Question

Escape velocity of an atmospheric particle which is 1000km above the earth's surface, is (radius of earth is 6400km and g=9.8m/s2

Answer 1

Escape velocity on surface of earth is
v = sqrt(2GM / R) = 11.2 km/s ……[2]

Escape velocity at height h is
v’ = sqrt[2GM / (R + h)] ……[1]

Divide equations [1] and [2]
v / v’ = sqrt[(R + h) / R)]
11.2 / v’ = sqrt[(6400 + 1000) / 6400]
v’ = 11.2 / 1.075
v’ = 10.4 km/s

Escape velocity at that height is 10.4 km/s

Answer 2

The escape velocity is 10.46km/s or ~ 10^4m/s
PLEASE MARK AS BRAINLIEST