Physics · Gravitation · NEET
Exactly zero. Escape velocity ve is defined as the minimum speed for which the body just reaches infinity with zero speed. Put u = ve into v(inf) = sqrt(u^2 - ve^2) and you get sqrt(0) = 0. The body escapes but arrives at infinity with no speed left. Only when u is greater than ve does it arrive with a real, non-zero speed.
Because energy adds as the square of speed (KE = (1/2)mv^2), not linearly. Conservation of energy gives (1/2)u^2 = (1/2)ve^2 + (1/2)v(inf)^2, so the SQUARES subtract: v(inf)^2 = u^2 - ve^2. If u = 2ve, the answer is sqrt(4-1)*ve = sqrt(3)*ve, which is about 1.73 ve, not (2-1) = 1 ve. Subtracting the plain speeds is the most common mistake.
Escape velocity ve is the launch speed you need at Earth's surface to just barely escape (about 11.2 km/s). Velocity at infinity v(inf) is the speed that is actually LEFT OVER when the body is very far away, after gravity has done negative work on it all the way out. If you launch at exactly ve, v(inf) = 0. If you launch above ve, v(inf) is positive.
When u is greater than ve, the total energy is positive, so the body never stops. It keeps slowing as it climbs, but its speed approaches a fixed non-zero value v(inf) = sqrt(u^2 - ve^2). It coasts away forever at that speed. Only for u equal to or below ve does gravity eventually win.
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v(inf) = sqrt(u^2 - ve^2), where u is the launch speed at Earth's surface and ve is the escape velocity (about 11.2 km/s). It is valid only when u is greater than ve.
v(inf)^2 = (3ve)^2 - ve^2 = 9ve^2 - ve^2 = 8ve^2, so v(inf) = sqrt(8)*ve = 2*sqrt(2)*ve, which is about 2.83 times the escape velocity.
Energy conservation is written in terms of kinetic energy (1/2)mv^2, which depends on v^2. The launch KE splits into the KE spent escaping (linked to ve^2) and the leftover KE at infinity (linked to v(inf)^2), so the squares subtract.
No. The mass m cancels from every term in the energy equation, so v(inf) depends only on u and ve, not on how heavy the body is.
They use the same energy equation, but here u is greater than ve so the body never comes back (there is no maximum height, only a leftover speed). Below ve, the body stops at a finite maximum height instead.
