Answer:
[tex]\frac{M_e}{M_s} = 3.07 \times 10^{-6}[/tex]
Explanation:
As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula
[tex]T = 2\pi \sqrt{\frac{r^3}{GM}}[/tex]
now for the time period of moon around the earth we can say
[tex]T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}[/tex]
here we know that
[tex]T_1 = 0.08 year[/tex]
[tex]r_1 = 0.0027 AU[/tex]
[tex]M_e[/tex] = mass of earth
Now if the same formula is used for revolution of Earth around the sun
[tex]T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}[/tex]
here we know that
[tex]r_2 = 1 AU[/tex]
[tex]T_2 = 1 year[/tex]
[tex]M_s[/tex] = mass of Sun
now we have
[tex]\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}[/tex]
[tex]\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}[/tex]
[tex]12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}[/tex]
[tex]\frac{M_e}{M_s} = 3.07 \times 10^{-6}[/tex]
