It collapses to a diameter of 10,000 kilometers, and it rotating at 10 revolutions per second.
What is the speed in kilometers per second at the equator?
2007-09-27 08:27:44 · 7 answers · asked by Feeling Mutual 7 in Science & Mathematics ➔ Astronomy & Space
Nacsez is right in principle, but not on facts. The speed of light is approximately 300,000 km/sec, not 30,000. However, it is still too fast.
Neutron stars and pulsars may rotate that fast for some time after their collapse, but they are much smaller.
Also, a red giant puffs away some of its mass before the white dwarf stage. A supernova that leaves a neutron star or pulsar does the same thing in the explosion.
2007-09-27 14:20:32 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Yes, angular momentum is conserved. Even in a black hole angular momentum is conserved (as is mass, charge, and strangeness, or so we currently think ☺)
If the diameter is 10,000 km, then the circumference is
10,000π km and if it's rotating 10 times per second the surface velocity is 100,000π km/sec (which is a kinda bad number since it's greater that the speed of light ☺)
Doug
2007-09-27 09:22:04 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
The above is real, angular momentum is conserved. in spite of the shown fact that, by using fact it is smaller in radius, the rotation fee will develop. From above, L = 2/5 * M * R^2 * w making use of the present mass, radius and rotation fee: M = a million.9891e30 kg R=6.955e8 m w = 2*pi/(30*24*60*60) rad/s L = 9.329e41 kg m^2 / s If the recent radius is 5% of R, r=0.05R, L = 2/5 * M * r^2 * w' the place w' is the recent rotation fee. on account that L = 2/5 * M * R^2 * w, w' = w / 0.05^2 w' = 400w w' = 9.696 rad/s it is one rotation each a million.8hr (108min). Rotational KE is defined as: KE = a million/2 * I *w^2 on account that I' = 0.05^2 * I and w' = 4 hundred*w, KE' = a million/2 * 0.05^2 * I * 4 hundred^2 * w^2 KE' = 4 hundred * a million/2 * I *w^2 KE' = 400KE it could have 4 hundred cases the rotational kinetic capability that it has right this moment. this added capability comes from changing gravitational skill capability into kinetic capability because it collapses.
2016-12-17 11:46:09 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Well, if the star is 10K in diameter, then it's circumference is pi X D, or 3.141529 X 10,000 = 31,415.29 kilometers.
Since it completes a revolution in 1/10 of a second, that means every second a point on the equator moves 10 X 31,415.29 km, or 314,152.9 Km/second
That's truckin'.
2007-09-27 08:33:53 · answer #4 · answered by quantumclaustrophobe 7 · 0⤊ 0⤋
Yes, angular momentum is conserved. However, given the impossibly high rate of revolution (faster than light at the equator!) you have postulated, this star would have thrown off some mass instead. But the net angular momentum of the sum of the pieces would remain the same.
2007-09-27 10:16:43 · answer #5 · answered by injanier 7 · 0⤊ 0⤋
last time i checked, the speed of light was 30,000 km a second, so i doubt that this is possible.
it should be noted that starts are generally fluid/plasma bodies and do not conserve angular momentum the same way as solid bodies would. that doesnt mean that angular momentum wouldnt be conserved, but the equator wouldnt be going faster than light.
who ever asked you this question didnt think it out very well
2007-09-27 08:43:47 · answer #6 · answered by nacsez 6 · 1⤊ 0⤋
Of COURSE!
2007-09-27 13:43:50 · answer #7 · answered by keylauder 2 · 0⤊ 0⤋