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Q: 8.20  Two stars each of one solar mass (\small =2 \times 10^3^0\hspace {1mm}kg) are approaching each other for a head on collision. When they are a distance  \small 10^9\hspace{1mm}km, their speeds are negligible. What is the speed with which they collide? The radius of each star is  \small 10^4\hspace{1mm}km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

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Total energy of stars is given by :

                                        E\ =\ \frac{-GMM}{r}\ +\ \frac{1}{2}mv^2

or                                            =\ \frac{-GMM}{r}\ +\ 0

or                                            =\ \frac{-GMM}{r}

Now when starts are just to collide the distance between them is 2R.

The total kinetic energy of both the stars is :  

                                                =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}mv^2\ =\ mv^2

And the total energy of both the stars is :

                                                =\ mv^2\ +\ \frac{-GMM}{2r}

Using conservation of energy we get :

                                       mv^2\ +\ \frac{-GMM}{2r}\ =\ \frac{-GMM}{r}

or                                              v^2\ =\ GM \left ( \frac{-1}{r}\ +\ \frac{1}{2R} \right )

or                                                      =\ 6.67\times 10^{-11}\times 2\times 10^{30} \left ( \frac{-1}{10^{12}}\ +\ \frac{1}{2\times 10^7} \right )

or                                                       =\ 6.67\times 10^{12}

Thus the velocity is  :              \sqrt{6.67\times 10^{12}}\ =\ 2.58\times 10^6\ m/s

Posted by

Devendra Khairwa

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