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) the orbital period of Deimos is 1.26 days

b) the height that the satellite would have to be above the surface of Mars is 2.44*10⁶ m.

Explanation:

a) First we need to calculate the mass of Mars. We do this by using the orbital motion formula:

T² / R³ = 4π² / GM

where

T is the orbital period R is the orbital radius G is the universal gravitational constant M is the mass of the planet

Thus,

Rearranging the formula we get:

M = 4π² R³ / GT²

= 4π² (9.4 x 10⁶ m) / [ (6.67 x 10⁻¹¹ kg⁻¹ m³ s⁻²) (0.32*24*3600) ]

= 6.43 x 10²³ kg

Thus, using the same orbital relation formula and rearranging it, we get

T = (4π² R³ / GM) ^ (1/2)

= (4π² (23.5 x 10⁸) ³ / [ (6.67 x 10⁻¹¹ kg⁻¹ m³ s⁻²) (6.43 x 10²³ kg) ]) ^ (1/2)

= 1.09*10⁸ s

To convert from seconds to days, you divide by 86400*10³

Thus,

T = 1.26 days

Therefore, the orbital period of Deimos is 1.26 days

b) A satellite that hovers above the surface of a planet and stays in the same location is called a geostationary satellite. The forces that the satellite experiences are the universal gravitational force and the centripetal force. In order for the satellite to stay in orbit, these two forces must be equal. Therefore,

F_g = GMm/r² (1)

F_c = mvr² (2)

where

G is the universal gravitational constant m is the mass of the satellite M is the mass of mars r is the radius of the orbit v is the speed of the satellite

Thus, equating equations (1) and (2), we get

GMm / r² = mvr²

GM / r = v

where v = 2πr / T

(T is the orbital period)

Thus,

GM / r = 4π²r² / T²

r³ = GM T² / 4π²

r = ∛[ GM T² / 4π² ]

where r = R_m + h

(R_m is the radius of Mars and h is the height at which the satellite is above the surface of Mars)

Thus,

R_m + h = ∛[ GM T² / 4π² ]

h = ∛[ GM T² / 4π² ] - R_m

Since 1 Martian day is equal to 1.02 Earth days, we can write that

T = 1.02*3600 = 3672 sec

Thus,

h = ∛[ (6.67 x 10⁻¹¹) (6.43 x 10²³) (3672) ² / 4π² ] - (3.397*10³)

h = 2.44 * 10⁶ m

Therefore, the height that the satellite would have to be above the surface of Mars is 2.44*10⁶ m.