An Earth satellite has time percircular, calculate its height. Given, radius Earth `= 6380 km, g` at the surface of Earth `= 9.8 ms^(-2)`.

Find the value of g at a height of 400 km above the surface of the earth . Given radius of the earth , R = 6400 km and value of g at the surface of the earth = 9.8 ms^(-2) .

Find the value of 'g(i)' at a height of 100 km (ii) at height 6400 km from the surface of the earth. (Radius of the earth = 6400 km, g on the surface of the earth = 9.8 ms^(-2) .

At what altitude, the acceleration due to gravity reduces to one fourth of its value as that on the surface of the earth ? Take radius of earth as 6.4 xx 10^(6) m , g on the surface of the earth as 9.8 ms^(-2) .

At what altitude, the acceleration due to gravity reduces to one fourth of its value as that on the surface of the earth ? Take radius of earth as 6.4 xx 10^(6) m , g on the surface of the earth as 9.8 ms^(-2) .

An earth satellite makes a complete revolution around the earth in 120 minutes. If the orbit is circular calculate the height of satellite above the earth. Radius of the earth = 6400 km g = 9.8 ms^(-2) .