Neutron Star Collision Homework: Layer Thickness & Gravitational Potential

[egsid]

Homework Statement

Suppose that an earth-mas object collides with a neutron star with radius 10 km and mass 1.4 $M_{sun}$. The material of the earth-like object would wrap around the neutron star and form a thin layer on top of the original neutron star surface. Assume the material gets converted to neutron star material so the density of the layer increases to become the same as the average density of the neutron star.
a) How thick would this layer be?
b) Assuming the object falls from infinity to the neutron star surface, how much gravitational potential energy is released? Compare this to the total nuclear energy released by the Sun during its main sequence lifetime.

Homework Equations

U = -GMm/r

The Attempt at a Solution

So I actually think I can solve part b of the question if I know what the answer to part a is. I would just take the thickness of the layer, add it to 10 km and that would be my r value in the equation above. I know the masses so I could calculate potential energy. The problem though is this thickness in the layer. Please help.

Thanks

 

[ShayanJ]

The volume of the neutron star before collision is $\frac{4}{3}\pi 10^3 km^3 \approx 4\times 10^3 km^3$ so the average density becomes $\frac{1.4}{4 \times 10^3}=3.5 \times 10^{-4} \frac{M_{sun}}{km^3}$.The mass of the Earth like object is approximately $3 \times 10^{-6} M_{sun}$ and the volume of the layer is$\frac{4}{3} \pi [(10+\delta)^3-10^3]\approx 4[\delta^3+30\delta^2+300\delta]$ so we should have $\frac{3\times 10^{-6}}{4[\delta^3+30\delta^2+300\delta]}=3.5 \times 10^{-4}$ from which you can find $\delta$ which is the thickness of the layer.

Anyway,never heard of such collisions.It was interesting.What's your book?

 

[egsid]

Thanks, I pretty much agreed with your answer. We don't have a book; our professor just comes up with problems by himself.

 

[berkeman]

The volume of the neutron star before collision is $\frac{4}{3}\pi 10^3 km^3 \approx 4\times 10^3 km^3$ so the average density becomes $\frac{1.4}{4 \times 10^3}=3.5 \times 10^{-4} \frac{M_{sun}}{km^3}$.The mass of the Earth like object is approximately $3 \times 10^{-6} M_{sun}$ and the volume of the layer is$\frac{4}{3} \pi [(10+\delta)^3-10^3]\approx 4[\delta^3+30\delta^2+300\delta]$ so we should have $\frac{3\times 10^{-6}}{4[\delta^3+30\delta^2+300\delta]}=3.5 \times 10^{-4}$ from which you can find $\delta$ which is the thickness of the layer.

Anyway,never heard of such collisions.It was interesting.What's your book?

Shyan -- please check your PMs. The student must do the bulk of the work on their homework. Please do not do their work for them.