Find the radius of the orbit measured from the center of mars

Scientists want to place a 2900.0 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.6 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:



mmars = 6.4191 x 10^23 kg
rmars = 3.397 x 10^6 m
G = 6.67428 x 10^-11 N-m2/kg2

What should the radius of the orbit be (measured from the center of Mars), if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

Approach:

Equate the gravitational force with the centripetal force:

(1) G * Mmars * Msat / (1.6 * Rmars)^2 = Msat * V1^2 / (1.6 * Rmars),

where V1 is the velocity of the satellite at a distance of 1.6 * Rmars.

Simplifying (1),

(2) V1^2 = G * Mmars / (1.6 * Rmars)

Let V2 = the velocity of the second orbit, which takes 8X longer than the first. Since:

(3) V2 = V1 / 8, then, substituting into (2):

(4) V2^2 = (V1 / 8)^2 = G * Mmars / (64 * 1.6 * Rmars) = G * Mmars / R2

Therefore:

(5) R2 = 64 * 1.6 * Rmars

= 64 * 1.6 * 3.397 e6

= 3.4785 e8 m = 3.4785 X 10^5 km

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HOLD THE PRESSES! It is the orbital period T2 which must be 8 X T1, NOT that the velocity V2 = V1/8. Therefore:
The period (time to complete one circular orbit) is given by:

(1) T = 2π * √ (r^3 / G * M) [see Source 1]

where:
r = the radius (semi-major axis = radius for a circular

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Therefore, let T1 be the period for the first orbit at 1.6Rmars:

(2) T1 = 2π * √ ( (1.6 * Rmars)^3 / G * Mmars)

Let T2 be the period for the second orbit: then:

(3) T2 = 8*T1

= 2π * √ (64 * (1.6 * Rmars)^3 / G * Mmars)

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But T2 also is:

(4) T2 = 2π * √ ( (R2)^3 / G * Mmars)

From (3) and (4), then we get:

(5) (R2)^3 = 64 * (1.6 * Rmars)^3 ==>

(6) R2 = 64^(1/3) * 1.6 * Rmars

= 64^(1/3) * 1.6 * 3.397 e6

= 2.1741e8 m = 2.1741e5 km

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Sources:

1. http://en.wikipedia.org/wiki/O…

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