Potential and Potential Energy

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. You carry a one kilogram bag from the Earth's surface all the way out to infinity. An endless journey, so the work must be infinite too, right? Hmm, we will see. And a second puzzle: in gravitation, potential energy comes out negative. Energy less than zero? We will settle both today. These ideas come up often in JEE gravitation. So, let's begin.

0:29Potential: work per unit mass, from infinity

So, a new word: potential. Take a small test mass far away at infinity; the sideways eight on screen is the infinity sign. An external agent, say your hand, brings it slowly to a point, so it picks up no speed, right? The agent's work per unit mass is the gravitational potential there; we call it capital V, okay? Now think. The big mass pulls the test mass towards itself. So does your hand push it in, or hold it back? Is that work positive or negative? Hold it back, yes, exactly! Gravity pulls it in, so to keep it slow, the agent pulls against the motion. Force opposite to displacement means negative work. With me? So, for a point mass capital M, the potential at a distance small r is minus capital G M over r. Capital G is the gravitational constant, and the minus sign is our negative work. The unit in brackets is joules per kilogram; the J is joules, and kilogram to the power minus one means per kilogram. Okay? Now put a second mass, small m, at that point. The potential energy of the pair, capital U, is minus capital G M m over r, and it is zero at infinite separation. Second puzzle solved: zero is at infinity, and attracting masses always sit below that zero. Right?

1:49From potential to field

Now, potential and field are closely linked. The field capital E, the force per unit mass from chapter one, equals minus d V over d r: the rate of change of potential with distance, with a minus sign. Clear? Let's test it on our point mass. Differentiate minus capital G M over r with respect to r, and put a minus sign in front. What do you get? Minus capital G M over r squared, correct! The double arrow means implies. Its size is the same field we found in chapter one, and the minus sign means it points towards the mass. Okay? In two or three dimensions, do the same along each axis. E x, with a small x written below, equals minus the partial derivative of V with respect to x. The curly d means partial: differentiate in x, keeping y and z fixed. Right? And from field back to potential: V B minus V A, the small letters below naming the points, equals minus the integral from A to B of E dot d r. The long S is the integral sign, the arrows mean vectors, and the dot is the dot product.

2:58Potential energy of a system

Got it? Now, more than two masses? Simple: take every pair. Masses m i and m j, a distance r i j apart, give minus capital G, m i, m j, over r i j; the small i and j below are just labels. Add this over every pair, and that is the potential energy of the system. Okay? Now tell me. How much work must you do to pull the system completely apart, every mass out to infinity? Look at the end first. With every mass at infinity, every pair term is zero, so U final, with the word final written below, is zero. With me? So the work is final minus initial: zero minus U, which is minus U, a positive amount. U is negative, so you must supply energy to break the system up. Clear?

3:48Raising a mass to height h

Now, lift a mass, small m, from the Earth's surface to a height h. We treat the Earth as a spherically symmetric body, of mass capital M and radius capital R, okay? The distance from the centre goes from R to R plus h. Delta U, the little triangle, means the change: final minus initial. So it is minus capital G M m over R plus h, minus, minus capital G M m over R, in brackets. Simplify, and it is capital G M m h, over R times, R plus h. With me? But questions usually give small g instead. So, from the last chapter, what is capital G M in terms of small g and R? Capital G M equals g R squared, yes! Put it in, cancel one R, then divide the top and the bottom by R, and delta U equals m g h, over one plus h over R, okay? This is exact, for any height. Right? Now if h is much less than R, h over R is nearly zero. The double less-than sign means much less than, and the wavy equals means approximately. So delta U is about m g h, the formula you already know! Got it? Now a quick one. You lift the mass to a height equal to the Earth's radius. The work is just m g R, right? No, no, wait! That is the trap. Here h is not small, so use the exact formula: one plus h over R is one plus one, two. So delta U is half of m g R, not the full m g R you guessed. Clear?

5:19Example 1: work to take 1 kg from Earth to infinity

Now, remember our first puzzle? Let's finish it. How much work is needed to take one kilogram from the Earth's surface to infinity? Pause the video, and try it first. Done? Okay, let's check. The distance is infinite. So the work is infinite too, right? No, no! Use energy. The work W is final minus initial potential energy. At infinity it is zero; at the surface it is minus capital G M times one, over R, okay? So W equals capital G M over R, for our one kilogram. Capital G M is g R squared, so this is just g times R, simple. Small g is nine point eight, and R is six point four times ten to the power six. Right? Multiply: W is about six point three times ten to the power seven joules. Sixty three million joules. Large, but finite! Mystery solved: the pull keeps getting weaker as you go out, so the total work stays finite.

6:21Example 2: potential energy of four masses on a square

Now a harder one, JEE style. Four particles, each of mass small m, sit at the corners of a square of side small a. Find the potential energy. Pause the video and try it. Hint: count the pairs carefully. Done? Okay, let's check. Four masses, four sides of the square, so four pairs, right? Oh no, careful! This is the trap. Every pair, remember? Opposite corners count too. So four sides of length a, and two diagonals of root two times a, by Pythagoras; the tick-shaped sign is the square root. Six pairs in all. With me? Now add. Four side pairs, each minus capital G m squared over a. Two diagonal pairs, each minus capital G m squared, over root two times a. Take out minus capital G m squared over a, and the bracket is four, plus two over root two. Okay? Two over root two is just root two. So U equals minus capital G m squared over a, times, four plus root two. Never forget the diagonals! That's it!

7:30Quick revision: V, U, systems, raising a mass to height h

So, quick revision. Potential and potential energy, minus capital G M over r and minus capital G M m over r, are both zero at infinity. Field is minus d V over d r; for a system, add over every pair, and separating it takes minus U, right? Raising a mass to height h takes m g h over one plus h over R; about m g h only when h is much less than R, okay? Now go and try the practice problems on this page. All the best!