time period of two planets around the sun are in the ratio of 1:8 then radii will be in the ratio of 1 : R find R
Kepler's third law of planetary motion.
The square of the period of a planet in a circular orbit is proportional to the cube of the orbital radius.
( T1/T2 )2 = (R1/R2)3
T1 =1 T2 = 8 and R1=1 and R2=R
Let us substitute T12=1 T22 = 64
then R13 =1 and R23 = R3
apply the law- ( T1/T2 )2 = (R1/R2)3
therefore 1/64 = 1/R3 so,
R3 = 64 = 43
therefore R = 4