Derive escape velocity from Earth's surface using energy conservation

Question

Using the principle of conservation of energy, derive the expression for escape velocity from the surface of the Earth. Calculate its value for Earth. (Take $g = 9.8\,\text{m/s}^2$ , $R_E = 6400\,\text{km}$ )

(NCERT Class 11, Chapter 8)

Solution — Step by Step

Escape velocity is the minimum speed at which an object must be launched from the surface so that it just escapes to infinity — reaching infinitely far with zero residual velocity.

At the surface (launch): $KE = \frac{1}{2}mv_e^2$ , $PE = -\frac{GMm}{R}$

At infinity (just escapes): $KE = 0$ , $PE = 0$

Conservation of energy:

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

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

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

Using $g = GM/R^2$ , so $GM = gR^2$ :

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

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

= \sqrt{125.44 \times 10^6} = \sqrt{1.2544 \times 10^8}

\approx 11.2 \times 10^3\,\text{m/s} = \mathbf{11.2\,\text{km/s}}

Why This Works

Energy conservation is the cleanest approach because gravity is a conservative force — only the starting and ending positions matter, not the path taken.

Notice that escape velocity is independent of the mass of the object. A marble and a rocket need the same speed to escape Earth’s gravity. This makes sense: both the kinetic energy ( $\propto m$ ) and gravitational PE ( $\propto m$ ) scale with mass, so $m$ cancels.

Also notice: $v_e = \sqrt{2} \times v_{\text{orbital}}$ . The orbital velocity at the surface would be $v_o = \sqrt{gR}$ , so escape velocity is just $\sqrt{2}$ times that. A satellite in low orbit only needs a 41% speed boost to escape.

Alternative Method — Force integration

W = \int_R^{\infty} \frac{GMm}{r^2}\,dr = \frac{GMm}{R}

Set this equal to $\frac{1}{2}mv_e^2$ : same result. This approach explicitly calculates the work done against gravity, which equals the initial KE needed.

For NEET and CBSE, $v_e = \sqrt{2gR}$ is the form to memorise. For JEE, also know $v_e = \sqrt{2GM/R}$ and the relation $v_e = \sqrt{2} \cdot v_{\text{orbital}}$ . Questions often ask: “What happens to escape velocity if the planet’s mass doubles and radius halves?” Just substitute into the formula.

Common Mistake

Students sometimes set the final KE to some positive value instead of zero. Escape velocity is the minimum launch speed — the object barely escapes. At infinity, it has zero velocity. If you set $v_{\infty} > 0$ , you’re calculating the launch speed for a specific final speed, not the escape velocity. The boundary condition is $v_{\infty} = 0$ and $r_{\infty} = \infty$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Force integration

Common Mistake

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