A particle is fired vertically upwards from the urface of earth and reaches a height 6400 km. Find the initial velocity of the particle if R = 6400 km and g at the surface of earth is 10 `m//s^2`.

Explanation

`1/2m upsilon^2=(mgh)/([1+(h//R)])`
here h=R=6400 km and g=10 `m//s^2`
so `upsilon^2=gh`, i.e., `upsilon =sqrt(10xx6400xx10^3)`= 8 km/s