. If the radius of the earth, assumed to be a perfect sphere, suddenly shrinks to half
its present value, the mass of the Earth remaining unchanged, what will be the
duration of one day?
2007-05-17 01:15:29 · 8 answers · asked by hafsa a 1 in Science & Mathematics ➔ Physics
Depends if you are assuming a constant rate of rotation. I'm assuming you are not, since the earth's rotation would theoretically increase if the size decreased. So, my rather unscientific answer is less than 24 hours. :) Of course, it would remain unchanged from the current 24 hour day if the rotational velocity was assumed to be a constant, but then, I'm sure you knew that, which would make the reason for asking this question suspect.
2007-05-17 01:30:07 · answer #1 · answered by Jeff 3 · 0⤊ 0⤋
This a question of angular momentum. Think of it like spinning in a rotating chair with arms extended, then pulling your arms in. The result is that you speed up.
I'm not 100% sure, but there is the principle of conservation of angular momentum. Given that the distance from the centre is reduced by one-half, the Earth would spin twice as fast in order to converse the angular momentum. Thus the sideral day would be one-half the current value. A solar day would be nearly 12 hours assuming that the orbit about the sun is unchanged. A perfect sphere need not be assumed.
Addendum: I think that Luis P might be right in saying that the angular momentum is inversely proportional to the square of the radius. However, angular physics functions were never my strong suit. I do know that the sphere would be spinning faster in connection with my example.
2007-05-17 01:27:47 · answer #2 · answered by Ѕємι~Мαđ ŠçїєŋŧιѕТ 6 · 1⤊ 0⤋
6 hours.
Assuming: the earth interior has constant density. A complicated deduction leads to the conclusion that the angular mass (I) is given by I = 3/20 * M * PI * R^2, with M the earth mass and R the radius.
The physics law we have to use is conservation of angular momentum:
Lfinal = Linit,
with L = I x W (W is the angular velocity). The duration of the day will be given in this angular velocity, currently Winitial=24 hours/day.
Some substitutions, Rfinal=Rinitial/2, Mfinal=Minitial, and we obtain,
Wfinal=Winitial/4.
Therefore the duration of the day will be 24 h / 4 = 6 hours.
2007-05-17 01:45:32 · answer #3 · answered by Luis P 2 · 4⤊ 0⤋
24hrs, same as it currently is, you can see this by drawing a circle with a straight line marking the radius (from centre to border) every point on that line takes the same amount of time to complete 360 degrees whilst you rotate your drawing round and round.
of course this answer assumes that the earth is still spinning at 15 degrees and hour
2007-05-17 01:32:53 · answer #4 · answered by hardcore_pawn 3 · 0⤊ 1⤋
2d or inertia for a sturdy sphere .. I = 0.40MR² .. I ? R² for consistent mass .. I variations with the aid of ingredient (½)² = a million/4 .. to shield angular momentum (I?) the ang vel (?) might enhance with the aid of a ingredient 4 .. the Earth might rotate 4x quicker .. so the day-length might decrease to a million/4 x (24hr) .. ?6 hr
2016-11-23 20:12:07 · answer #5 · answered by kasee 4 · 0⤊ 0⤋
The earth isn't a perfect sphere, it bulges in the middle because it's spinning
2007-05-17 01:30:08 · answer #6 · answered by loathsomedog 3 · 0⤊ 1⤋
The duration of one day would remain the same unless the rate of rotation has changed aswell.
2007-05-17 01:24:05 · answer #7 · answered by Mike 5 · 0⤊ 1⤋
how does the mass stay the same if the radius changes?
2007-05-17 01:25:11 · answer #8 · answered by jammys 3 · 0⤊ 1⤋