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.(a) Determine the height of the satellite above the earth.s surface . (b) If the satellite is stopped suddenly in its orbit and allowed to fall freely into the earth , find the speed with which its hits the surface of the earth. [g= 9.8 ms^(-2) and R_(E) = 6400km]

An artificial satellite is moving in a circular orbit around the earth

1.	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.(a) Determine the height of the satellite above the earth.s surface . (b) If the satellite is stopped suddenly in its orbit and allowed to fall freely into the earth , find the speed with which its hits the surface of the earth. [g= 9.8 ms^(-2) and R_(E) = 6400km]
Answer» Solution :(a) We know that for satellite motion `v_(0)=SQRT((GM)/r) =Rsqrt((G)/((R+h)))[ "as"g=(GM)/R^2and r = R+h]` In this problem `v_(0)=1/(2)v_(e)=1/2sqrt(2gR)["as "v_(e)=sqrt(2gR)]` So , `(R^2g)/(R+h)=1/2 gR` i.e , 2R =h + Rorh = R = 6400 km (B) By conservation of ME `0+[-(GMM)/r]=1/2mv^2+[-(GMm)/R]" or " v^2=2GM[(1)/R-(1)/(2R)]` [as r = R + h = R + R = 2R] `v =sqrt((GM)/R) = sqrt(gR) = sqrt(10 xx6.4 xx10^(6))= 8km//s `