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.

Explanation

(i) Given, orbital velocity `= (1)/(2) xx` escape velocity
`:. sqrt((GM)/(R + h)) = (1)/(2) sqrt((2GM)/(R )) = sqrt((GM)/(2R))`
or `R + h = 2R`
or `h = R = 6400 km`
(ii) According to law of conservation of energy,
total energy of satellite at height `h`
`=` total energy of satellite at earth's surface
i.e., `0 - (GM m)/((R + h)) = (1)/(2) m upsilon^(2) - (GM m)/(R )`
or `(1)/(2) m upsilon^(2) = (GM m)/(R ) - (GM m)/(R + R) = (GM m)/(2R)`
`[:' h = R]`
or `upsilon = sqrt((GM)/(R )) = sqrt((gR^(2))/(R )) = sqrt(gR)`
`= sqrt(9.8 xx 6400 xx 10^(3)) = 7.92 xx 10^(3) ms^(-1)`
`= 7.92 km s^(-1)`