Calculate the time taken by a satellite for one revolution revolution revolving at a height of 6400 km above the earth's surface with velocity 5.6 km/s .

Satellite revolving around the earth at a height of 400 Km above the earth's surface has a time period of about

Calculate the speed and period of revolution of a satellite orbiting at a height of 700 km above the earth's surface. Assume the orbit to be circular. Take radius of earth as 6400 km and g at the centre of earth to be 9.8 m//s^2

Hubble telescope revolves around the earth in a orbit at a height 540 km above the earth 's surface. Find the velcoity of the telescope.

Suppose the satellite is revolving at a height of 35780 km above the earth's surface with velocity 3.08 km/s . How much time the satellite will take to complete one revolution around the earth ?

Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth's surface. M_E = 6 xx 10^24 kg R_E = 6.4 10^6 m

Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth

The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around of the earth at a height h above the earth's surface is (R = radius of the earth).

Solve the following examples / numerical problems: A satellite of mass 100 kg performs uniform circular motion around the earth at a height of 6400 km from the earth's surface . Find its gravitational potential energy. [g=9.8 m/s^2, R=6400 km]

Calculate the value of acceleration due to gravity at a point a. 5.0 km above the earth's surface and b. 5.0 km below the earth's surface. Radius of earth =6400 km and the value of g at the surface of the earth is 9.80 ms^2