Question

A satellite orbits Earth 249 km above its surface. If
the radius of the Earth is 6.38 x 106 m and the mass of
the Earth is 5.97 x 1024 kg, what is the satellite’s orbital
speed and period?

Answer 1

QUESTION:-

A satellite orbits in circular motion at an altitude of 5000 kilometers above the surface of the Earth. Calculate the orbital (tangential) velocity of the satellite.

SOLUTION:-

Recall the orbital relation

v = √(GM/r)

G = 6.67 x 10-11 Nm2/kg2

r = (5 x 106 m) + (6.371 x 106 m) = 1.1371 x 107 m

[By the way, you did not state the units of Re.]

v = √[(6.67 x 10-11)(5.972 x 1024) / (1.1371 x 107)] m/s ≅ 5919 m/s