34. Mars has two moons, Phobos and Deimos. Phobos has an orbital radius of 9.4 x 10° m and an

orbital period of 0.32 days. Deimos has an orbital radius of 23.5 x 108 m.

a. What is the orbital period of Deimos?

b. At what height above the surface of Mars would a satellite have to be placed so that it

remains above the same location on the surface of Mars as the planet rotates below it.

A Martian day is equal to 1.02 Earth days.

a) By Newton's law of gravitation and uniform circular motion we know that the period of rotation ([tex]T[/tex]), in days, is directly to the orbital radius ([tex]r[/tex]), in meters, up to [tex]\frac{2}{3}[/tex]. Then, the orbital period of Phobos ([tex]T_{P}[/tex]), in days, is calculated by this expression:

[tex]T_{D} = T_{P}\cdot \left(\frac{r_{D}}{r_{P}} \right)^{\frac{3}{2} }[/tex] (1)

Where [tex]D[/tex] is associated to Deimos.

If we know that [tex]T_{P} = 0.32\,d[/tex], [tex]r_{D} = 9.4\times 10^{6}\,m[/tex] and [tex]r_{P} = 23.4\times 10^{6}\,m[/tex], then the orbital period of Phobos is:

[tex]T_{P} = (0.32\,d)\cdot \left(\frac{23.4\times 10^{6}\,m}{9.4\times 10^{6}\,m} \right)^{\frac{3}{2} }[/tex]

[tex]T_{P} = 1.256\,d[/tex]

The orbital period of Deimos is 1.256 days. [tex]\blacksquare[/tex]

b) Mars has a mass of [tex]6.39\times 10^{23}\,kg[/tex]. By Newton's law of gravitation and uniform circular motion we have the following expression for the radius of the satellite ([tex]r[/tex]), in meters:

[tex]r = \sqrt[3]{G\cdot M}\cdot \left(\frac{T}{2\pi} \right)^{\frac{2}{3} }[/tex] (2)

Where:

[tex]G[/tex] - Gravitational constant, in newton-square meters per square kilogram.
[tex]M[/tex] - Mass of Mars, in kilograms.
[tex]T[/tex] - Rotation period of Mars, in seconds.

If we know that [tex]T = 88128\,s[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex] and [tex]M = 6.39\times 10^{23}\,kg[/tex], then the radius of rotation of the satellite is:

[tex]r = \left[\sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.39\times 10^{23}\,kg)}\right] \cdot \left(\frac{88128\,s}{2\pi} \right)^{\frac{2}{3} }[/tex]

[tex]r\approx 20.319\times 10^{6}\,m[/tex]

The orbit of the satellite in geostationary position is approximately [tex]20.319\times 10^{6}\,m[/tex]. [tex]\blacksquare[/tex]

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