Practice portal › Gravitation › Escape Velocity

A particle of mass m is projected with a velocity v=kVₑ (k<1) from the surface of the earth. (Vₑ = escape velocity) The maximum height above the surface reached by the particle is

Asked in NEET 2021 · Projection below escape speed

Answer: (4) (Rk²)/(1-k²)

Step-by-step solution

Conserve mechanical energy between the surface and the highest point, where the speed is zero.

1/2mv²-(GMm)/R=-(GMm)/(R+h).

So (v²)/2=GM(1/R-1/(R+h))=(GMh)/(R(R+h)).

Put v=kVₑ with Vₑ²=(2GM)/R: the left side becomes (k²GM)/R, giving k²=h/(R+h).

Solving, k²R+k²h=h, so h=(Rk²)/(1-k²).

Why the other options are wrong

More Escape Velocity questionsAll Escape Velocity questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer