Question

The period of a satellite moving very close to the surface of the earth of radius R is 84 minute. what will be the period of the same satellite, If it is taken at a distance of 3R from the surface of the earth?

OPTIONS

A 84 min

B 84 × 4 min

C 84 × 8 min

D 84√8 min

Answer 1

A. 84 min

Explanation:

Kepler's Third Law states that period squared is directly related to the radius cubed:
`T^2 = (4π^2)/(GM) R^3`
where T is the period, G is the universal gravitational constant, M is the mass of the earth (in this case), and R is the distance from the centers of the 2 bodies.

From that we can get the equation for the period:
`T=2pisqrt(R^3/(GM))`
It would appear that if the radius is tripled (3R), then T would increase by a factor of `sqrt(3^3)=sqrt27`

However, the distance R must be measured from the centers of the bodies. The problem states that the satellite flies very close to the surface of the earth (very small difference), and because the new distance 3R is taken at the surface of the earth (very small difference * 3), the radius hardly changes. This means that the period should stay at around 84 min. (choice A)

It turns out that if it were possible to fly a satellite (theoretically) exactly at the surface of the earth, the radius would equal the radius of the earth, and the period would be 84 minutes (click here for more info). According to this problem then, the change in distance from the surface 3R is effectively `0*3=0`, so R stays the same.