The Moon's much lower escape velocity is why a small spacecraft can liftoff easily — and why the Moon has no atmosphere (gas molecules escape too easily).

Step 6 — Doesn't depend on the object's mass.

Notice that the mass $m$ of the escaping object cancels out in the derivation. A ping-pong ball needs the same escape velocity as a battleship — 11.2 km/s for both.

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Answer: Earth's escape velocity from the surface is

$\boxed{\;v_{\text{esc}} = \sqrt{2gR} \approx 11.18\;\text{km/s} \approx 25{,}000\;\text{mph}\;}$

Anything launched at exactly this speed (with no further propulsion) will coast outward, slowing down forever, but never quite stopping or returning. Faster than this, and it leaves on a hyperbolic trajectory.

Try It

• Adjust g and R to simulate different planets — the HUD shows preset escape velocities for the Moon, Mars, Jupiter, etc.
• Use the launch speed widget to see whether your object escapes (green path) or falls back (red path).
• Notice how dramatically escape speed grows for larger or denser planets.