A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius `(1.01)` R. The period of the second satellite is larger than the first one by approximately

Options

A. `1.5%`
B. `3.0%`
C. `1.0%`
D. `0.5%`

Answer

A

Explanation

According to Kepler's law `T^(2)prop R^(3)` or `T^(2)=kR^(3)`
where k is a constant
Taking logarithms on both sides,
2 log T = log k + 3 log R
Differentiating, we get
`2((delta T)/(T))=0 + 3 ((delta R)/(R ))`
`therefore (delta T)/(T)=(3)/(2)((delta R)/(R ))=(3)/(2)((1.01R-R)/(R ))xx100`
`=(3)/(2)xx100xx(1)/(100)=(3)/(2)=1.5%`