Question

If you weigh 660 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 ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s2 .

Answer 1

Answer: $W_{n}=5.724 (10)^{13})N$

Explanation:

The weight $W$ of a body or object is given by the following formula:

$W=m.g$ (1)

From here we can find the mass of the body:

$m=\frac{W}{g}=67.346 kg$ (2)

Where $m$ is the mass of the body and $g$ is the acceleration due gravity in an especific place (in this case the earth).

In the case of a neutron star, the weight $W_{n}$ is:

$W_{n}=m.g_{n}$ (3)

Where $g_{n}$ is the acceleration due gravity in the neutron star.

Now, the acceleration due gravity (free-fall acceleration) $g$ of a body is given by the following formula:

$g=\frac{GM}{r^{2}}$ (4)

Where:

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$M$ the mass of the body (the neutron star in this case)

$r$ is the distance from the center of mass of the body to its surface. Assuming the neutron star is a sphere with a diameter $d=24km$, its radius is $r=\frac{d}{2}=12.5km=12500m$

Substituting (4) and (2) in (3):

$W_{n}=m(\frac{GM}{r^{2}})$ (5)

$W_{n}=(67.346kg)(\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.99(10)^{30}kg)}{(12500m)^{2}})$ (6)

Finally:

$W_{n}=5.724 (10)^{13})N$