The ratio of cube of circumferences of the orbit of a satellite to the volume of the earth is `6xx10^(10)(g//R)`, where `g` is the acceleration of gravity. Find the time period of satellite (in secods). `R` is the radius of earth

The height at which the acceleration due to gravity becomes (g)/(9) (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is

The height at which the acceleration due to gravity becomes (g)/(9) (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :

A satellite is orbiting around the earth at a mean radius of 16 times that of the geostationary orbit. What is the period of the satellite ?

The period of revolution of a satellite orbiting Earth at a height 4R above the surface of Warth is x hrs, where R is the radius of earth. The period of another satellite at a height 1.5R from the surface of the Earth is

A satellite of time period 2A h is orbiting the earth at a height 6R above the surface of earth, where R is radius of earth. What will be the time period of another satellite at a height 2.5 R from the surface of earth ?

Find the binding energy of a satellite of mass m in orbit of radius r , (R = radius of earth, g = acceleration due to gravity)

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 of mass is in a stable circular orbit around the earth at an altitude of about 100 km. If M is the mass of the earth, R its radius and g the acceleration due to gravity, then the time period T of the revolution of the satellite is

A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.