Question

You are given the following data: $g=9.81$ $m{s}^{-2}$, radius of earth $=6.37\times {10}^6$m, the distance the Moon from the earth $=3.84\times {10}^8$m and the time period of the Moon's revolution $=27.3$days. Obtain the mass of the earth in two different ways. $G=6.67\times {10}^{-11}$N $m^2$ $k{g}^2$.

Answer 1

Here, $g=9.81$m $s^{-2}$
$R=6.37\times {10}^6$m, $r=3.84\times {10}^8$m
$T=27.3$days $=27.3\times 24\times 60\times 60$s
$\therefore M_e=\frac{g{R}_e^2}{G}=\frac{9.81\times (6.37\times {10}^6{)}^2}{6.67\times {10}^{-11}}=5.97\times {10}^{24}$kg
Alternative method: Since the gravitational pull provides the required centripetal force, so
$mr\left(2\pi /T^2\right)=G{M}_{e}m/r^2$ or $M_e=\frac{4{\pi }^2{r}^3}{G{T}^2}$
$=\frac{4\times (22/7{)}^2\times (3.84\times {10}^8{)}^3}{6.67\times {10}^{-11}\times (27.3\times 24\times 60\times 60{)}^2}$
$=6.02\times {10}^{24}$kg.