I guess by falling it means tangential velocity
F = ma = mv^2/r where m is mass of body n orbit
= G m M/r^2 where M is mass of central body, in this case earth
so v^2 = GM /r
call moon orbit r = 1 and sun r = 372
so for moon 1.4^2 = G M/1
and sun v^2 = G M/372
v^2 = 1.42^2/372
v = 1.42/sqrt(372)
If the sun is 372 times farther from the Earth than the Moon, and the Moon is falling around the earth at 1.4 mm/s, what is the rate that the Sun is "falling?"
Is it 1/(372^2) or 7.22*10^-6 mm/s?
3 answers
1.4mm/s=
.0014m/s(60sec/min)*60min/hr(24hr/day)(14day/halfMoonCycle)=1690m about a land mile.
Hmmmm. So I wonder in the hypothesis, what direction is the 1.4mm/s?
.0014m/s(60sec/min)*60min/hr(24hr/day)(14day/halfMoonCycle)=1690m about a land mile.
Hmmmm. So I wonder in the hypothesis, what direction is the 1.4mm/s?
Direction? The Moon revolves the Earth counter-clockwise. The Sun is rotating counter-clockwise. But these are not related to the question as it does not ask for direction.