Asked by Ruth

1)What must be the orbital speed of a satellite in a circular orbit 740 above the surface of the earth?

2)What is the period of revolution of a satellite with mass that orbits the earth in a circular path of radius 8000 (about 1630 above the surface of the earth)?

Answers

Answered by drwls
You must provide units with your numbers. You should have learned that by now
Answered by tchrwill
Since 8000 - 1630 = 6370, the radius of the earth in km, I believe I am on safe ground assuming that you mean "740km" above the earth in your first question.

The velocity required to maintain a circular orbit derives from Vc = sqrt(µ/r) where Vc = the velocity in m/s, µ = the earth's gravitational constant, 3.9863x10^14 m^3/sc^2 and r = the orbital radius in meters.

Therefore,
Vc = sqrt[3.9863x10^14/(6370+740)1000]

The period of a particular orbit derives from
T = 2(Pi)sqrt(r^3/µ)

I'll let you complete this one.
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