Scientists want to place a 3500 kg satellite in orbit around Mars. They plan to

Question

Scientists want to place a 3500 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 1023 kg mars-3.397 x 106 m G 6.67428 x 101 N-m2/kg 1) What is the force of attraction between Mars and the satellite? N Submt 2) What speed should the satellite have to be in a perfectly circular orbit? 6947.6 m/s Submit Your submissions: 6947.61 value: 6947.6 Submitted: Thursday, September 28 at 9:09 PM Computed

Explanation / Answer

Given,

m(mars) = 6.4191 x 10^23 ; r(mars) = 3.397 x 10^6 m ; m(sat) = 3500 kg

1)We know that the force of gravity between two masses is given by:

F = G m1 m2/r^2

r = 3.397 x 10^6 + 1.6 x 3.397 x 10^6 = 8.8322 x 10^6 m

F = 6.67428 x 10^-11 x 6.4191 x 10^23 x 3500/(8.8322 x 10^6) = 1.6978 x 10^10 N

Hence, F = 1.6978 x 10^10 N

2)the velocity will be given by:

v = sqrt (G M/r)

v = sqrt (6.67428 x 10^-11 x 6.4191 x 10^23/(8.8322 x 10^6)) = 2202.44 m/s

Hence, v = 2202.44 m/s

3)time required will be:

t = d/v

d = 2 pi r

t = 2 x 3.14 x 8.322 x 10^6/2202.44 = 2.3729 x 10^4 s = 6.5914 hrs

Hence, t = 6.5914 hrs

4)the mass of the planet ; radius of the orbit

5)t = 8 x 2.3729 x 10^4 = 1.918 x 10^5 s

d = v t

d = 2202.44 x 1.918 x 10^5 = 4.223 x 10^8

r = 4.223 x 10^8/2 x 3.14 = 6.726 x 10^7 m

Hence, r = 6.726 x 10^7 m

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