Escape Velocity of Earth (11.2 km/s) and Its Meaning

Physics · Gravitation · NEET

The escape velocity of Earth is about 11.2 km/s (11,200 m/s). It is the minimum speed a body needs at Earth's surface to move away forever and never fall back, with no extra push after launch. Memory hook: "11.2 lets you leave" — v_e = sqrt(2gR) = sqrt(2 x 9.8 x 6.4x10^6).
EarthR=6.4e6 mu < 11.2: falls backu = 11.2 km/s: escapesu > 11.2: leaves withv_inf = sqrt(u^2 - v_e^2)Escape speedv_e = sqrt(2gR)= sqrt(2 x 9.8 x 6.4e6)= 1.12e4 m/s= 11.2 km/s
Below 11.2 km/s a body falls back; at exactly 11.2 km/s it just escapes; above it, leftover speed at infinity is v_inf = sqrt(u^2 - v_e^2). The value comes from v_e = sqrt(2gR).

Your doubts, answered

Is escape velocity (11.2 km/s) the same as orbital velocity?

No. Orbital velocity is the speed to go around Earth in a circle near the surface, about 7.9 km/s. Escape velocity is the speed to leave Earth forever, about 11.2 km/s. The link is v_e = sqrt(2) x v_orbital, so escape velocity is about 1.414 times the orbital velocity. NEET often mixes these two, so keep both numbers ready: 7.9 to orbit, 11.2 to escape.

Does escape velocity depend on the mass of the object being launched?

No. From v_e = sqrt(2GM/R) = sqrt(2gR), only Earth's mass M and radius R appear, not the object's mass m. A cricket ball and a satellite both need the same 11.2 km/s. The mass m cancels out because both kinetic energy and gravitational PE contain m. This is a very common NEET trap.

Why exactly is Earth's escape speed 11.2 km/s?

Put the numbers into v_e = sqrt(2gR): g = 9.8 m/s^2 and R = 6.4x10^6 m. So v_e = sqrt(2 x 9.8 x 6.4x10^6) = sqrt(1.254x10^8) = 1.12x10^4 m/s = 11.2 km/s. That is why the value is fixed for Earth's surface — it comes straight from g and R.

Does the direction of launch change the escape velocity?

No, the magnitude 11.2 km/s is the same in every direction (ignoring air and Earth's rotation). Escape velocity is really an escape speed. The energy equation uses only the size of the velocity, not its direction, because gravitational PE depends only on distance r, not on angle. That is why NCERT calls it 'escape speed'.

What if the speed is more than or less than 11.2 km/s?

If speed is less than 11.2 km/s, the body rises, stops, and falls back to Earth. If it is exactly 11.2 km/s, it just barely reaches infinity with zero speed left. If it is more than 11.2 km/s, it escapes and still has leftover speed at infinity, found from v_infinity = sqrt(u^2 - v_e^2).

⚠️ The NEET trap
A heavy rocket needs more than 11.2 km/s and a light ball needs less, because heavier things are harder to launch.
Both need the same 11.2 km/s. Escape velocity v_e = sqrt(2gR) does not contain the object's mass m — it cancels out. Mass only affects the energy or fuel needed, not the required speed.
🧠 Escape SPEED ignores the launched mass — only Earth's g and R decide it.

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Frequently asked

What is the escape velocity of Earth?

About 11.2 km/s, which equals 11,200 m/s. It comes from v_e = sqrt(2gR) with g = 9.8 m/s^2 and R = 6.4x10^6 m.

What is the formula for escape velocity?

v_e = sqrt(2GM/R), which also equals sqrt(2gR) at the surface, since g = GM/R^2. For Earth this gives 11.2 km/s.

Is escape velocity a vector or scalar?

Physically it is a speed (scalar). Its value is the same in all directions, which is why NCERT prefers the name escape speed.

How is escape velocity related to orbital velocity?

v_e = sqrt(2) x v_orbital. So escape velocity (11.2 km/s) is about 1.414 times the orbital velocity near the surface (7.9 km/s).

Why is Moon's escape velocity smaller than Earth's?

The Moon has smaller g and smaller radius R, so v_e = sqrt(2gR) is smaller — about 2.4 km/s. That is why gases escaped and the Moon has almost no atmosphere.