The weight of a star is usually balanced by two forces: the gravitational force,
ID: 2265138 • Letter: T
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 13.7 km. (The mass of the Sun is 1.99
Explanation / Answer
(a)
2?R/v = T [v is velocity, T is time period]
v = 2?R/T = (2?*13.7/1.45) km/s = 59.38 km/s.
(b)
g = GM/R^2 [M is mass of star]
= (6.673*10^-11)*(2*1.99*10^30)/(13.7*1000)^2
= 1.415*10^12 m^2/s
(c)
Weight of 1.2 kg mass on earth= 1.2*9.8 = 11.76 N
Weight of 1.2 kg mass on the star= 1.2*1.8132*10^12 N= 2.176*10^12 N
This is 1.85*10^11 times the weight on earth. It is however the same size on the star.
(d)
Balancing the forces at distance 2R from the centre of the star,
GM/R^2 = v^2/R [the mass of the satellite has been divided from both sides]
v = sqrt(GM/R)
v = sqrt(6.67*10^-11*1.99*10^30/13.7*10^3)
v = 0.97*10^8 m/s.
No. of revolutions per second = v/2?R [R here is the radius of orbit]
No. of revolutions per second = 0.97*10^8/2?*13.7*10^3
No. of revolutions per second = 3540 revolutions per seconds
(e)
Geostationary radius is given by R = cuberoot(GM/w^2) [? is the angular velocity, equal to 7.2921*10^-5 rad/s]
R = cuberoot[6.67*10^-11 * 1.99 * 10^30/(7.2921*10^-5)^2]
= 24.96* 10^6 km.
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