Assuming that the moon moves around the earth in a circular orbit of radius `3.8xx10^5` km in 27 days and the earth moves around the sun in a circular orbit of radius of radius `1.5xx10^8` km in 365 days, find the ratio of the masses of the sun and the earth.

Explanation

From Kepler.s law `(T^2)/(r^3)`= constant
where T=time period , r=radius of the orbit.
if mass of the sun is `M_0` and radius of the earth.s orbit around the sun is `r_0`, orbital speed,
`v_0=sqrt((GM_0)/(r_0))`
`therefore` Time period `T_0` of the earth
`=(2pi r_0)/(v_0) =2pi r_0sqrt((r_0)/(GM_0))=2pi sqrt((r_0^3)/(GM_0))`
Similarly, if mass of the earth is M, radius of the moon.s orbit is r, time period,
`T=2pi sqrt((r^3)/(GM))`
`therefore T/(T_0) =sqrt((r/(r_0))^3(M_0)/M) or, ((T)/(T_0))^2=((r)/(r_0))^3*(M_0)/(M)`
`or, (M_0)/(M)=(T/(T_0))^2*((r_0)/r)^3=((27)/(365))^2*((1.5xx10^8)/(3.8xx10^5))^3= 3.37xx10^5=337000`
Hence, the sun is 337000 times more massive than the earth.