A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. What is the time period of another satellite orbiting at a height of 2.5 R from the surface of the earth ?

Options

A. 6.2 hour
B. 8.48 hour
C. 9.5 hour
D. 11.6 hour

Answer

B

Explanation

`T^(2)prop R^(3)` (Kepler's law)
`therefore (T_(2))/(T_(1))=((R_(2))/(R_(1)))^(3//2)`
where `R_(2)=2.5R+R=3.5R` and `R_(1)=7R`
`therefore T_(2)=((3.5R)/(7R))^(3//2)xx T_(2)=((1)/(2))^(3//2)xx 24=8.48` hour