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March 29, 2015

March 29, 2015

Posted by **mac** on Saturday, February 26, 2011 at 11:54am.

- physics -
**tchrwill**, Saturday, February 26, 2011 at 12:08pmThe velocity required to maintain a circular orbit around the Earth may be computed from the following:

Vc = sqrt(µ/r)

where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the Earth, ~3.986365x10^14 ft.^3/sec.^2, and r is the distance from the center of the earth to the altitude in question in feet.

Therefore,

V = sqrt[3.986365x10^14/7.51x10^7]

- physics -
**Jane**, Saturday, February 26, 2011 at 12:49pmI am confused because the answer should be in mi/s.

- physics -
**tchrwill**, Saturday, February 26, 2011 at 1:25pmMy apologies for the typo.

The velocity required to maintain a circular orbit around the Earth may be computed from the following:

Vc = sqrt(µ/r)

where Vc is the circular orbital velocity in meters per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the Earth, ~3.986365x10^14 met.^3/sec.^2, and r is the distance from the center of the earth to the altitude in question in meters.

Therefore,

V = sqrt[3.986365x10^14/7.51x10^7]

= 2303.926 met./sec.x3.281ft./met.

= 7559 ft./sec.x1mile/5280ft.

= 1.4316miles/sec.x3600sec./1 hour

= 5154 mi/hr.

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