Scientists want to place a 3800 kg satellite in orbit around Mars. They plan to

Question

Scientists want to place a 3800 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.1 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-m'/kg 1) What is the force of attraction between Mars and the satellite? 3199.13433 N Submit 2) What speed should the satellite have to be in a perfectly circular orbit? 3) How much time does it take the satellite to complete one revolution? hrs Submit 4) Which of the following quantities would change the speed the satellite needs to orbit at? the mass of the satellite the mass of the planet the radius of the orbit Submit 5) 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? Submit

Explanation / Answer

1)

force of attraction F = G*Mmars*Msatellite/r^2

r = rmars + 2.1*rmars = 3.1*rmars = 3.1*3.397*10^6

F = 6.67428*10^-11*6.4191*10^23*3800/(3.1*3.397*10^6)^2

F = 1468.07 N

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2)

centripetal force = gravitational force

Msatellite*vo^2/r = G*Mmars*Msatellite/r^2

orbital speed vo = sqrt(G*Mmars/r)

vo = sqrt(6.67428*10^-11*6.4191*10^23/(3.1*3.397*10^6))

vo = 2017.02 m/s

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3)

time T = 2*pi*r/vo = 2*pi*r^(3/2)/sqrt(G*Mmars)

T = 2*pi*(3.1*3.397*10^6)^(3/2)/sqrt(6.67428*10^-11*6.4191*10^23)

T = 32803.97 s

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4)

the radius of the orbit

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5)

T2/T1 = (r2/r1)^(3/2)

T2 = 8T1

8 = (r2/r1)^(3/2)

r2 = 4*r1

the radius is 4 times longer

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