Escape Speed

What escape speed means

Near the Earth's surface a projectile carries two kinds of energy: positive kinetic energy $\tfrac12 m v^2$ and negative gravitational potential energy $-\,GM_E m / R_E$. Their sum is the total mechanical energy, and gravity is a conservative force, so this sum stays fixed throughout the flight. As the object rises it trades kinetic energy for potential energy. If it begins too slowly, the kinetic energy runs out before it reaches infinity; it stops, turns, and falls back.

The escape speed is the minimum launch speed at which the object is just able to reach infinity — arriving there with zero leftover speed. At that critical launch speed the total mechanical energy is exactly zero. Launch any faster and the object reaches infinity with speed to spare; launch any slower and it returns. NCERT calls this quantity the escape speed, and notes it is sometimes loosely called the escape velocity.

Derivation from energy conservation

Follow NCERT §7.8. Suppose the object is thrown from the surface ($h=0$) with speed $v_i$ and reaches infinity with final speed $v_f$. Take the gravitational potential energy at infinity to be zero. Total energy at infinity is

$$E_\infty = \tfrac12 m v_f^2 .$$

Total energy at the surface, at distance $R_E$ from the centre, is kinetic plus the negative potential energy:

$$E_{\text{surface}} = \tfrac12 m v_i^2 - \frac{G M_E m}{R_E} .$$

Energy conservation sets these equal:

$$\tfrac12 m v_f^2 = \tfrac12 m v_i^2 - \frac{G M_E m}{R_E} .$$

The left-hand side is a kinetic energy and so cannot be negative; its smallest possible value is zero. The object can therefore just reach infinity provided

$$\tfrac12 m v_i^2 \;\ge\; \frac{G M_E m}{R_E} .$$

The minimum launch speed corresponds to equality (the object arrives at infinity with $v_f = 0$). Setting $\tfrac12 m v_e^2 = G M_E m / R_E$ and cancelling $m$ gives the escape speed:

$$\boxed{\,v_e = \sqrt{\dfrac{2 G M_E}{R_E}}\,}$$

Using the surface relation $g = G M_E / R_E^{2}$ — so that $G M_E = g R_E^{2}$ — the same result rewrites in terms of $g$:

$$v_e = \sqrt{\dfrac{2 g R_E^{2}}{R_E}} = \sqrt{2 g R_E}.$$

The value ≈ 11.2 km/s for Earth

Putting Earth's numbers into $v_e = \sqrt{2gR_E}$ with $g = 9.8~\text{m s}^{-2}$ and $R_E = 6.4 \times 10^{6}~\text{m}$:

$$v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}} = \sqrt{1.254 \times 10^{8}} \approx 1.12 \times 10^{4}~\text{m s}^{-1} = 11.2~\text{km s}^{-1}.$$

The same number follows from $v_e = \sqrt{2GM_E/R_E}$ using $M_E = 6.0 \times 10^{24}~\text{kg}$ and $G = 6.67 \times 10^{-11}~\text{N m}^2\text{kg}^{-2}$. NCERT records this surface escape speed for Earth as $11.2~\text{km s}^{-1}$.

Equation $v_e = \sqrt{2gR}$ applies equally to any body. For the Moon, both $g$ and the radius are smaller, and the escape speed comes out to about $2.3~\text{km s}^{-1}$ — roughly five times smaller than Earth's. This is why the Moon has no atmosphere: gas molecules at its surface routinely exceed $2.3~\text{km s}^{-1}$ and so escape its gravity.

Independence from the projectile's mass

In the derivation the projectile mass $m$ appeared in both the kinetic energy and the potential energy, and cancelled exactly. The final formula $v_e = \sqrt{2GM/R}$ contains only the planet's mass $M$, its radius $R$, and $G$ — nothing about the projectile. A pebble and a loaded spacecraft must be launched at the same speed to escape. NCERT exercise 7.7 makes the point directly: escape speed does not depend on the mass of the body, nor on the direction of projection.

Dependence on the planet's M and R

Since $v_e = \sqrt{2GM/R}$, escape speed rises with the planet's mass and falls with its radius. A more useful NEET form expresses it through the mean density $\rho$. Writing $M = \tfrac{4}{3}\pi R^3 \rho$ and substituting,

$$v_e = \sqrt{\frac{2G}{R}\cdot \frac{4}{3}\pi R^3 \rho} = R\sqrt{\frac{8\pi G \rho}{3}}\;\;\propto\; R\sqrt{\rho}.$$

So for a fixed density, escape speed is proportional to radius; and for a fixed radius, it is proportional to $\sqrt{\rho}$. This density form is the engine behind several NEET items that compare two planets — keep it ready.

Relation to orbital velocity

A satellite circling just above the surface needs centripetal force supplied entirely by gravity: $\dfrac{m v_o^2}{R} = \dfrac{G M m}{R^2}$, giving the orbital velocity $v_o = \sqrt{GM/R}$. Compare it with the escape speed:

$$v_o = \sqrt{\frac{GM}{R}}, \qquad v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2}\;v_o .$$

Escape speed is exactly $\sqrt{2}$ times the surface orbital velocity, independent of the planet. For Earth the near-surface orbital speed is about $7.9~\text{km s}^{-1}$, and $\sqrt{2} \times 7.9 \approx 11.2~\text{km s}^{-1}$ — the escape speed, as expected. Physically, a satellite already in low orbit has half the kinetic energy needed to escape; only a $\sqrt{2}$ boost in speed (a doubling of kinetic energy) takes it to infinity.

Why a black hole has v_e ≥ c

The formula $v_e = \sqrt{2GM/R}$ carries a striking implication. For a fixed mass $M$, squeezing it into a smaller radius $R$ raises the escape speed without limit. Compress the mass enough and the required escape speed reaches the speed of light $c$. The critical radius at which $v_e = c$ follows from $c = \sqrt{2GM/R}$, giving $R = 2GM/c^2$.

If a body is compressed below this radius, its escape speed exceeds $c$. Because nothing — not even light — can travel faster than $c$, nothing launched from such an object can ever reach infinity. That object is a black hole: gravity so strong that light itself cannot escape. The Newtonian estimate $R = 2GM/c^2$ coincides, remarkably, with the Schwarzschild radius from general relativity, though a full treatment requires the relativistic theory.

Worked examples