Question

A $9.4\times {10}^{21}kg$moon orbits a distant planet in a circular orbit of radius $1.5\times {10}^{8}m$. It experiences a $1.1\times {10}^{19}N$gravitational pull from the planet. What is the moon’s orbital period in earth days?

Short answer

The moon's orbital period in earth days is $=26$days

Step-by-step solution

Given information 

Moon orbit's mass $=9.4\times {10}^{21}kg$

Moon orbit's radius$=1.5\times {10}^{8}m$

Gravitational pull$=1.1\times {10}^{19}N$

Explanation

Gravitational pull $=F=\frac{GMm}{r^2}$

localid="1649397302533" $F=1.1\times {10}^{19}N$

localid="1649397341555" $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

$m=9.4\times {10}^{21}\mathrm{kg}$

localid="1649397362538" $r=1.5\times {10}^{8}m$

Substitute the values

localid="1649397565238" $1.1\times {10}^{19}N=\left(\frac{\left(6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}\right)\left(9.4\times {10}^{21}kg\right)}{{\left(1.5\times {10}^{8}m\right)}^2}\right)$

$\Rightarrow M=$Mass of the distant planet $=3.947\times {10}^{23}\mathrm{kg}$

Moon's orbital period in earth days is given by the expression,

$T=2\pi \sqrt{\frac{r^3}{G(M+m)}}$

localid="1649397468079" $=2\times 3.14\sqrt{\left(\frac{{\left(1.5\times {10}^{8}m\right)}^3}{\left(6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}\right)\left[\left(3.947\times {10}^{23}kg\right)+\left(9.4\times {10}^{21}kg\right)\right]}\right)}$

$=2248240\mathrm{s}$

localid="1649397499772" $=\left(\frac{2248240s}{3600s\times 24s}\right)\text{days}$

$=26$days

Final answer

The moon’s orbital period in earth days is

$=26$days