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November 27, 2015
Posted by **Faith** on Thursday, January 13, 2011 at 1:09pm.

- Physics -
**tchrwill**, Friday, January 14, 2011 at 9:54amThe velocity required to keep the earth in a circulor orbit of radius 4 x(93,000,000) miles derives from

V = sqrt[µ/r] where µ = the Sun's gravitational constant = 4.68772x10^21 ft.^3/sec.^2 and r = the radius of the increased orbit in feet = 4.9104x10^11 feet.

Therefore, the orbital velocity becomes

V = sqrt[4.68772x10^21ft.^3/sec.^2/4.901x10^11] = 48,853 ft./sec. = mph. = 9.25 mph. compared to 97,706fps and 18.5 mph in its existing orbit

More symplictically, V = 97,700sqrt[r/4r] = 97,706[1/4) = 97,706/2 = 48,853 fps or 9.25 mph.