Question

Two stars each of one solar mass $(=2\times {{10}^{30}}kg)$ are approaching each other for a head on collision. When they are at distance ${{10}^{9}}km$, their speeds are negligible. What is the speed with which they collide? The radius of each star is ${{10}^{4}}km$. Assume the stars to remain undistorted until they collide. (Use the known value of G)

Answer 1

Hint: Using the law of conservation of energy the total energy of the system (Kinetic energyPotential energy) remains conserved until external force does work on the system.
The total energy of the system is the sum of kinetic energy and the potential energy.
Use the concept of conservation of momentum.

Complete step by step solution:
It is given that mass of each star is$M=2\times {{10}^{30}}kg$
Radius of each star is$R={{10}^{4}}km$
We know that$1km=1000m$
Hence, the radius is$R={{10}^{4}}\times 1000m={{10}^{7}}m$
Initial distance between the centers of the two stars is$r={{10}^{9}}km={{10}^{12}}m$
As initially the two stars started moving toward each other, therefore initially the total energy of the system consisted of only gravitational potential energy.
Let initial total energy is${{E}_{i}}$
$\begin{align}
  & {{E}_{i}}={{\left( KE \right)}_{i}}+{{\left( PE \right)}_{i}} \\
 & =0+\left( -\dfrac{GMM}{{{r}^{2}}} \right) \\
 & =\left( -\dfrac{G{{M}^{2}}}{{{r}^{2}}} \right)\ldots \ldots \left( i \right)
\end{align}$
Let the velocities of the stars when they collide is${{v}_{1}}$and${{v}_{2}}$
Using conservation of momentum,
Total initial momentum of the system$=$total final momentum of the system
$\begin{align}
  & M\left( 0 \right)+M\left( 0 \right)=M{{v}_{1}}+M{{v}_{2}} \\
 & \left| {{v}_{1}} \right|=\left| {{v}_{2}} \right|=v
\end{align}$
Final energy of the system is${{E}_{f}}$
${{E}_{f}}={{\left( KE \right)}_{f}}+{{\left( PE \right)}_{f}}$
When they collide then the distance between their centers is $2R$
$\begin{align}
  & {{E}_{f}}=\dfrac{M{{v}^{2}}}{2}+\dfrac{M{{v}^{2}}}{2}+\left( -\dfrac{G{{M}^{2}}}{{{\left( 2R \right)}^{2}}} \right) \\
 & =M{{v}^{2}}+\left( -\dfrac{G{{M}^{2}}}{4{{R}^{2}}} \right)\ldots \ldots \left( ii \right)
\end{align}$
From equation$\left( i \right)$and$\left( ii \right)$
\[\begin{align}
  & M{{v}^{2}}+\left( -\dfrac{G{{M}^{2}}}{2R} \right)=-\dfrac{G{{M}^{2}}}{r} \\
 & {{v}^{2}}=\dfrac{GM}{2R}-\dfrac{GM}{r} \\
 & v=\sqrt{\dfrac{GM}{2R}-\dfrac{GM}{r}} \\
 & =\sqrt{6.67\times {{10}^{-11}}\times 2\times {{10}^{30}}\left( \dfrac{1}{2\times {{10}^{7}}}-\dfrac{1}{{{10}^{12}}} \right)} \\
 & =2.58\times {{10}^{6}}m/s
\end{align}\]

Hence, the velocity of the star when they collide is \[2.58\times {{10}^{6}}m/s\].

Note: We use conservation of total energy when the system is isolated.
To find the gravitational potential energy we take distance as center to center distance.