Question

The moon revolves round the earth $13$ times in one year. If the ratio of sun-earth distance to earth-moon distance is $392$, then the ratio of masses of sun and earth will be

• A. $365$
• B. $356$
• C. $3.56\times {10}^5$
• D. $1$

Answer 1

Answer: (C)

$3.56\times {10}^5$

From Kepler's law of periods, time period of revolution for Earth around Sun is
$T_e^2=\frac{4{\pi }^2}{G}\frac{a_e^3}{M_s}$; where $a_e$ is the distance between Sun and Earth.
Similarly for Moon and Earth'
$T_m^2=\frac{4{\pi }^2}{G}\frac{a_m^3}{M_e}$
Given
$T_m=\frac{1}{13}$, $T_e=1$ and $\frac{a_e}{a_m}=392$
$\therefore {\left(\frac{T_e}{T_m}\right)}^2=\frac{a_e^3}{a_m^3}\frac{M_e}{M_s}$
$\Rightarrow \frac{M_s}{M_e}={\left(\frac{T_m}{T_e}\right)}^2\frac{a_e^3}{a_m^3}=\frac{{392}^3}{{13}^2}=356427.7=3.56\times {10}^5$(approx)