A satellite is launched into a circular orbit of radius `R` around the earth. While a second is lunched into an orbit of radius `1.01R` The period of the second satellite is longer than the first one by approximately:

Options

A. `1.0%`
B. `3.0%`
C. `1.5%`
D. `0.7%`

Answer

C

Explanation

From Kepler's third law of motion
`T^(2)=kR^(3)`
where `k` is constant.
Therefore, `(T_(2))/(T_(1))=((R_(2))/(R_(1)))^(3//2)=((1.01R)/R)^(3//2)`
`implies (T_(2))/(T_(1))=(1+1.01)^(3//2)=1+3/2xx0.01`
Percentage increases in time period is
`(DeltaT)/Txx100=((T_(2))/(T_(1))-1)xx100`
`=3/2xx0.01xx100=1.5%`