Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Two students solve the same JEE Main question: find small g, the acceleration due to gravity, at a height of one thousand six hundred kilometres. Both are quick, and both are confident. One writes four point nine metres per second squared. The other writes six point two seven. Who is right? Hold on to your answer. By the end of this video you will know, and you will know exactly why. This video is your map for the mixed problems of gravitation. So, let's begin.

0:34What JEE Main asks in gravitation

So, first, what does JEE Main actually ask in gravitation? Recent questions are short, and very often they are ratios. Okay? Now think. Capital G, the gravitational constant, sits in almost every formula of this unit. But in a ratio, what happens to it? It cancels, yes! Capital G cancels, and so does everything that stays the same, top and bottom. That is why these questions are short. Right? So, what are the types? Type one: small g at heights and depths. Clear? Type two: escape velocity of a planet with a scaled mass, radius or density. Capital M is mass, capital R is radius, and the Greek letter rho is density. Got it? Type three: satellites, their periods and their energies, together with Kepler's laws. Three types, and you have already learnt all of them in chapters one to four. With me?

1:31How this chapter is built

Now, where are the real exam questions? The JEE Main questions from twenty twenty one to twenty twenty six are at the end of chapters one to four; the dash on screen means to. Okay? This chapter is different. Its problems are original, written in the same style as those real questions. So, fresh practice. Right? There are three sets, each with eight solved problems and three practice problems on the page. Both of today's examples come from set one. Clear? And unless a question says otherwise, take small g as nine point eight metres per second squared, and capital R, the Earth's radius, as six thousand four hundred kilometres. Got it?

2:12Toolkit: g with height and depth

Now, your toolkit, starting with small g. Look at the graph: small g against small r, the distance from the centre of the Earth. So tell me, where is small g the greatest? At the centre? No, no, careful! At the centre, small g is zero. It is greatest at the surface, the peak of the graph, marked g nought. There, small g equals capital G M over R squared, with capital M the mass of the Earth. Okay? Go up to a height small h, along the green curve outside. g h, with a small h written below, equals small g over the bracket, one plus h over R, whole squared. With me? There is also a shortcut: g h is approximately g times one minus two h over R; the wavy equals sign means approximately. But only when h is much less than R; the double less than sign means much less than. Remember this condition, right? And down to a depth small d, along the straight line inside, where small g is proportional to r; the sign like a stretched alpha means proportional to. g d, with a small d written below, equals g times one minus d over R, for a uniform Earth. Clear?

3:26Toolkit: escape and satellites

Next, escape velocity. v e, with a small e written below, equals the square root of two G M over R, or the square root of two g R; for the Earth, that is about eleven point two kilometres per second. Okay? For planets, compare, do not compute. When densities are given, v e is proportional to R times the square root of rho; that is the same stretched alpha sign. Got it? Now, satellites. Take one at a height small h above the Earth. In the orbit formulas, is small r just equal to h? No, no, careful! Small r is measured from the centre of the Earth, so r equals R plus h. This trap hides in many problems of this unit. Right? For a circular orbit, capital T squared, the period squared, is proportional to r cubed. That is Kepler's third law. For an ellipse, use the semi-major axis, small a, instead. With me? And the total energy in a circular orbit: capital E equals minus G M m over two r, with small m the satellite's mass. Always negative. Clear?

4:32Example 1: g at a height of 1600 km, exact formula

Now, back to our puzzle. Example one, an easy one from set one. Find small g at a height of one thousand six hundred kilometres. Take small g as nine point eight metres per second squared, and capital R as six thousand four hundred kilometres, the chapter's values. Pause the video, and try it first. Done? Okay, let's check. The first student used the shortcut: nine point eight, times one minus two times one thousand six hundred over six thousand four hundred. That gives four point nine. But is the shortcut allowed here? No, no, careful! Here h over R is one thousand six hundred over six thousand four hundred, which is zero point two five. That is not much less than one, so no shortcut. Right? Use the full formula. One plus h over R is one point two five. So g h equals g over one point two five squared, which is zero point six four times g. Got it? Zero point six four times nine point eight: about six point two seven metres per second squared. So the second student was right. Puzzle solved! The four point nine came from a shortcut used outside its condition.

5:43Example 2: speed of a body falling from height 3R

Okay? Now a harder one, a medium problem, also from set one. Here it is: a body is released from rest at a height three R above the Earth's surface. Find its speed on reaching the surface. Pause the video, and try it. Hint: think energy. Done? Okay, let's check. It falls from rest, so v squared equals two g h, with h equal to three times capital R, right? No, no, wait! That formula needs small g to stay constant all the way down, which is fine only when h is small compared with capital R, okay? Over a height of three R, small g changes a lot, so we use energy conservation instead. Clear? First, where does the body start? Height three R above the surface, so its distance from the centre is R plus three R: four R, not three R, right? Now, kinetic energy gained equals potential energy lost, after dividing out the body's mass, small m. One half v squared equals G M times, one over R minus one over four R, which is three quarters of G M over R, with me? Now the trick from chapter two: G M equals g R squared. So one half v squared is three quarters g R; the double arrow means so, and v equals the square root of three halves g R, okay? Numbers, with R in metres: one point five times nine point eight times six point four times ten to the power six, under the root, in metres per second. About nine point seven kilometres per second. One energy line, and done. That's it!

7:19Quick revision: r = R + h, shortcut limits, use energy

So, quick revision. Height h means r equals R plus h, and use the shortcut for small g only if h is much less than capital R, okay? In a ratio, capital G and everything unchanged cancel. And over large heights, use energy, not constant g formulas. Now go and try the three sets of this chapter. With this toolkit, you are ready for them. All the best!