An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of esacpe velocity from the earth the height of the satellite satellite above the earth surface will be

Options

A. 6000 km
B. 5800 km
C. 7500 km
D. 6400 km

Answer

D

Explanation

Given that the orbital velocity of satellite `=("Esacpe velocity")/(2) rarr v_(0)=(v_(e ))/(2)`
but we know that `v_(0)=sqrt(gR^(2))/(R+h)` and `v_(e)=sqrt(2gR)`
on putting these values in Eq I we get
`sqrt(gR^(2))/(R+h)=sqrt(2gR)/(2)`
on squaring both side we obtain
`rarr 2R =R+h rarr R=h`
`rarr h=R =6400 km`