Bodies Moving under Mutual Attraction

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Out in space, two stars hold each other by gravity and go round and round. We call this a binary star. One star is heavy, the other is light. Now tell me: which one has more kinetic energy? The heavy one, obviously, right? More mass, more energy. Hmm, hold on to that answer. By the end of this video, you may want to change it! And look, this set-up comes up often in JEE gravitation questions. So, let's begin.

0:35The idea: both bodies circle the centre of mass

So, here is our set-up. Two point masses, capital M one and capital M two; the small one and two written below just number the bodies, okay? They are a distance d apart, and nothing else acts on them, only their mutual attraction. Clear? Now tell me. Does the light body go round the heavy one, while the heavy one just sits still? Not quite! Both of them move. Each body moves on a circle about their common centre of mass, written C M for short, okay? Both circles share that centre. And the thing that keeps each body on its circle is gravity itself. The net gravitational force on each body provides its centripetal force, the force towards the centre that circular motion needs. This one equation is the heart of this topic. Got it? One more thing. The two bodies always stay on opposite sides of the C M, on one straight line through it. So they go round together: the same omega, the Greek letter for angular speed, and the same period, capital T, okay?

1:36Where is the centre of mass?

Now, the radii. Call the distances from the C M r one and r two; these are the radii of the circles. The C M balances the masses, so M one r one equals M two r two. And the two radii together make the full gap: r one plus r two equals d. With me? So, solve these two equations. What is r one, in terms of the masses and d? So: r one equals M two d, over M one plus M two, and r two equals M one d, over M one plus M two. Look carefully: each radius has the other body's mass on top. Interesting, right? So the heavier body sits closer to the C M, on the smaller circle, and the lighter body swings round on the bigger circle. Keep this in mind; it will answer our puzzle. Clear?

2:21The key result for two stars

Now, the key result of this concept. For two stars a distance d apart, moving on circular orbits: omega squared equals capital G, times M one plus M two, over d cubed. Capital G here is the gravitational constant. We will prove it in Example two, okay? And the period: capital T equals two pi over omega, which is two pi times the square root of d cubed, over capital G times M one plus M two. The long tick is the square root sign. Right? Look at this formula closely. Only two things decide the period: the total mass and the separation. How the mass is shared between the two stars does not matter at all! Got it? Now a quick trick. For the same two stars, capital T is proportional to d to the power three over two; that symbol like a stretched alpha means is proportional to. So, if we double the separation, does the period double? No, no, careful! It is not two times. Two to the power three over two is two root two; the wavy equals sign means about, so about two point eight times. For ratios, just use the proportion. Perfect!

3:34Example 1: speed and period of two equal masses d apart

Alright, Example one. Read it with me. Two particles, each of mass small m, are a distance d apart, and they move in a circle about their centre of mass. Find their speed and their period. Pause the video, and try it first. Done? Okay, let's check. First, the circle. Each particle goes round the C M, okay? Many students take its radius as d, the distance between them. The radius is d, right? No, no, careful! That is the trap. The masses are equal, so the C M is exactly at the midpoint. Each particle goes round on a circle of radius d over two. Use d in the force, but d over two in the circle. Clear? So, capital G m squared over d squared equals m v squared, over d over two. Cancel one m and solve for v. The double arrow means this gives: v equals the square root of capital G m over two d. Okay? Now the period: one round, divided by the speed. So capital T equals two pi times d over two, over v. Put in v and simplify: capital T equals two pi, times the square root of d cubed over two capital G m. With me? Quick check with the key result: the total mass is two m, so inside the root we get d cubed over capital G times two m. Exactly the same period! Very good!

4:59Example 2: binary star period and kinetic energy ratio

Now, remember our puzzle? Let's finish it. Two stars of masses capital M one and capital M two, a distance d apart, revolve about their centre of mass. Find the period, and the ratio of their kinetic energies. Pause the video, and try it. Done? Okay, let's check. Take star one. Its circle is centred at the C M, so its radius is r one, which is M two d over M one plus M two, just as we found. Right? Now the gravitational force on star one. Which distance goes in it: d, or r one? The distance d, exactly! The force uses the distance between the stars; r one is only the radius. So capital G M one M two over d squared equals M one, omega squared, times M two d over M one plus M two. Okay? Now cancel M one and M two from both sides. The double arrow gives: omega squared equals capital G, times M one plus M two, over d cubed. And so capital T, two pi root d cubed over capital G times M one plus M two: our key result, now proved. That's it! Now the kinetic energies. Both stars have the same omega, and v equals omega r. So v is proportional to r, and r is proportional to one over capital M, because each radius has the other mass on top. Lighter star, bigger circle, higher speed. With me? So now, back to our puzzle. Which star has more kinetic energy, the heavy one or the light one? Kinetic energy, capital K, is half capital M v squared; the dot here just means times. The mass gives one factor of capital M, but v squared gives one over capital M squared. So capital K is proportional to one over capital M, okay? So capital K one to capital K two equals capital M two to capital M one; the colon is read as to. The masses, upside down! The lighter star has more kinetic energy! So if you said the heavy one at the start, that was the trap: the lighter star moves so much faster. Perfect!

7:09Quick revision: force uses d, circle uses radius to CM

So, quick revision. Each body circles the common centre of mass, and the net gravitational force on it is its centripetal force. Use d, the separation, in the force, and the radius to the C M in the circle. And the period is two pi root d cubed, over capital G times the total mass, so capital T is proportional to d to the power three over two. Now go and try the practice problems on this page. All the best!