Question

The radius of a planet is $74\times {10}^{3}km$. A satellite of this planet completes one revolution in $16.7$ days. If the radius of the orbit of satellite is $27$ times the radius of planet, then the mass of planet is

• A. $2.4\times {10}^{26}kg$
• B. $2.4\times {10}^{30}kg$
• C. $2.4\times {10}^{23}kg$
• D. $2.4\times {10}^{42}kg$

Answer 1

The correct option is A $2.4\times {10}^{26}kg$

$R=74\times {10}^{6}m$

$T=16.7\times 60\times 24\times 60sec=1.44288\times {10}^{6}sec$

$r=27R=27\times 74\times {10}^{6}m=1.998\times {10}^{9}m$

$T=2\pi \sqrt{\frac{r^3}{GM}}\Rightarrow \frac{T^2}{4{\pi }^2}=\frac{r^3}{GM}$

$\Rightarrow$ mass $m=\frac{r^{3}4{\pi }^2}{G{T}^2}=\frac{(2\times {10}^9{)}^3\times 4\times 9.86}{6.67\times {10}^{-11}\times (1.4{)}^2\times {10}^{12}}$

$\Rightarrow m=2.4\times {10}^{27+11-12}kg$

$m=2.4\times {10}^{26}kg$