A satellite having time period same as that of the earth's rotation about its own axis is orbiting the earth at a height 8R above the surface of earth. Where R is radius of earth. What will be the time period of another satellite at a height 3.5 R from the surface of earth ?

Options

A. `2 sqrt(2)` h
B. `4 sqrt(2)` h
C. `6 sqrt(2)` h
D. `3 sqrt(2)` h

Answer

C

Explanation

`T^(2) propR^(3)` (By Kepler's third law)
`:. (T_(1)^(2))/(T_(2)^(2))=(R_(1)^(3))/(R_(2)^(3))rArr((T_(2))/(T_(1)))^(2)=((R_(2))/(R_(1)))^(3)`
`rArr T_(2)^(2)=T_(1)^(2)[(4.5R)/(9R)]^(3)rArrT_(2)^(2)=(T_(1)^(2))/(8)rArr T_(2)=(T_(1))/(2 sqrt(2))`
`T_(2)=(24)/(2sqrt(2))" "(because T_(1)=24h)`
`:. T_(2)=6 sqrt(2)h`.