The weight of a star is usually balanced by two forces: the gravitational force,

Question

The weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reaction, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy - a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth.

a) Consider a neutron star whose mass is twice the mass of the Sun and whose radius is 10.1 km. If it rotates with a period of 2.51 s, what is the speed of a point on the Equator of this star?

b) What is the value of g at the surface of this star?

c) If a satellite is to circle 10.1 km above the surface of such a neutron star, how many revolutions per minute will it make?

d) What is the radius of the geostationary orbit for this neutron star?

Explanation / Answer

a) Set the centripetal force equal to the gravitational force. mv^2/r = mMG/r^2 v^2 = MG/r v = sqrt(MG/r) T = 2.51 s = v*2*pi*r T = 2.51 s = 2*pi*v*r r*v = 2.51/(2*pi) mv^2/r = GmM/r^2 v^2 = GM/r (vr)^2 = GMr = (2.51/(2*pi))^2 r = (1/(GM))*(2.51/(2*pi))^2

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