Question

Escape velocity of a particle which is 1000 km above earths surface radius of earth 6400km and g is 9.8

Answer 1

Escape velocity is given by
v = sqrt(2GM / R)

At Earth’s Surface
v = sqrt(2GM / R)
= sqrt(2GM / 6400)
= 11.2 km/s ……[1]

At height h
v’ = sqrt(2GM / (R + h))
= sqrt(2GM / (6400 + 1000))
= sqrt(2GM / 7400) ……[2]

Divide equations [1] and [2]
v / v’ = sqrt(7400 / 6400)
11.2 / v’ = sqrt(7400 / 6400)
v’ = 11.2 × sqrt(6400 / 7400)
v’ = 10.41 km/s

Escape velocity of particle at that height is 10.41 km/s