Certain neutron stars (extremely dense stars) are believed to be rotating at about 50 rev/s. If such a star has a radius of 15 km, what must be its minimum mass so that material…

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asked Feb 4, 2020 51.2k views

1 Answer

STerliakov · Answer 1 · 2020-02-08T04:34:35+0000

Answer:
M=49.95* 10^(26) kg

Step-by-step explanation:

Given

N=50 rev/s

$\omega =2\pi N$

$\omega =2\pi \cdot 50=314.2 rad/s$

radius of neutron star
r=15 km

Centripetal force on material having mass m is given by

F_c=(mv^2)/(r)

Gravitational Force between neutron star and mass m is

F_g=(GMm)/(r^2) ,where M=mass of Neutron star

equating centripetal Force and Gravitational Pull

(mv^2)/(r)=(GMm)/(r^2)

M=(v^2r)/(G)

$M=(\omega ^2r^3)/(G)$

M=((314.2)^2* (15000)^3)/(6.67* 10^(-11))

M=(0.4995* 10^17)/(10^(-11))

M=49.95* 10^(26) kg