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

`v=516526.9m/s`

Step-by-step 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=(GMm)/(r^2)`

Where
`G=6.67*10^(-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=(GMm)/(r^2)`

Which means:

`a=(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=(2GMd)/(r^2)`

`v=\sqrt{(2GMd)/(r^2)}`

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with
`r=5*10^3m`. With our values then we have:

`v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s`