Question

A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth’s mass is 6 × 1024 kg and its radius is 6.4 × 106 m

Answer 1

When a satellite orbits the Earth, it has a tangential velocity of 5000 m/s.

Since the mass of Earth is 1024 x 6 kg and the radius is 106 m x 6.4 then we get to see that by standard notation the distance that exists between the satellite and the Earth would be equal to 10^6 x 2.6.

Answer 2

ME x G/ (rE)² = ω²r

ME  is the mass of the earth

G = gravitational constant = 6.6742 x 10-11 m3 s-2 kg-1

Re is the radius of the earth

ω is the angular velocity

and r is the orbit's radius

In order to convert to angular velocity:

Tangential velocity = rω

ω = 5000/r

(6x10²⁴) x (6.6742 x 10⁻¹¹) /(6.4 x 10⁶)²

= (5000/r)²r

r = 2557110.465 m

So the distance is  2.6 x 10^6 m