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 at a speed of 2492 m/s in a perfectly circular orbit. Here is some information that may help solve this problem:

mmars = 6.4191 x 1023 kg

rmars = 3.397 x 106 m

G = 6.67428 x 10-11 N-m2/kg2 1)

1) What radius should the satellite move at in its orbit? (Measured frrom the center of Mars.)

2) What is the force of attraction between Mars and the satellite?

3) What is the acceleration of the satellite in orbit?

4) What should the speed of the orbit be, if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

Explanation / Answer

1) In a circular orbit the centripetal force mv^2/r is equal to the force of gravity GmM/r^2.
The radius to use is (A) x 3.397 x 10^6 m.
Therefore r = (GM/v^2) = (6.67428 x 10^-11 x 6.4191 x 10^23 / 2492^2 ) = 6898941.9 m

So A = 2.03

2) F = GmM/r^2 = (6.67428 x 10^-11 x 6.4191 x 10^23 x 3800)/6898941.9 ^2 = 3420.56 N

3) Acc. = a = F/m = 3420.56/3800 = 0.9 m/s

4) Period = circumference/velocity = 2 pi x 2..03 x 3.397 x 10^6 / 2492 = 1.74 x 10^4 sec = 4.83 hour

Kepler's third law tells you that distance cubed is proportional to period squared => distance is proportional to period^(2/3) = 2.86

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