Asked by Alvin
Find the orbital speed of a satellite in a circular orbit 3.70×107 m above the surface of the Earth.
v = 3030 m/s
So I've been trying to solve this problem for some time now and haven't figured it out. What I did is find orbital radius, use that to find the circumference of orbit, and divided it by the seconds in a day.
I get 3150 m/s (sig. figs.) but it is not close enough. My question is how they arrived at 3030 m/s.
v = 3030 m/s
So I've been trying to solve this problem for some time now and haven't figured it out. What I did is find orbital radius, use that to find the circumference of orbit, and divided it by the seconds in a day.
I get 3150 m/s (sig. figs.) but it is not close enough. My question is how they arrived at 3030 m/s.
Answers
Answered by
bobpursley
v^2/(re+h)=9.8(re/(re+h))^2
v^2=9.8*re^2/(re+h)
v^2=9.8* (6.37E6^2/(6.37E6+ 3.7E7)
v^2== 9168863.73
v=3028 m/s rounded to three sig figures is 3030m/s
If I had used 9.81 for g...
v^2=9.8*re^2/(re+h)
v^2=9.8* (6.37E6^2/(6.37E6+ 3.7E7)
v^2== 9168863.73
v=3028 m/s rounded to three sig figures is 3030m/s
If I had used 9.81 for g...
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