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. `24 h`
B. `6//2.5 h`
C. `2.5//6 h`
D. `6 sqrt(2) h`

Answer

D

Explanation

Here `T = 24 hr`
According to kepler's law `T^(2) prop r^(3)`,
`:. T_(2)^(2) prop r_(2)^(3)` and `T_(1)^(2) prop r_(1)^(3)`
Let `r_(2)= 2.5 R + R = 3.5 R, r_(1) = 6R + R = 7R`
`(T_(2))/(T_(1)) = ((r_(2))/(r_(1)))^(3//2) = ((3.5R)/(7R))^(3//2) = ((1)/(2))^(3//2)`
or `(T_(2))/(T_(1)) = (1)/(2sqrt(2))`
or `T_(2) = (1)/(2sqrt(2)) T_(1) = (1)/(2sqrt(2)) xx 24 = 6 sqrt(2) h`.