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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.

Respuesta :

Answer:

M = 4.7 10²⁴ kg

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

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