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Mars rotates on its axis once every 24.8 hours. It's radius is 3.37 x 10^6m, and it's mass is...

Question:

Mars rotates on its axis once every 24.8 hours. It's radius is 3.37 x 10{eq}^{6} {/eq} m, and it's mass is 6.42 x 10{eq}^{23} {/eq} kg.

What is the speed of a geosynchronous satellite orbiting Mars?

The Orbital speed of a satellite orbiting Mars:

A satellite that is orbiting the Mars at its orbit has an orbital speed.

The orbital speed of the satellite orbiting Mars is given by;

{eq}v = \sqrt{\dfrac{GM_m}{r}} {/eq}

where, {eq}M_m {/eq} is the mass of the Mars, {eq}r {/eq} is orbiting radius of the satellite and {eq}G = 6.67 \times 10^{-11} \ Nm^2/kg^2 {/eq} is the gravitational constant.

Answer and Explanation:

Given:

  • A satellite orbiting Mars.
  • The period of rotation of the Mars is {eq}T = 24.8 \ Hours \ = 24.8 \times 3600 \ s. {/eq}
  • The mass of the Mars is {eq}M_m = 6.42 \times 10^{23} \ kg {/eq}
  • The radius of the Mars is {eq}R_m = 3.37 \times 10^{6} \ m {/eq}

The radius of the satellite from the center of the Mars is given by;

{eq}\begin{align} r^3 &= \dfrac{GM_m}{4 \pi^2} T^2 \\ &= \dfrac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{4 \pi^2} (24.8 \times 3600)^2 \\ \implies r &= 2.0524 \times 10^7 \ m \\ \end{align} {/eq}

Therefore, the speed of a geosynchronous satellite orbiting Mars is;

{eq}\begin{align} v &= \dfrac{2\pi r}{T}\\ &= \dfrac{2\pi \times 2.0524 \times 10^7 }{24.8 \times 3600}\\ &=1.44 \times 10^3 \ m/s \\ \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
502

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