the earth's orbit around the sun is very cirular, with an average radius of 1.5 x 10^8km. Assume the mass of hte earth is is 6 x 10^ 24 kg
2007-03-06 11:54:28 · 4 answers · asked by a cool person 3 in Science & Mathematics ➔ Physics
You don't need the mass of the Earth.
Radius = 1.5E8 km
Circumference = 2 pi R = 9.42E+08 km
Time for one orbit = 365.25 days
Speed = distance / time
= 9.42E8 km / 365.25 days
= 2,580,000 km / day
= 107,500 km / hour
2007-03-06 12:00:00 · answer #1 · answered by morningfoxnorth 6 · 0⤊ 0⤋
The mass of the earth has nothing to do with it. Just figure the circumference of the orbit, and divide by the number of (pick your unit) in a year. In English units, it's about 66,000 mph.
2007-03-06 20:05:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
i'm assuming this is from a circular motion unit.
v = (2 pi r) /T
v= (2 pi 1.5 x 10^8km) / 365days
= 2583km/day = 30m/s
2007-03-06 20:00:48 · answer #3 · answered by doverbeach 3 · 0⤊ 0⤋
A rough average is 30,000 meters per second. It varies by ~3% or so.
2007-03-06 20:00:38 · answer #4 · answered by jim m 5 · 0⤊ 0⤋