cientists want to place a 2700.0 kg satellite in orbit around Mars. They plan to
ID: 2166146 • Letter: C
Question
cientists want to place a 2700.0 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.4 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-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
The radius of the orbit is (2.4 + 1)* radius of mars = 11549800 meters (a) force of attraction = GMm/r^2 = = 6.67x10^-11 * 6.4191x10^23 * 2700 / (11549800)^2 = = 866.6 Newtons (b) for circ orbit speed = sqrt (GM/r) = = sqrt(6.67x10^-11 * 6.4191x10^23 / 11549800) = = 1925 m/s (c) distance for one rev = 2pi r = 2pi * 11549800 = 72569703 meters TIME for one orbit = distance / speed = 72569703 / 1925 = = 37691 seconds = 10.47 hours (d) using Kepler's third law, we know that r^3 is prop to time squared. So if the time is multiplied by 8, we know 8 squared is 64 and the cube root of that is 4. So the radius should be 4 times greater or... 4 * 11549800 = 4.620 x 10^7 meters Get homework help More than 200 experts are waiting to help you now... ADVERTISEMENT
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