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Certain neutron stars (extremely dense stars) are believed to be rotating at about 0.77 rev/s. If such a star has a radius of 10 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

2007-11-03 06:45:03 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

Fg>=Fc
F=ma

Fc= mV^2/R =m w^2 R

Fg= GMm/R^2
M- mass of the neutron star
m - pas of the object
R - radius of the neutron star
G - gravitational constant

GMm/R^2=m w^2 R
GM=w^2R^3
M(min)= w^2R^3/G

M(min)= (0.77 x 2 pi)^2 (10E3)^3 / 6.67300 E-11 m^3/ kg1 s^2

M(min)=0.351 E+24 kg

in comparison to our Sun m=1.98892 E+30 kg

it is 5.6 E+6 smaller then our sun

in comparison to our Earth m=5.9742E+ 24
it is 17 times less massive than Earth?

It does not sound like a neutron star I ever heard.

"A typical neutron star has a mass between 1.35 and about 2.1 solar masses, with a corresponding radius between 20 and 10 km"(see ref below)

2007-11-03 06:57:31 · answer #1 · answered by Edward 7 · 0 0

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