The orbit velocity of a satellite at a height `R` above the surface of Earth is `v`. The escape velocity from the location is

The orbital velocity of a body at height h above the surface of Earth is 36% of that near the surface of the Earth of radius R. If the escape velocity at the surface of Earth is 11.2 km s_(-1) , then its value at the height h will be

The orbital velocity of a satellite very near to the surface of earth is v. What will be its orbital velocity at an altitude 7 times the radius of the earth?

If v_(0) be the orbital velocity of a satellite in a circular orbit close to the earth's surface and v_(e) is the escape velocity from the earth , then relation between the two is

sqrt((GM)/(R+h)) : critical velocity of a satellite at height h from surface of planet : : ……………… : escape velocity on the surface of planet

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 satelite is revolving in a circular orbit at a height h above the surface of the earth of radius R. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. The relation between h and R is

The orbital velocity of an satellite in a circular orbit just above the earth's surface is v . For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is