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

Two stars each of one solar mass (= 2 × 10³⁰ kg) are approaching each other for a head on collision. When they

Exercises · 7.20

= 2 \times 10^{30}\text{ kg}

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Concept understanding — Conservation Of Mechanical Energy

Conservation of Mechanical Energy

The Intuition First

Imagine you're holding a heavy stone at shoulder height. Your arm is tired. That stone has potential energy — energy stored because of its position. Now let it go. As it falls, it speeds up. The potential energy is turning into kinetic energy — the energy of motion. Just before it hits the ground, all the original potential energy has become kinetic energy.

Now imagine the reverse: you throw a ball straight up. It leaves your hand fast (lots of kinetic energy), rises, slows down, stops for an instant at the top (zero kinetic energy), then falls back. At the top, all the kinetic energy you gave it has turned back into potential energy.

This back-and-forth transformation — potential ↔ kinetic — is the heart of the idea. Energy doesn't disappear; it just changes form. That's conservation.

The Precise Statement

Important

Conservation of Mechanical Energy: In an isolated system where only conservative forces (like gravity or an ideal spring) do work, the total mechanical energy of the system remains constant.

Total mechanical energy is the sum of kinetic energy ( $K$ ) and potential energy ( $U$ ):

E_{\text{mech}} = K + U

The law says:

K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}

Or, in symbols:

E_{\text{mech, initial}} = E_{\text{mech, final}}

What This Means in Practice

Let's go back to the falling stone. Suppose you hold it 5 metres above the ground. Its mass is 2 kg. Take $g = 10\ \text{m/s}^2$ .

At the top (initial):

$K_i = 0$ (not moving)

$U_i = mgh = 2 \times 10 \times 5 = 100\ \text{J}$

$E_{\text{mech}} = 0 + 100 = 100\ \text{J}$
Just before hitting ground (final):

$U_f = 0$ (height = 0)

$K_f = \frac{1}{2}mv^2$

Conservation says $K_f = 100\ \text{J}$ , so $\frac{1}{2} \times 2 \times v^2 = 100$ , giving $v = 10\ \text{m/s}$ .

You never needed to know the time of fall or acceleration. Energy conservation gave you the speed directly.

The Two Critical Conditions

Watch out

Mechanical energy is not always conserved. It is conserved only when:

No non-conservative forces (like friction, air resistance, or applied pushes/pulls) do work.
The system is isolated — no external forces transfer energy in or out.

If friction is present, some mechanical energy turns into heat (thermal energy). The total energy of the universe is still conserved, but mechanical energy alone is not.

A Simple Example to Cement It …

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Applying conservation of mechanical energy to the two-star system, with the stars starting essentially at rest at a huge separation and gravity doing all the work as they fall together, gives a collision speed of about $2.58 \times 10^6\ \text{m/s}$ for each star.

Only gravity acts on the two stars (they are isolated, and gravity is a conservative force), so the total mechanical energy — kinetic plus gravitational potential — of the two-star system stays constant from the moment they start ("at rest," for practical purposes) until the instant their surfaces touch.

Setting up the problem

Let each star have mass $M = 2 \times 10^{30}\ \text{kg}$ (one solar mass) and radius $R = 10^4\ \text{km} = 10^7\ \text{m}$ .

Initial state: centre-to-centre separation $r_i = 10^9\ \text{km} = 10^{12}\ \text{m}$ , speeds negligible, so initial kinetic energy $K_i \approx 0$ .

Final state (just before collision): the stars are undistorted spheres, so they touch when their surfaces meet — that is, when the centres are separated by

r_f = R + R = 2R = 2\times10^7\ \text{m}

Watch out

A common mistake is to take the final separation as zero (treating the stars as points). Because the stars have a real, finite radius and "remain undistorted until they collide," the collision happens at $r_f = 2R$ , not $r_f = 0$ .

The gravitational potential energy of the two-star system at separation $r$ is

U(r) = -\frac{GM^2}{r}

(taking $U \to 0$ as $r \to \infty$ ).

By symmetry (equal masses, starting from rest, pulled together by a mutual force along the line joining them), the two stars always move with equal speed $v$ in opposite directions about their common centre of mass. So the total kinetic energy just before collision is

K_f = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2

Applying conservation of energy

K_i + U(r_i) = K_f + U(r_f)

0 - \frac{GM^2}{r_i} = Mv^2 - \frac{GM^2}{r_f}

Solving for $v^2$ :

Mv^2 = \frac{GM^2}{r_f} - \frac{GM^2}{r_i} = GM^2\left(\frac{1}{r_f}-\frac{1}{r_i}\right)

v^2 = GM\left(\frac{1}{r_f}-\frac{1}{r_i}\right)

Substituting numbers

\frac{1}{r_f} = \frac{1}{2\times10^7} = 5\times10^{-8}\ \text{m}^{-1}, \qquad \frac{1}{r_i} = \frac{1}{10^{12}} = 10^{-12}\ \text{m}^{-1}

Since $10^{-12}$ is about $40{,}000$ times smaller than $5\times10^{-8}$ , the initial separation contributes almost nothing — practically all the kinetic energy at collision comes from the last stretch of the fall, close to $r_f$ :

\frac{1}{r_f}-\frac{1}{r_i} \approx 4.999\times10^{-8}\ \text{m}^{-1}

Now compute $GM$ : …

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Setting up the problem

Applying conservation of energy

Substituting numbers

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