91Ó°ÊÓ

The given data can be listed below as:

• Weigh of individual on the earth is,$W=675\text{ N}=675\text{ N}×\left(\frac{\text{1 k²µ}}{\text{9.8 N}}\right)=68.9\text{ k²µ}$.
• The diameter of the star is,$D=2r_n=20\text{ k³¾Ã—}\left(\frac{{\text{10}}^{\text{3}}\text{″¾}}{\text{1 k³¾}}\right)={\text{20×10}}^{\text{3}}\text{″¾}$.
• The Mass of the neutron star is, $M_n=M_s=1.99×{10}^{30}\text{ k²µ}$.

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$1.99×{10}^{30}\text{ k²µ}$, $r_n$is the radius of the planet, $g_n$is the acceleration due to gravity and$G$is the gravitational constant with value$6.67×{10}^{−11}{\text{ N.³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$.

According to the question,$D=2r_n$and $M_n=M_s=1.99×{10}^{30}\text{ k²µ}$.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{g}}_{\mathrm{n}}=\frac{6.67×{10}^{−11}{\text{ N.³¾}}^{\text{2}}/{\text{kg}}^2×1.99×{10}^{30}\text{ k²µ}}{{\left({10}^4\right)}^2{\text{m}}^{\text{2}}}\\ =\frac{1.32×{10}^{20}{\text{N.m}}^{\text{2}}/\text{kg}}{{10}^8{\text{″¾}}^{\text{2}}}\\ =1.32×{10}^{12}\text{ N/°ì²µ}\\ =1.32×{10}^{12}\text{ N/°ì²µ}×\text{kg}â‹\dots \text{m/kg}â‹\dots {\text{s}}^{\text{2}}\\ =1.32×{10}^{12}{\text{″¾/s}}^{\text{2}}\end{array}$

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.

$\begin{array}{c}{W}_1=68.9\text{ k²µ}×1.32×{10}^{12}{\text{″¾/s}}^{\text{2}}\\ =9.15×{10}^{13}\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}\\ =9.15×{10}^{13}\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}×\left(\frac{1\text{ N}}{1\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}}\right)\\ =9.15×{10}^{13}\text{N}\end{array}$

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