A satellite is in a circular orbit around the Earth. If it is at an altitude equal to twice the radius of the Earth, 2RE, how does its speed v relate to the Earth's radius RE, and the magnitude g of the acceleration due to gravity on the Earth's surface?

Explanation:

height of satellite, h = 2 Re

where, Re is the radius of earth

The centripetal force is equal to the gravitational force between the earth and the satellite.

where, m is the mass of satellite and M is the mass of earth, and r is the distance between the centre of earth and the satellite.

r = Re + h = Re + 2Re = 3 Re

where, G M = gRe²

The above expression is the orbital velocity of the satellite at a height.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.