When a large star becomes a supernova, its core, with a mass about equal to that of our sun, may be compressed so tightly that it becomes a neutron star (composed of only neutrons). Suppose the radius of the neutron star is 20 km (about the size of Rochester) and that it rotates once every 1.2 s.
(a) What is the speed of a neutron on the surface of the neutron star at its equator? m/s
(b) What is the magnitude of the acceleration of this neutron? m/s2
2006-12-14 09:26:34 · 2 answers · asked by ckielblock18 1 in Science & Mathematics ➔ Physics
Let's see here...
The neutron goes all the way around the circumference of the star every 1.2 s.
The circumference is 20000 m * 2 * pi = 125664 m
So the speed is 125664 m / 1.2 s = 104720 m/s
Acceleration = (104720 m/s)^2 / (20000 m) = 548311 m/s^2
2006-12-14 09:44:07 · answer #1 · answered by Bill C 4 · 0⤊ 0⤋
(a) The distance a newtron overcomes every rotation at the equator of a star with the radius R=20km is
L = 2PiR = 2*3.14*20km = 125.6 km
Hence, the speed is
V = L/Tr = 125.6 km/1.2 s = 104.67 km/s = 104 667 m/s
We can see that the speed is significantly less than the speed of light 300 000 000 m/s, and this means that we can use usual Newton space - time classical equations.
(b) The magnitude of acceleration is
a = V^2/R = 10955111111 (m^2/s^2)/20000 (m) = 547755 (m/s^2)
2006-12-14 17:45:47 · answer #2 · answered by Oakes 2 · 0⤊ 0⤋