Asked by katy
if the radius of the earth's orbit around the sun is 93,000,000 miles, what is the speed of the earth in it's orbit in miles per hour?
Answers
Answered by
tchrwill
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.
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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