Potential of a Shell and of a Solid Sphere

Transcript

0:00Introduction

Hello everyone! Okay, a puzzle to start. Imagine a smooth tunnel drilled straight through the Earth, along a diameter, right through the centre. You drop a stone into it, from rest. Now, at the centre of the Earth, gravity pulls the stone with zero force. So how fast is the stone moving there? Zero? Or very fast? Hold on to your answer. To solve this, we need the potential inside a sphere, and that is today's topic. Questions on it come up often in JEE gravitation. So, let's begin.

0:34Thin shell: constant potential inside

So, look at the graph first. The vertical axis is capital V, the gravitational potential, and the horizontal axis is small r, the distance from the centre. The mark at capital R is the radius. And first, a thin shell: all its mass, capital M, sits on a hollow sphere of radius capital R, okay? Outside the shell, r is greater than R; the sign opening to the left means greater than. From chapter one, a shell acts on an outside point as if all its mass were at its centre. So V equals minus capital G M over small r, just like a point mass, with capital G the gravitational constant. Clear? Now step inside the shell. From chapter one, the field E inside a thin shell is zero. So the potential inside is also zero, right? No, no, careful! This is the classic trap. From the last concept, E equals minus d V over d r, the rate at which V changes with r. So zero field only means V does not change; the double arrow means so. Constant, not zero! Got it? And which constant? The value on the shell itself. So inside, for r less than R, the sign pointing the other way, V equals minus capital G M over capital R, the same at every inside point, even the centre. On the graph, that is the dashed line, marked shell: constant, okay?

1:58Uniform solid sphere: inside and outside

Now, a uniform solid sphere: mass spread evenly through the whole ball, like the Earth if we treat it as uniform. Outside, the same story. For r greater than R, V equals minus G M over r, as before. On the graph, the right-hand part is common to both. Right? Inside, the formula changes. V equals minus capital G M, over two R cubed, times the bracket, three R squared minus small r squared. The little three on R means cubed, and the little two means squared. On the graph, it is the solid curve, marked solid sphere. Clear? Now, a topper's habit. Check any inside formula at the surface. Tell me, what must it give at r equals capital R? The outside value, exactly! Three R squared minus R squared is two R squared. Two R squared over two R cubed is one over R, so we get minus G M over R, the same as outside. The two formulas meet at the surface. With me? Now tell me: for the solid sphere, where is the potential lowest, at the surface, or at the centre? At the centre, yes! Put small r equal to zero. The bracket becomes three R squared, and three R squared over two R cubed is three over two R, right? So V at the centre is minus three G M over two R, the lowest label on the graph. Clear? That is one point five times the surface value; the cross means times. More negative means lower, so the potential is lowest at the centre. Got it? But look at the field E, from chapter one: zero at the centre, and greatest at the surface. Lowest potential and zero field, at the same point! Remember this one; our tunnel puzzle needs it.

3:45Example 1: potential at R/2 inside a solid sphere

Example one, from your course. Find the potential at a distance R over two from the centre of a uniform solid sphere. Pause the video, and try it first. Done? Okay, let's check. Quick way: minus G M over r, with r equals R over two, gives minus two G M over R, and we are done, right? No, no, careful! Minus G M over r is only for points outside. Here r is R over two, which is less than R, so the point is inside the sphere. We must use the inside formula. Clear? So, small r squared is R over two, the whole squared, which is R squared over four. Three R squared minus R squared over four is eleven R squared over four. Okay? Now multiply by minus G M over two R cubed; the dot means times. R squared over R cubed leaves one over R, and two times four is eight. So V equals minus eleven G M over eight R, and that's it! Quick check. On the same scale, the surface value is minus eight G M over eight R, and the centre value is minus twelve G M over eight R, right? Eleven sits between eight and twelve, just as it should. Got it?

5:02Example 2: speed at the centre of a tunnel through Earth

Now, remember our tunnel puzzle? Let's finish it. Example two, from your course. Here, a body is dropped from rest into a smooth tunnel along a diameter of the Earth. Find its speed at the centre. Pause the video, and try it. Done? Okay, let's check. Treat the Earth as a uniform solid sphere, and smooth means no friction. V surface, with the word surface written small below, is minus G M over R, and V centre is minus three G M over two R, from the formula we just saw. Right? Now think. As the body falls in, where is the potential higher: at the surface, or at the centre? At the surface, correct! It is less negative. So the potential energy falls by small m times, V surface minus V centre, where small m is the body's mass. Minus G M over R, plus three G M over two R, gives G M over two R, so the loss is G M m over two R, as you see on screen. Clear? Now a topper's trick for Earth questions. Small g equals G M over R squared, so G M equals g R squared. Then G M m over two R becomes one half m g R, which is much easier to use, okay? This lost potential energy becomes kinetic energy. So, one half m v squared equals one half m g R, and we cancel one half m from both sides. Then v equals the square root of g R, nice and simple! The mass cancels, so any body, heavy or light, gets the same speed. With me? Now the course values: g is nine point eight, and R is six point four times ten to the power six metres. m s to the power minus one means metres per second, and the wavy equals sign means approximately. So v is about seven point nine kilometres per second! At the centre, the pull is zero, but the body is moving very fast. Zero force does not mean zero speed. Puzzle solved!

7:02Quick revision: shell and sphere formulas, zero field trap

So, quick revision. Shell: minus G M over capital R, constant everywhere inside, and minus G M over small r outside. Uniform solid sphere, inside: minus G M over two R cubed, times three R squared minus r squared, and at the centre, minus three G M over two R, the lowest point. And zero field means constant potential, not zero potential. Now go and try the practice problems on this page. All the best!