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harina [27]
3 years ago
12

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

m, what must be its minimum mass so that material on its surface remains in place during the rapid rotation in kilograms?
Physics
1 answer:
Aleks [24]3 years ago
4 0

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

\frac{GM_{ns}}{R^2}= \omega^2 R

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

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3 years ago
Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets
schepotkina [342]
I don't actually understand what your question is, but I'll dance around the subject
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I hope I told you something that you were actually looking for.
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