Question

A binary star system has two stars, each with the same mass as our sun, separated by $1.0*{10}^{12}m.$A comet is very far away and essentially at rest. Slowly but surely, gravity pulls the comet toward the stars. Suppose the comet travels along a trajectory that passes through the midpoint between the two stars. What is the comet’s speed at the midpoint?

Short answer

The comet’s speed at the midpoint is$32.57\times {10}^{3}m/s.$

Step-by-step solution

Given information 

A binary star system has two stars, each with the same mass as our sun, separated by$1\times {10}^{12}m.$

Calculation

When the comet is far away from the binary system, there is no kinetic and potential energy. Therefore, the initial kinetic and potential energy of the system is zero.

In the final state, the comet passes through the midpoint of the binary star system, it will have both kinetic energy and potential energy

Now, one can find the velocity of the comet in the midpoint of the binary stars system using the conservation of energy principle.

Initial energy = Final energy.

$$\begin{gathered} 0=K{E}_{\text{final}}+P{E}_{\text{final}} \\ 0=\frac{m{v}^2}{2}-\frac{GMm}{r}-\frac{GMm}{r} \\ \frac{2GMm}{r}=\frac{m{v}^2}{2} \\ {v}^2=\frac{4GM}{r} \\ v=\sqrt{\frac{4GM}{r}} \end{gathered}$$

Here 'r' is half the distance between the stars, so let's take it as 'd/2', then,

$v=\sqrt{\frac{8GM}{d}}$

Where 'v' is velocity, 'G' is the universal gravitational constant, and 'M' is the mass of the sun and 'd' is the distance between the stars.

Continuation of calculation 

Substituting G, M, d values in the above equation and solving for required v we get,

$$\begin{gathered} v=\sqrt{\frac{8GM}{d}} \\ v=\sqrt{\frac{8(6.67\times {10}^{-11}N{m}^2/k{g}^2)(1.989\times {10}^{30}kg)}{(1.0\times {10}^{12}m)}} \\ v=32.57\times {10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$

Final answer

At a speed of $32.57\times {10}^3\mathrm{m}/\mathrm{s},$the comet travels towards the midpoint of the binary star system.