Physics · Gravitation · NEET
In the energy equation, the mass m of the body appears in BOTH terms: kinetic energy ½mv_e² and potential energy GMm/R. When you set ½mv_e² = GMm/R and solve for v_e, the m cancels on both sides. So v_e = √(2GM/R) depends only on the planet's mass M and radius R, not on the body. A feather and a rocket need the same 11.2 km/s from Earth.
For the body to just barely escape, it must reach infinity (r = ∞) with the smallest possible speed, which is zero. At infinity, potential energy = 0 and kinetic energy = 0, so total energy = 0. Since total mechanical energy is conserved, the total energy at the surface must also be 0: ½mv_e² + (−GMm/R) = 0. This gives the minimum launch speed.
Orbital velocity v_o = √(GM/R) keeps a satellite going in a circle near the surface; escape velocity v_e = √(2GM/R) lets a body leave forever. Compare the formulas: v_e = √2 × v_o. So escape speed is about 1.414 times orbital speed. For Earth, v_o ≈ 7.9 km/s and v_e ≈ 11.2 km/s.
At the surface, acceleration due to gravity is g = GM/R². So GM = gR². Substitute this into v_e = √(2GM/R): v_e = √(2·gR²/R) = √(2gR). This form is very useful in NEET numericals because g and R are usually given directly.
No. Escape velocity is a scalar minimum speed; it is derived only from energy conservation, which does not involve direction. Whether you launch straight up, at an angle, or sideways, the same minimum speed √(2GM/R) is needed (ignoring air drag and the planet's rotation). Direction only affects the path, not the escape speed.
The escape velocity of a body from the earth's surface is 11.2 km/s. If the same body is projected upward with velocity 22.4 km/s, its velocity at infinite distance from the centre of the earth will be:
The ratio of escape velocity at earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice that of earth is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
About 11.2 km/s (11,200 m/s), obtained by putting M = 5.97×10²⁴ kg and R = 6.4×10⁶ m (or g = 9.8 m/s² and R) into v_e = √(2gR). It does not depend on the mass of the launched body.
Yes, in NCERT and NEET they are used interchangeably. Strictly it is a speed (a scalar minimum value) because its derivation uses only energy, not direction, so 'escape speed' is the more accurate term.
Three equivalent forms: v_e = √(2GM/R), v_e = √(2gR) using g = GM/R², and v_e = R√(8πGρ/3) using mean density ρ. All come from the same energy-conservation derivation.
Gravitation gives 1 NEET question almost every year, and escape velocity is a favourite because it links energy conservation, orbital velocity, density scaling and velocity-at-infinity in one topic. Knowing the derivation lets you handle all these variations.
It escapes and still has speed left at infinity. From v∞² = u² − v_e², any launch speed u greater than v_e leaves a non-zero speed v∞ at infinity, so the body never returns.
