Near the earth's surface time period of a satellite is 4 hrs. Find its time period if it is at the distance '4R' from the centre of earth :-

Near the earth's surface time period of a satellite is 1.4 hrs. Find its time period if it is at the distance '4R' from the centre of earth :-

The time period of a satellite is related to the density of earth (p) as

A geostationary satellite is orbiting the earth at height 11R above surface of the earth. R being radius of earth. If the time period of this satellite is 24hr find out time period of another satellite which is revolving at 2R from surface of earth

What is time period of a second's pendulum at the centre of the earth?

A geostationary satellite is orbiting the earth at a height of 6R above the surface oof earth where R is the radius of the earth .The time period of another satellite at a distance of 3.5R from the centre of the earth is ….. hours.

A satellite with a time period of 24 hours is orbiting around the earth at a height of 4R above the earth's surface. Calculate the time period of another satellite at a height of 2R from the surface of earth. Here R is radius of earth.

Time period of pendulum, on a satellite orbiting the earth, is

A geostationary satellite is orbiting the Earth at a height of 6 R above the surface of Earth, where R is the radius of the Earth. The time period of another satellites is 6 sqrt(2) h . Find its height from the surface of Earth.

A satellite S_(1) of mass m is revolving at a distance of R from centre of earth and the other satellite S_(2) of mass 4m is revolving at a distance of 4R from the centre of earth. The time periods of revolution of two satellites are in the raito 1 : n. Find the value of n.