The time period of a satellite of earth is 5 hours. If the separation between the centre of earth and the satellite is increased to 4 times the previous value, the new time period will become-

Options

A. 10 h
B. 80 h
C. 40 h
D. 20 h

Answer

C

Explanation

Accroding to kepler third law `T^(2) prop r^(3)`
or `5^(2) prop r^(3)`
and `(T)^(2) prop (4r)^(3)`
from Eqs I and ii we get
`(25)/(T)^(2)=(r^(3))/(64r^(3)) rarr T =sqrt(1600)=40 h`