Approximately one billion years ago, the Moon orbited the Earth much closer than it does today. The radius of the orbit was only 24 400 km. Today, the radius is 385 000 km. The orbital period was only 23 400 s. The present period is 2.36 x 10 to the 6th power s. Assume that the orbit of the Moon is circular.
2007-06-27 16:12:43 · 2 answers · asked by Tam 1 in Science & Mathematics ➔ Physics
Answer 1 made a fundamental error. Angular rate ω = v/r, not v*r.
So period P = 2*pi*r/v, and v = 2*pi*r/P, units in this case being km/s.
Ratio vthen / vnow = (24400/23400) / (385000/2.36E6) = 6.3918, vnow/vthen = 0.15645.
2007-06-28 03:09:06 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
P = 2Ï/Ï = 2Ï/rv
v = 2Ï/rP
v(now)/v(then) =
24,400*23,400/(385,000*2.36*10^6)
v(now)/v(then) â 6.2840*10^-4
2007-06-28 02:14:17 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋