Question

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answer 1

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=\frac{GMm}{r^2}$

Where $G=6.67\times10^{-11}Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=\frac{GMm}{r^2}$

Which means:

$a=\frac{GM}{r^2}$

The object departs from rest ($v_0=0m/s$) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$, which for our case will be:

$v^2=2ad=\frac{2GMd}{r^2}$

$v=\sqrt{\frac{2GMd}{r^2}}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5\times10^3m$. With our values then we have:

$v=\sqrt{\frac{2GMd}{r^2}}=\sqrt{\frac{2(6.67\times10^{-11}Nm^2/Kg^2)(2\times10^{30}Kg)(0.025m)}{(5\times10^3m)^2}}=516526.9m/s$