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Mars has two moons, Phobos and Deimos (Fear and Panic, the companions of Mars, the god of war)....

Question:

Mars has two moons, Phobos and Deimos (Fear and Panic, the companions of Mars, the god of war). Deimos has a period of 30 h, 18 min and a mean distance from the centre of Mars of 2.3 x {eq}10^4 {/eq} km. If the period of Phobos is 7 h, 39 min, what is the mean distance (in km) from the centre of Mars?

Period of a Satellite:

{eq}\\ {/eq}

A satellite is an object that orbits around a celestial body by the virtue of the gravitational force exerted on it by the celestial body. Examples of natural satellites are- Moons orbiting about their respective planets, Planets around the sun, etc.

The period of a satellite is equal to the time needed to complete its orbit. If the celestial body is spherical in shape, and the motion of the satellite is circular, then the period of the satellite depends on:

  • The mass of the celestial body.
  • The distance of the satellite from the center of the celestial body.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The period of Deimos, {eq}T_d=30\;\rm h\;18\;\rm min=109080\;\rm s {/eq}
  • The mean distance of Deimos from the center of Mars, {eq}r_d=2.3\times 10^{4}\;\rm km=2.3\times 10^{7}\;\rm m {/eq}
  • The period of Phobos, {eq}T_p=7\;\rm h\;39\;\rm min=27540\;\rm s {/eq}


The time period of the motion of a satellite around a planet depends only on the mass, {eq}M {/eq}, of the planet and its distance, {eq}r {/eq}, from the center of the planet by the following equation:

{eq}T=2\pi\sqrt {\dfrac{r^3}{GM}} {/eq}

Here,

  • {eq}G=6.67\times 10^{-11}\;\rm m^3/kg.s^2 {/eq} is the gravitational constant.


From the above equation, it is clear that for a fixed planet,

{eq}T^2\propto r^3 {/eq}


Therefore, for the two moons, we have:

{eq}\begin{align*} \dfrac{T_d^2}{T_p^2}&=\dfrac{r_d^3}{r_p^3}\\ \Rightarrow r_p&=\left ( \dfrac{T_p}{T_d} \right )^{2/3}r_d\\ &=\left ( \dfrac{27540}{109080} \right )^{2/3}\times 2.3\times 10^{7}\\ &\approx 0.4\times 2.3\times 10^{7} \\ &=\boxed{9.2\times 10^{6}\;\rm m} \end{align*} {/eq}



Learn more about this topic:

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Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
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