An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth.
(i) Determine the height of the satellite above the earth's surface.
(ii) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed with which it hits the surface of the earth.

Answer

A

Explanation

(a) orbital speed of a satellite of a distancek r from centre of earth,
`nu_(0) = sqrt((GM)/(r)) = sqrt((GM)/(R+h))` …(i)
Given, `nu_(0) = nu_(e)/(2) = sqrt(2GM//R)/(2) = sqrt((GM)/(2R))` …(ii) Froms Eqs. (i) and (ii), We get
`h = R= 6400km` (b) Decrease in potential energy = increase in kenetic energy
or `(1)/(2)m nu^(2) = Delta U`
:.`nu = sqrt((2(DeltaU))/(m)) = sqrt((2((mgh)/(1+h//R)))/(m)) = sqrt(gR)` `(h =R)`
`= sqrt (9.8 xx 6400 xx 10 ^(3))`
`= 7919m//s = 7.9 km//s`