Scientists want to place a 4400 kg satellite in orbit around Mars. They plan to

Question

Scientists want to place a 4400 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2 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   rmars = 3.397 x 106 m   G = 6.67428 x 10-11 N-m2/kg2

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

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?

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

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?

Explanation / Answer

(a) we use radius r = 3 Rmars= 3 (3.397 x 10^6 m)=1.019x10^7 m

F = GmM/r^2. = 6.67x10^-11 (4400) (6.4191 x 10^23 )/(1.019x10^7)2 = 1814.25 N

(b) In a circular orbit the centripetal force mv^2/r is equal to the force of gravity GmM/r^2.
Therefore v=sqrt (GM/r) = sqrt (6.67428 x 10^-11 x 6.4191 x 10^23 / (1.019x10^7) =2050 m/s

(c) Period = circumference/velocity = 2 pi x (1.019x10^7) / 2050 = 2.667 x 10^4 sec =8.67 hours

(d)  v=sqrt (GM/r)

the velocity will change by changing mass of planet and radius of orbit.

(e) Kepler's third law tells you that distance cubed is proportional to period squared => distance is proportional to period^(2/3) = 4


So the orbit radius should be four times as much

so radius of orbit= 4 (1.019x10^7)=4.076 x10^7 m

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