Question

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation? Hint: The equator of the star is where the maximum centripetal acceleration occurs.

Answer 1

M = 4.7 10²⁴ kg

Step-by-step explanation:

For this exercise we must use the law of universal gravitation

F = G m M / r²

Newton's second law with centripetal acceleration

a = v² / r

v = w r

a = w² r

F = m a

G m M / r2 = m w2 r

G M = w² r³

M = w² r³ / G

Let's reduce the magnitudes to the SI system

w = 1 rev / s (2π rad / 1 rev) = 2π rad / s

r = 20 km (1000m / 1 km) = 2 10⁴ m

Let's calculate

M = (2π)²2 (2 10⁴)³ / 6.67 10⁻¹¹

M = 47.35 10²³ kg

M = 4.7 10²⁴ kg