The weight of an object at the surface of a planet is given by:
[tex]F=mg[/tex]
where m is the mass and g is the gravitational acceleration, given by
[tex]g= \frac{GM}{r^2} [/tex]
where G is the gravitational constant, M is the planet's mass, and r the radius of the planet.
At Earth's surface, [tex]g=9.81 m/s^2[/tex]. Since we know the weight of the person at Earth's surface, F=670 N, we can find his mass:
[tex]m= \frac{F}{g}= \frac{670 N}{9.81 m/s^2}=68.3 kg [/tex]
Now we have to calculate the value of g at the surface of the neutron star. The mass of the neutron star is 25 times the Sun's mass:
[tex]M=25 \cdot 1.99 \cdot 10^{30}kg=4.98 \cdot 10^{31} kg[/tex]
While its radius is:
[tex]r= \frac{d}{2}= \frac{25.0 km}{2}=12.5 km = 12500 m [/tex]
Therefore, the value of g at the neutron star surface is
[tex]g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11}Nm^2/kg^2)(4.98 \cdot 10^{31} kg)}{(12500 m)^2}=2.12 \cdot 10^{13} m/s^2 [/tex]
Therefore, the weight of the person at the surface of the neutron star would be
[tex]F' = mg = (68.3 kg)(2.12 \cdot 10^{13} m/s^2 )=1.45 \cdot 10^{15} N[/tex]
