Question #115139
Phobos, the larger of Mars’ two moons, orbits 5.99Γ—10 3 km above the planet surface.
Given that the mass of Mars is 6.42Γ—10 23 kg and the radius of Mars is 3.39Γ—10 3 km, what are
the orbital speed and period of Phobos?
1
Expert's answer
2020-05-12T09:57:02-0400

Let denote the height at which Phobos is revolving from the surface of the Mars hh .Radius of Mars is RR, mass of Mars and Phobos are M&mM \&m respectively,vv is the orbital velocity of Phobos and TT is the orbital period of Phobos.

We have given that,

h=5.99Γ—103km=5.99Γ—106mM=6.42Γ—1023kgR=3.39Γ—103km=3.39Γ—106mh=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=GMm(R+h)2(1)F=G\frac{Mm}{(R+h)^2} \hspace{1cm} (1)

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

F=mv2(R+h)(2)F=\frac{mv^2}{(R+h)} \hspace{1cm}(2)

Therefore, from equation 1 and 2 we get,


GMm(R+h)2=mv2(R+h)β€…β€ŠβŸΉβ€…β€Šv=GMR+h(♣)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=6.67Γ—10βˆ’11β‹…6.42Γ—10239.38Γ—106v=2.136Γ—103msβˆ’1v=\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=2.136Γ—103msβˆ’1v=2.136\times 10^3ms^{-1} .


Now, we also know that

v=(R+h)Ο‰β€…β€ŠβŸΉβ€…β€Šv=(R+h)2Ο€Tβ€…β€ŠβŸΉβ€…β€ŠT=2Ο€(R+h)v(β™ )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 (β™ )(\spades) we get,


T=2π×9.38Γ—1062.136Γ—103β€…β€ŠβŸΉβ€…β€ŠT=27.591Γ—103sT=\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=27.591Γ—103sT=27.591\times10^3 s .


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Comments

Asfaque
10.06.20, 12:45

I have really enjoyed such an elangant solution and free too! Thanks assignmentexpert.com

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