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A satellite is placed in equatorial orbit above Mars, which has a radius of 3397 km and a mass MM...

Question:

A satellite is placed in equatorial orbit above Mars, which has a radius of 3397 km and a mass MM = 6.40 x 10{eq}^{23} {/eq} kg. The mission of the satellite is to observe the Martian climate from an altitude of 488 km. What is the orbital period of the satellite?

Kepler's Third Law.

Suppose that a satellite is orbiting a planet M on a circular path of radius R. In this case, uniform circular motion and universal gravitation are intrinsically linked to obtaining the third law of Kepler.

Answer and Explanation:

{eq}\text{Known data:}\\ r = 3397 \,km\\ M = 6.40\times{10^{23}}\,kg\\ h = 488\,km\\ G = 6.674\times{10^{ - 11}}\dfrac{N\,m^2}{kg^2} = 6.674\times{10^{ - 11}}\dfrac{m^3}{kg\,{s^2}} \,\,\, \text{(Universal gravitational constant)}\\ \text{Unknowns:}\\ T = ? \\ {/eq}

The orbital radius of the satellite is:

{eq}R = r + h =3397 + 488 = 3885\,km = 3.885\times{10^6} \,m\\ {/eq}

We are going to determine the orbital period of the satellite from the following expression:

{eq}T = \sqrt {\dfrac{{4{\pi ^2}{R^3}}}{{G\,M}}} {/eq}

Replacing known data:

{eq}T = \sqrt {\dfrac{{4{\pi ^2}{{\left( {3.885 \times {{10}^6}m} \right)}^3}}}{{\left( {6.674 \times {{10}^{ - 11}}{m^3}/kg\,{s^2}} \right)\left( {6.40 \times {{10}^{23}}kg} \right)}}} = 7362\,s\\ \color {blue} {T = 7362\,s} {/eq}


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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