A body of mass m is projected with velocity λ vₑ in vertically upward direction from the surface of the earth into space. It is given that vₑ is escape velocity and λ<1. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be :
(R : radius of earth)

Solution

Physics sir · 4 steps

Write energy conservation and let vₑ² = 2GM/R do the work. Everything except r cancels neatly.

1. Energy at launch. At the surface the body has kinetic energy 1/2mλ² vₑ² and potential energy -GMm/R. At the highest point, at distance r from the centre, it stops, so only -GMm/r is left.

   1/2mλ² vₑ² - GMm/R = -GMm/r

2. Use escape velocity. Since vₑ² = 2GM/R, the kinetic energy term becomes (λ² GMm)/R. Cancel GMm throughout to get λ²/R - 1/R = -1/r.
3. Solve for r. This gives 1/r = (1-λ²)/R, so r = R/(1-λ²). Because λ < 1, this is a finite distance larger than R, as it should be.
4. So. The maximum distance from the centre of the earth is R/(1-λ²), which is option (b).

The idea to keep. Launch speed as a fraction of vₑ? Use energy conservation with vₑ² = 2GM/R. And always check whether they ask for distance from the centre or height above the surface.

Answer from the NTA final answer key. Solution by PrepWiser.