Asked by Sam
(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 153 km?
m/s
(b) What is the period of revolution?
min
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m/s
(b) What is the period of revolution?
min
Read the eBook
Answers
Answered by
tchrwill
= te gravitational constantThe speed required to hold a satellite in a circular orbit derives from V = sqrt(µ/r) where V = the orbital speed in feet/sec., r = the orbital radius in feet and µ = the gravitational constant of the earth = 1.407974x10^16, ft.^3/sec.^2 or 3.986365x10^14 m^2/sec.^2.
Using the earth radius of 3963 miles, or 6378 km, r becomes 6378 + 153 = 6531 km. or 6531000m. Then V = sqrt(3.986365x10^14/6531000) = 7812.65m/s or.
The orbital period derives from T = 2(Pi)sqrt(r^3/µ).
Using the earth radius of 3963 miles, or 6378 km, r becomes 6378 + 153 = 6531 km. or 6531000m. Then V = sqrt(3.986365x10^14/6531000) = 7812.65m/s or.
The orbital period derives from T = 2(Pi)sqrt(r^3/µ).
Answered by
Sam
thank you!!!
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