91Ó°ÊÓ

The separation of two neutron stars is${\text{1.0×10}}^{\text{10}}\text{m}$

The mass of each Neutron star is$1.0×{10}^{30}\text{kg}$

The radius of each star is$1.0×{10}^5\text{m}$

The momentum of the system is conserved because both stars have the same mass, same speed, and same kinetic energy. Therefore, using the principle of conservation of energy, find the initial and final potential and kinetic energy of the system. Hence, find the speed of two neutron stars when their separation decreases to one-half its initial value and when they are about to collide.According to the law ofconservation of energy, energy can neither be created nor be destroyed.

The formula is as follows:

$U_i+K_i=U_f+K_f$

where U is potential energy and K is kinetic energy.

Now,

$$\begin{gathered} U_i+K_i=U_f+K_f \\ {U}_i=-\frac{G{M}^2}{r_i} \end{gathered}$$

${\text{r}}_{\text{i}}$ is their initial center-to-center separation.

Initially, both stars are at rest,${\text{K}}_{\text{i}}=0$.

As the final separation is$\frac{{\text{r}}_{\text{i}}}{2}$.

$U_f=-\frac{2G{M}^2}{r_i}$

As each star has,

$$\begin{gathered} K=\frac{1}{2}M{v}^2 \\ {K}_f=M{v}^2 \\ -\frac{G{M}^2}{r_i}=-\frac{2G{M}^2}{r_i}+M{v}^2 \\ M{v}^2=\frac{G{M}^2}{r_i} \\ v=\sqrt{\frac{GM}{r_i}} \end{gathered}$$

As

$\begin{array}{rcl}{r}_i& =& 1.0×{10}^{10}\text{m},andM=1.0×{10}^{30}\text{kg}\\ v& =& \sqrt{\frac{{\displaystyle \left(\text{6.67}×{\text{10}}^{\text{- 11}}\frac{{\displaystyle {\text{m}}^{\text{3}}}}{{\displaystyle {\text{s}}^{\text{2}}·\text{kg}}}\right)\left(\text{1.0}×{\text{10}}^{\text{30}}\text{kg}\right)}}{{\displaystyle \text{1.0}×{\text{10}}^{\text{10}}\text{m}}}}\\ & =& 8.2×{10}^{\text{4}}\text{m/s}\\ & & \\ & & \end{array}$

Therefore, the speed of two neutron stars when their separation has decreased to one-half its initial value is $\text{8.2}×{\text{10}}^{\text{4}}\text{m/s}$.

Now,

$U_i+K_i=U_f+K_f$

As, the separation between the centers is $r_i=2R=2×{10}^{\text{5}}\text{m}$, where R is the radius of each star,

$U_f=-\frac{G{M}^2}{r_i}$

Therefore, according to energy conservation law,

$$\begin{gathered} -\frac{G{M}^2}{r_i}=-\frac{G{M}^2}{r_f}+M{v}^2 \\ M{v}^2=-\frac{G{M}^2}{r_i}+-\frac{G{M}^2}{r_f} \\ M{v}^2=G{M}^2\left(\frac{1}{r_f}-\frac{1}{r_i}\right) \end{gathered}$$

$\begin{array}{rcl}v& =& \sqrt{\left(\text{6.67}×{\text{10}}^{\text{- 11}}\frac{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{2}}·\text{kg}}\right)\left(\text{1.0}×{\text{10}}^{\text{30}}\text{kg}\right)\left(\frac{\text{1}}{\text{2}×{\text{10}}^{\text{5}}\text{m}}-\frac{\text{1}}{{\text{10}}^{\text{10}}\text{m}}\right)}\\ & =& 1.8×{10}^7\text{m/s}\\ & & \end{array}$

Hence, the speed when they are about to collide is .$1.8×{10}^7\text{m/s}$

Therefore, using the law of conservation of energy and the formula for gravitational potential and kinetic energy, the speed of stars can be found.