certain neutron stars(extremely dense stars) are believed to be rotating at about one revolution per second . if such a star has a radius of 20km, what must its minimum mass be so that objects on its surface will still be attracted to the star and not be "thrown off" by the rapid rotation?
2006-09-23 05:02:29 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
It will be the mass such that the centripetal acceleration is the same as the graviational acceleration. The centripetal acceleration is rw^2, but you have to express r (the equatorial radius, which you were given) in m, and w (the rotational velocity) in rad/s instead of rev/s as you were given (with 1 rev = 2pi rad). The gravitational acceleration is GM/r^2, with r again in m and M the unknown mass of the star in kg. G is the gravitational constant, 6.67 x 10^-11 Nm^2/kg^2. Set the two equal, and M is your only unknown.
2006-09-23 05:28:55 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
So any point on the surface will be traveling (20 x 3.14) km/s. So the escape velocity has to be slightly less than that. What mass gives that escape velocity/gravity?
2006-09-23 05:07:45 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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2006-09-25 04:35:16 · answer #3 · answered by Anonymous · 0⤊ 0⤋
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2006-09-23 05:15:28 · answer #4 · answered by flying peanuts 3 · 0⤊ 1⤋