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)?
2 answers
You must provide units with your numbers. You should have learned that by now
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.
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.
17 answers