Problems

#03

The gravitational acceleration and escape speed equations are $g=GM/R^2$ and $v_{esc} = \sqrt{2GM/R}$, for $M$ the object mass and $R$ its spherical radius. Here we have a $1.4 M_\odot$ neutron star with a radius of 10 km, or $10^4$ m. Converting to kg, the mass is $2.8\times 10^{30}$ kg. The gravitation acceleration is then $g=6.67\times 10^{-11} \times 2.8\times 10^{30}/(10^4)^2 = 1.9\times 10^{12}$ m/s$^2$. The escape speed is $v_{esc} = 190,000$ km/s (notice that I did the unit conversion from m to km). If the radius were 4 km, the escape speed goes up by the factor of $\sqrt{10/4} = 1.6$, giving a value of just about the speed of light itself. This suggests that 4 km is the smallest size for a neutron star of this mass.

#07

Speed is distance over time. In this case we need the circumference divided by the rotation period. At 1000 times/sec, the period must be 0.001 sec. The circumference is $2\pi r$, and we are told $r=10$ km, so the circumference is about 60,000 m. Dividing these two gives 60 million m/s, or 60,000 km/s, which is about 20% of the speed of light (or $0.2c$). The circular orbit speed is $v_c =\sqrt{GM/R} =\sqrt{6.67\times 10^{-11} \times 2.8\times 10^{30}/10^4}=136,000$ km/s, which is about $0.45c$.