Question

If you weigh 670 n on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 km ? take the mass of the sun to be m s = 1.99×1030 kg , the gravitational constant to be g = 6.67×10−11 n⋅ m 2 /k g 2 , and the acceleration due to gravity at the earth's surface to be g = 9.810 m/ s 2 .

Answer 1

The weight of an object at the surface of a planet is given by:
$F=mg$
where m is the mass and g is the gravitational acceleration, given by
$g= \frac{GM}{r^2}$
where G is the gravitational constant, M is the planet's mass, and r the radius of the planet.

At Earth's surface, $g=9.81 m/s^2$. Since we know the weight of the person at Earth's surface, F=670 N, we can find his mass:
$m= \frac{F}{g}= \frac{670 N}{9.81 m/s^2}=68.3 kg$

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:
$M=25 \cdot 1.99 \cdot 10^{30}kg=4.98 \cdot 10^{31} kg$

While its radius is:
$r= \frac{d}{2}= \frac{25.0 km}{2}=12.5 km = 12500 m$

Therefore, the value of g at the neutron star surface is
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11}Nm^2/kg^2)(4.98 \cdot 10^{31} kg)}{(12500 m)^2}=2.12 \cdot 10^{13} m/s^2$

Therefore, the weight of the person at the surface of the neutron star would be
$F' = mg = (68.3 kg)(2.12 \cdot 10^{13} m/s^2 )=1.45 \cdot 10^{15} N$