Minimum velocity to escape Earth's gravity. Why it's independent of mass.

Escape Velocity of Earth — Derive vₑ = √(2gR)

Question

Derive the expression for escape velocity from Earth’s surface. Show that it equals $v_e = \sqrt{2gR}$ , and explain why it does not depend on the mass of the projected object.

Solution — Step by Step

We want the minimum launch speed so the object just barely escapes Earth’s gravity — meaning it reaches infinity with zero kinetic energy left. Apply conservation of mechanical energy between the surface and infinity.

KE_{\text{surface}} + PE_{\text{surface}} = KE_{\infty} + PE_{\infty}

At infinity, both $KE$ and $PE$ are zero (we define $PE = 0$ at $r = \infty$ ).

Let $m$ = mass of object, $M$ = mass of Earth, $R$ = radius of Earth.

\frac{1}{2}mv_e^2 + \left(-\frac{GMm}{R}\right) = 0

The gravitational PE at the surface is $-\frac{GMm}{R}$ — negative because gravity is attractive.

Rearranging:

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

v_e^2 = \frac{2GM}{R}

Notice $m$ cancels — this is why escape velocity is mass-independent.

\boxed{v_e = \sqrt{\frac{2GM}{R}}}

We know surface gravity: $g = \frac{GM}{R^2}$ , so $GM = gR^2$ .

Substitute:

v_e = \sqrt{\frac{2gR^2}{R}} = \sqrt{2gR}

Using $g = 9.8\ \text{ms}^{-2}$ and $R = 6.4 \times 10^6\ \text{m}$ :

v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2\ \text{km/s}

Why This Works

The core idea is that escape doesn’t mean “going beyond gravity.” Gravity extends to infinity — it just gets weaker. What we really need is for the object’s kinetic energy to exactly offset the gravitational potential energy well at the surface.

Because both $KE = \frac{1}{2}mv_e^2$ and $PE = -\frac{GMm}{R}$ are proportional to $m$ , the object’s mass cancels out cleanly. A feather and a rocket need the same escape velocity — what differs is the thrust required to reach that velocity.

This result comes directly from Newton’s law of gravitation + energy conservation. No calculus of orbits needed. For JEE, the energy method is far faster than force-integration, and both NEET and board papers have repeatedly tested exactly this derivation.

Alternative Method

Using integration of work done against gravity:

We can calculate the work done in moving mass $m$ from $R$ to $\infty$ against gravity:

W = \int_R^{\infty} \frac{GMm}{r^2}\, dr = \left[-\frac{GMm}{r}\right]_R^{\infty} = \frac{GMm}{R}

Setting initial KE equal to this work:

\frac{1}{2}mv_e^2 = \frac{GMm}{R} \implies v_e = \sqrt{\frac{2GM}{R}}

Same result, different path. The energy conservation method is cleaner for exams; the integration form shows you exactly where the $\frac{1}{r}$ potential comes from.

Common Mistake

A very common error is setting the condition as “velocity becomes zero at some finite height $H$ ” — and then confusing escape velocity with orbital velocity or maximum height problems. Escape velocity specifically requires the object to reach $r = \infty$ . If you write $v = 0$ at $r = R + H$ for finite $H$ , you’re solving a different problem (maximum height). The condition for escape is $v \to 0$ as $r \to \infty$ .

Remember: $v_{\text{orbital}} = \sqrt{\frac{GM}{R}}$ and $v_{\text{escape}} = \sqrt{\frac{2GM}{R}}$ , so $v_e = \sqrt{2} \cdot v_{\text{orbital}}$ . This ratio $\sqrt{2}$ appears every year in MCQs — JEE Main 2024 asked it directly as a one-liner.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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