phobos orbits mars in 27,553 s at a distance of 9.378×10⁶ m. what is the mass of mars?\n2.58×10¹¹…

Explanation:

Step1: Recall the formula for orbital motion

The centripetal - force for the orbiting body (Phobos) is provided by the gravitational force between Phobos and Mars. The formula for the period of a circular orbit is $T = 2\pi\sqrt{\frac{r^{3}}{GM}}$, where $T$ is the period of the orbit, $r$ is the radius of the orbit, $G$ is the gravitational constant ($G=6.67\times 10^{- 11}\ m^{3}\ kg^{-1}\ s^{-2}$), and $M$ is the mass of the central body (Mars). We can re - arrange this formula to solve for $M$: $M=\frac{4\pi^{2}r^{3}}{GT^{2}}$.

Step2: Identify the given values

We are given that $T = 27553\ s$ and $r = 9.378\times 10^{6}\ m$, and $G = 6.67\times 10^{-11}\ m^{3}\ kg^{-1}\ s^{-2}$.

Step3: Substitute the values into the formula

First, calculate $r^{3}=(9.378\times 10^{6})^{3}=9.378^{3}\times10^{18}\approx824.9\times 10^{18}\ m^{3}$. Then, calculate $T^{2}=(27553)^{2}=7.592\times 10^{8}\ s^{2}$. [ \begin{align*} M&=\frac{4\pi^{2}r^{3}}{GT^{2}}\ &=\frac{4\times(3.14)^{2}\times824.9\times 10^{18}}{6.67\times 10^{-11}\times7.592\times 10^{8}}\ &=\frac{4\times9.86\times824.9\times 10^{18}}{6.67\times 10^{-11}\times7.592\times 10^{8}}\ &=\frac{4\times9.86\times824.9}{6.67\times7.592}\times10^{18 + 11-8}\ &=\frac{32597.9}{50.64}\times10^{21}\ &\approx6.43\times 10^{23}\ kg \end{align*} ]

Answer:

$6.43\times 10^{23}\ kg$ (the third option)