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March 1, 2015

Posted by **vera** on Saturday, August 4, 2012 at 7:32pm.

- math -
**vera**, Saturday, August 4, 2012 at 10:12pmThe answer should be 7.54 hours. What is the formula I need to get this answer?

- math -
**tchrwill**, Sunday, August 12, 2012 at 11:43amThe formula that will clear up your problem is

......Vc = sqrt(µ/r)

where Vc = the velocity required to keep a body in a circular orbit, in feet/sec., r = the orbital radius in feet and µ = the Earth's gravitational constant or 1.407974x10^16 ft^3/sec^2.

The numbers you offer are inconsistent and unreal. Also, the radius of the earth is 3963 miles or 20,924,640 feet.

The real circular velocity for an orbit of 820 miles altitude is

Vc = sqrt(1.407974x10^16/(3963+820)5280

Vc = 23,612 fps or 16,099 mph.

The orbital period is therefore

Tc = (3963+820)5280(2)3.14/23,612 hr

Tc = 112 min.

The orbital period may also be derived from

Tc = 2(Pi)sqrt(r^3/µ)

Tc = 6.28sqrt(25,254,240^3/1.407974x10^16)

Tc = 6720 sec. or 1.866 hr or 112 min.

For the 1400 mile altitude,

Tc = 15,203mph and Tc = 132.98 min.

Tc(1400)/Tc(840) = r(1400)^(3/2)/r(840)^(3/2)

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