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. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.

Options

A. 10 hour
B. `(6 // sqrt2)` hour
C. 6 hour
D. `6 sqrt2` hour

Answer

D

Explanation

According to Kepler.s third law `T^2 prop R^3`
` therefore (T_2/T_1)^2 = (R_2/R_1)^3`
Given `R_1 = 6R + R = 7R, R_2 = 2.5 R + R = 3.5 R`
`T_1 = 24` hour and `T_2 = ?`
` therefore ((T_2)/(24))^2 = ((3.5 R)/(7R))^3 = 1/8`
`T_2^2 = ((24)^2)/(8) " or " T_2 = (24)/(sqrt8) = (12)/(sqrt2) = 6 sqrt2 ` hour