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 onto the earth, find the speed with which it hits the surface of the earth. [g = 9.8 ms^(-2) "and " R_(E) = 6400 km]

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 onto the earth, find the speed with which it hits the surface of the earth. [g = 9.8 ms^(-2) "and " R_(E) = 6400 km]
Answer» Solution :(a) We know that for satellite motion `v_(0) = SQRT((GM)/(R)) = Rsqrt((g)/((R+H)))` `["as g" = (GM)/(R^(2)) "and r = R + h"]` In this problem `v_(0) = (1)/(2)v_(e) = (1)/(2) sqrt(2gR)` `["as " v_(e) = sqrt(2gR)]` So `(R^(2)g)/(R+h) = (1)/(2)gR`, i.e., `2R = h+R` or `h = R = 6400 km` (b) By conservation of ME `0 + [-(GMm)/(r)] = (1)/(2) mv^(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 xx 6.4 xx 10^(6)) ~= 8km//s`