Question

Time period of satellite of the earth is 5 hrs.if the seperation b/w the earth and the satellite is increased to 4 times the previous value ,then what is the new time period of the satellite?

Answer 1

According to Kepler's third law the time period (T) - radius (R) relation is given as

T = R^3/2 (1)

so,

T₁ / T₂ = R₁^3/2 / R₂^3/2

or

T₂ = T₁.(R₂^3/2 / R₁^3/2) = T₁.(R₂/ R₁)^3/2 (2)

now as per given

T₁ = 5 hours

and if

R₁ = R

then

R₂ = 4R₁= 4R

so, we get

T₂ = 5 x (4R/R)^3/2

or

T₂ = 5 x 4^3/2

thus, the time period will be

T₂ = 5 x 8 = 40 hours