Posted by katy on Saturday, September 10, 2011 at 7:39pm.
See if any of the following links will help you:
http://www.google.com/search?q=if+the+radius+of+the+earth%27s+orbit+around+the+sun+is+93%2C000%2C000+miles%2C+what+is+the+speed+of+the+earth+in+it%27s+orbit+in+miles+per+hour%3F+&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a
Sra
One year = 365.25 days = 8766 hours
The circumference of the orbit is 93,000,000(2)3.14 = 584,336,234 miles
The orbital velocity is therefore
V = 584,336,234/8766 = 66,660 mph.
It could also be derived knowing the Sun's gravitational constant GM = 4.68772x10^21 ft.^3/sec.^2.
The velocity required to keep the earth in a circular orbit derives from
Vc = sqrt(GM/r), r in feet.
Therefore, Vc = sqrt[(4.687872x10^21(93,000,000)5280] = 97,706 fps = 66,618 mph.
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