The weight of a star is usually balanced by two forces: the gravitational force,
ID: 1467278 • 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 weight 100 million metric ton on the Earth. A) Consider a neutron star whose mass is twice the mass of the Sun and whose radius is 14.3 km. If it rotates with a period of 1.57s, 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) Compare the weight of a 1.50-kg mass on the Earth with its weight on the neutron star. How many times bigger is this mass on the neutron star than on Earth? D)If a satellite is to circle 14.3 km above the surface of such a neutron star, how many revolutions per minute will it makes? E) What is the radius of the geostationary orbit for this neutron star?
Explanation / Answer
here,
mass of neutronstar , m = 2*ms
m = 4 * 10^30 kg
r = 14.3 km
time period , T = 1.57 s
A)
let the speed be v
v = r*w
v = r * 2*pi/T
v = 14300 2*pi/1.57
v = 57200 m/s
B)
value of g , g' = G*m/r^2
g' = 6.67 * 10^-11 * 4 * 10^30 /( 14300)^2
g' = 1.3 * 10^12 m/s^2
thenaccelration due to gravity on this star is 1.3 * 10^12 m/s^2
C)
g'/g = 1.3 * 10^12 m/s^2 /9.8m/s^2
g'/g = 1.33 * 10^11
the g on the star is 1.33 * 10^11 times the g on earth
D)
height of satellite , h = 14.3 km
T = 2*pi*sqrt( (r + h)^3 / G*m)
T = 2*pi*sqrt( ( 14300 + 14300)^3 / 6.67 * 10^-11 * 4 * 10^30)
T = 0.00186 s
w = 2*pi/T
w = 6.28/(0.00186)
w = 3376.34 rad/s
w = 32241.67 rpm
the number of oscillation per minute is 32241.67 rpm
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