Consider a compact star with a mass equal to the sun(1.99*10^30kg), a radius of
ID: 1408630 • Letter: C
Question
Consider a compact star with a mass equal to the sun(1.99*10^30kg), a radius of only 10km and a a rotation period of 1.0s.
a. what is the speed of a point on the equator of the star?
b. what is g at the surface of the compact star?
c. a stationary 1.0kg mass has a weight on earth of 9.8N. what would be its weight on the compact star?
d. how many revolutions per minute are made by a satelite orbiting 1.0km above the star's surface?
e. what is the radius of a geosynchronous orbit about the compact star?
Explanation / Answer
a) If the star rotates once, a point on the equator travels a distance of 2 * pi * 10 km. The neutron star rotates with a period of 1 second
so the speed of a point on the equator is 2 * pi * 10000 / 1 m/s = 62832 m/s (approximately)
b) Use the universal law of gravitation:
F = Gm1m2 / r²
F = force
G = gravitational constant = 6.674 x 10^-11 N·m²/kg²
m1 = mass of first body
Ms = mass of star = 1.99 x 10^30 kg
r = radius = 10000 m
We're looking for the acceleration due to gravity at the surface of this star. Use Newton's second law, F=ma, so solve for the acceleration a of an object with mass m:
a = F/m
Plug this back into our first equation. Since it's the acceleration due to gravity, we'll call it g:
a = g = F/m1 = Gm2 / r²
g = (6.674 x 10^-11 N·m²/kg²)(1.99 x 10^30 kg) / (10000 m)²
g 1.33 x 10^12 m/s²
c) Weight = mg = 1*1.33 x 10^12 = 1.33 x 10^12 N
d) Radius of the orbit R = 10000 + 1000 = 11000 m
T = R^1.5 * sqrt(K)
K = 4pi^2 / GMs = 2.99*10^-19
T = 6.31*10^-4 sec
f= 1/T = 1585 rev/sec = 95100 rev/min
e) Period of the satellite = period of the star
T^2 = KR^3
1 = 2.99*10^-19 * R^3
R = (3.344*10^18)^1/3 = 1.5*10^6 m
1.49*10^6 m above the surface
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