Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller diameter. If you weigh $\mathbf{675}\mathbf{}\mathbf{N}$on the earth, what would you weigh at the surface of a neutron star that has the same mass as our sun and a diameter of $\mathbf{20}\mathbf{km}$?

The given data can be listed below as:

• Weigh of individual on the earth is,$W=675\text{\hspace{0.17em}N}=675\text{\hspace{0.17em}N}\times \left(\frac{\text{1\hspace{0.17em}kg}}{\text{9.8\hspace{0.17em}N}}\right)=68.9\text{\hspace{0.17em}kg}$.
• The diameter of the star is,$D=2r_n=20\text{\hspace{0.17em}km×}\left(\frac{{\text{10}}^{\text{3}}\text{\hspace{0.17em}m}}{\text{1\hspace{0.17em}km}}\right)={\text{20×10}}^{\text{3}}\text{\hspace{0.17em}m}$.
• The Mass of the neutron star is, $M_n=M_s=1.99\times {10}^{30}\text{\hspace{0.17em}kg}$.

The acceleration due to gravity is the acceleration acquired by an object because off the gravitational forces. The acceleration due to gravity is directly proportional to the mass of a planet and inversely proportional to the radius of the planet.

The expression for acceleration due to gravity on the surface of a planet star is expressed as:

${\mathrm{g}}_{\mathrm{n}}=\frac{{\mathrm{GM}}_{\mathrm{n}}}{{{\mathrm{r}}_{\mathrm{n}}}^2}$

Here,$M_n$is the mass of the planet with value$1.99\times {10}^{30}\text{\hspace{0.17em}kg}$, $r_n$is the radius of the planet, $g_n$is the acceleration due to gravity and$G$is the gravitational constant with value$6.67\times {10}^{-11}{\text{\hspace{0.17em}N.m}}^{\text{2}}/{\text{kg}}^{\text{2}}$.

According to the question,$D=2r_n$and $M_n=M_s=1.99\times {10}^{30}\text{\hspace{0.17em}kg}$.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{g}}_{\mathrm{n}}=\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}N.m}}^{\text{2}}/{\text{kg}}^2\times 1.99\times {10}^{30}\text{\hspace{0.17em}kg}}{{\left({10}^4\right)}^2{\text{m}}^{\text{2}}}\\ =\frac{1.32\times {10}^{20}{\text{N.m}}^{\text{2}}/\text{kg}}{{10}^8{\text{\hspace{0.17em}m}}^{\text{2}}}\\ =1.32\times {10}^{12}\text{\hspace{0.17em}N/kg}\\ =1.32\times {10}^{12}\text{\hspace{0.17em}N/kg}\times \text{kg}\cdot \text{m/kg}\cdot {\text{s}}^{\text{2}}\\ =1.32\times {10}^{12}{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$

The equation of my weight on the surface of the neutron star is expressed as:

$W_1=W{g}_n$

Here,$W$ ismy weigh on the earth and $g_n$is the acceleration due to gravity.

Substitute the values in the above equation.

$\begin{array}{c}{W}_1=68.9\text{\hspace{0.17em}kg}\times 1.32\times {10}^{12}{\text{\hspace{0.17em}m/s}}^{\text{2}}\\ =9.15\times {10}^{13}\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\\ =9.15\times {10}^{13}\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\times \left(\frac{1\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =9.15\times {10}^{13}\text{N}\end{array}$

Thus, the weight at the surface of the neutron star is $9.15\times {10}^{13}\text{N}$.