Answer to Question #115139 in Astronomy | Astrophysics for sara

Question

Phobos, the larger of Mars’ two moons, orbits 5.99×10 3 km above the planet surface.
Given that the mass of Mars is 6.42×10 23 kg and the radius of Mars is 3.39×10 3 km, what are
the orbital speed and period of Phobos?

Answer 1

Let denote the height at which Phobos is revolving from the surface of the Mars $h$ .Radius of Mars is $R$, mass of Mars and Phobos are $M \&m$ respectively,$v$ is the orbital velocity of Phobos and $T$ is the orbital period of Phobos.

We have given that,

$$h=5.99\times 10^3km=5.99\times 10^6m\\ M=6.42\times10^{23}kg\\ R=3.39\times 10^3km=3.39\times 10^6m$$

Now, we know that from Newton's law of Gravitation, The gravitational force(centripetal force)of Mars is the only force which responsible for revolution of Phobos around Mars,thus

Force between Mars and Phobos is

$$F=G\frac{Mm}{(R+h)^2} \hspace{1cm} (1)$$

where, $R+h$ is the distance from center of Mars to Phobos.hence, the same force $F$ can also be written as

$$F=\frac{mv^2}{(R+h)} \hspace{1cm}(2)$$

Therefore, from equation 1 and 2 we get,

$$G\frac{Mm}{(R+h)^2}=\frac{mv^2}{(R+h)}\\ \implies v=\sqrt{\frac{GM}{R+h}} \hspace{1cm} (\clubs)$$

Now, from equation $(\clubs)$ ,

$$v=\sqrt{\frac{6.67\times 10^{-11}\cdot 6.42\times 10^{23}}{9.38 \times 10^6}}\\ v=2.136\times 10^3ms^{-1}$$

Therefore, orbital speed of Phobos is $v=2.136\times 10^3ms^{-1}$ .

Now, we also know that

$$v=(R+h)\omega\\ \implies v= (R+h) \frac{2\pi}{T}\\ \implies T=\frac{2\pi (R+h)}{v} \hspace{1cm}(\spades)$$

Thus, from equation $(\spades)$ we get,

$$T=\frac{2\pi \times 9.38\times 10^6}{2.136\times10^3}\\ \implies T=27.591\times10^3 s$$

Therefore, period of Phobos is $T=27.591\times10^3 s$ .