The planet Mars has two moons. If one of them | vedclass

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(D) According to Kepler's third law of planetary motion, the square of the orbital period $T$ is proportional to the cube of the orbital radius $r

Vedclass · Accepted Answer

$6.00 	imes 10^{23}\, 	ext{kg}$

Vedclass · Answer

(D) According to Kepler's third law of planetary motion, the square of the orbital period $T$ is proportional to the cube of the orbital radius $r$:
$T^{2} = \frac{4 \pi^{2}}{G M} \cdot r^{3}$
Rearranging the formula to solve for the mass of Mars $M$:
$M = \frac{4 \pi^{2}}{G} \cdot \frac{r^{3}}{T^{2}}$
Given values:
$T = 7\, 	ext{hours}, 30\, 	ext{minutes} = 7.5\, 	ext{hours} = 7.5 	imes 3600\, 	ext{s} = 2.7 	imes 10^{4}\, 	ext{s}$
$r = 9.0 	imes 10^{3}\, 	ext{km} = 9.0 	imes 10^{6}\, 	ext{m}$
$\frac{4 \pi^{2}}{G} = 6 	imes 10^{11}\, 	ext{N}^{-1} 	ext{m}^{-2} 	ext{kg}^{2}$
Substituting these values into the equation:
$M = (6 	imes 10^{11}) \cdot \frac{(9.0 	imes 10^{6})^{3}}{(2.7 	imes 10^{4})^{2}}$
$M = (6 	imes 10^{11}) \cdot \frac{729 	imes 10^{18}}{7.29 	imes 10^{8}}$
$M = (6 	imes 10^{11}) \cdot (100 	imes 10^{10}) = 6 	imes 10^{23}\, 	ext{kg}$

The planet Mars has two moons. If one of them has a period of $7\, \text{hours}, 30\, \text{minutes}$ and an orbital radius of $9.0 \times 10^{3}\, \text{km}$, find the mass of Mars. $\left\{\text{Given}: \frac{4 \pi^{2}}{G} = 6 \times 10^{11}\, \text{N}^{-1} \text{m}^{-2} \text{kg}^{2}\right\}$

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