Asked by paula
Scientists want to place a 3300 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.2 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?
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?
Answers
Answered by
Damon
F = G M1 M2 / d^2
d = 2.2+1 = 3.2 mars radius=10.9*10^6 meters
= 6.67*10^-11 (3300)(6.42*10^23)10^-12/(10.9)^2
= 1194 * 10^0
= 1194 Newtons
d = 2.2+1 = 3.2 mars radius=10.9*10^6 meters
= 6.67*10^-11 (3300)(6.42*10^23)10^-12/(10.9)^2
= 1194 * 10^0
= 1194 Newtons
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.