Earth satellite – 1
A satellite moves on a circle orbit 640 km above the Equator on the Earth’s surface. Its tangential speed is 27000km/h. Assuming the Earth’s radius of 6378km, what are:
(1) the period of motion of the satellite?
(2) what is the free fall acceleration at the orbit height?
Data and unknown:
h = 640km – distance from the Earth’s surface to the satellite
R = 6378 – radius of the Earth
v = 27000 km/h – tangential speed of the satellite
T = ? – period of motion
gor = ? free fall (gravitational) acceleration at the orbit height.
Solution.
The satellite moves in a circular orbit with a radius equal to the sum of the Earth’s radius and of its height above the Earth’s surface.
Rs = R + h (1)
A period of motion, that is the time required to complete one full revolution, is equal to the ratio
length of one orbit / speed of satellite
T = 2πRs/v< (2)
So
T = 2π(R + h)/v (3)
The gravitational acceleration acting on satellite must be equal to the centripetal acceleration due to uniform circular motion. This is a condition which keeps the satellite at a given orbit. Otherwise the satellite would move towards the Earth or in the opposite direction. So we can write
gor = v2/Rs = v2/(R +h) (4)
Substituting numbers given in the problem into equations (2) and (4) we get
T = 5879 s , gor = 8.0m/s2.