Question

34.The time period of satellite of earth is 5hr.If the seperationbetween earth and the satellite is increased to 4 times the previous value,the new time period will become ----------------------

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

T 2 = 5 x 8 = 40 hours