Question

A neutron star has a mass about \(1.5\) times that of the sun; its radius is \(10 [\text{km}]\). A mass, m, falls onto the neutron star from a very large distance (\(1000 [\text{km}]\)). What is the approximate speed at which the mass reaches the surface of the neutron star? Relativistic effects should be ignored here.
You might need to use some of the following constants:
\[R_\odot=7.0\times10^8[\text m]\\
G=6.67\times10^{−11}[\text m^3/\text{kg}\cdot\text s^2] \\
M_\odot=1.9891\times10^{30}[\text{kg}]\]

\[
•\quad1\times10^8 [\text {m}/\text s]\\
•\quad1.4\times10^8 [\text {m}/\text s]\\
•\quad2\times10^8 [\text {m}/\text s]\\
•\quad2.8\times10^8 [\text {m}/\text s]\\
•\quad4\times10^8 [\text {m}/\text s]
\]

Answer 1

\[U_\text{max}=G\frac{mM}R\\KE_\text{max}=\tfrac12mv^2\]
\[mv^2=G\frac{mM}R\\\quad v=\sqrt{\frac{2GM}R}\]

Answer 2

\[U_\text{max}=G\frac{mM}r\\KE_\text{max}=\tfrac12mv^2\]
\[mv^2=G\frac{mM}r\\\quad v=\sqrt{\frac{2GM}r}\]

Answer 3

\[v=\sqrt{\frac{2\times6.67\times10^{-11}\left[\text {m}^3/\text{kg}\cdot\text s^2\right]\times1.5\times1.9891\times10^{30}[\text{kg}]}{1000[\text {km}]}}\\
\quad=\sqrt{2\times6.67\times1.5\times1.9891\times10^{-11+30-6}\left[\text {m}^2\cdot\text s^2\right]}\\\quad\approx\sqrt{39.8\times10^{13}}\left[\text {m}/\text s\right]\\\quad\approx2\times10^7\left[\text {m}/\text s\right]\]

Answer 4

My result isn't close enough to any of the options, where am i going wrong?

Answer 5

*typo\[\tfrac12mv^2=G\frac{mM}R\]

Answer 6

see attached, the full answer is 198504085.83 or 2x10^8