Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle to start. You climb straight up, high above the Earth, until your weight drops by thirty six percent. How high are you now? Many students grab a short formula here, and fall straight into a trap. We will settle it in example one. This video is Part A of the unit problem bank, JEE Main level. Your course says: attempt Part A like a thirty minute test, after chapters one to five. Today we solve four of these problems together, and catch the traps on the way. So, let's begin.

0:36Idea 1: orbital speed at a height

So, here is idea one. Read it with me: find the orbital speed at a height three R above the Earth, where capital R is the Earth's radius. Okay? For a circular orbit, the orbital speed v o, with a small o written below, is the square root of capital G M over small r; the tick-shaped sign is the square root. Capital G is the gravitational constant, capital M is the Earth's mass, and small r is the radius of the orbit, measured from the Earth's centre. Got it? Now, the height is three R, okay? So we simply put small r equal to three R, right? No, no, careful! Height is measured from the surface, but small r from the centre. So small r is R plus three R, which is four R, not three. Your course lists this as a trap: height h always means small r equals R plus h. Clear? Now, the question gives no capital G or capital M, but we know small g, the acceleration due to gravity at the surface. So we use G M equals g R squared; the double arrow means so. Put it in: g R squared over four R, which is g R over four, under the root. With me? Now the numbers: small g is nine point eight metres per second squared, and capital R is six point four times ten to the power six metres. Take g R over four, then the root: about three point nine six kilometres per second. The wavy equals sign means approximately, and s to the minus one means per second. That's it!

2:14Idea 2: the period at a height

Now, idea two. Read it with me: a near-surface satellite has a period of eighty four point six minutes. Find the period at a height capital R, okay? The period, capital T, is two pi times the root of r cubed over G M; pi is the Greek letter. Two pi and G M stay fixed, so capital T is proportional to r to the power three over two; the little fish-shaped sign means proportional to. Clear? Near the surface, small r is approximately capital R, the radius itself. And at a height R, small r is R plus R, that is two R, so the radius doubles. Got it? So the radius doubles. Then the period simply doubles too, right? No, no, wait! Capital T goes as r to the power three over two, not as r. So the factor is two to the power three over two, which is two root two. Two root two, times eighty four point six: about two hundred and thirty nine minutes. Same idea as before: height R means radius two R, and then the right power. With me?

3:19Example 1: height where a body loses 36% of its weight

Now, remember our opening puzzle? Here it is, example one. At what height does a body lose thirty six percent of its weight? Pause the video, and try it first. Done? Okay, let's check. Weight is m g, and the mass m does not change. So losing thirty six percent of the weight means small g drops to sixty four percent of its surface value. g h, with a small h written below, is zero point six four g. Right? Now the quick way. Use the short formula: g times, one minus two h over R, equals zero point six four g. Then h comes out as zero point one eight times R, and we are done, aren't we? No, no, careful! That was the trap from the start. Your course says this short formula is only for h much less than R; the double less-than sign means much less than. Losing thirty six percent is a big drop, so the height is not small compared with R, and the approximation fails. Okay? So, use the exact form, valid at any point outside the Earth: g h equals g over, one plus h over R, the whole squared. That is the same as g times, R over R plus h, the whole squared. Clear? Set it equal to zero point six four g, and cancel g. R over R plus h, squared, is zero point six four. Take the square root of both sides: R over R plus h is zero point eight. With me? Flip it: R plus h is one point two five times R, so h is R over four. With R as six thousand four hundred kilometres, h is one thousand six hundred kilometres. Puzzle solved! The short formula would have given you the wrong height. That's it!

5:09Example 2: energy for a satellite at radius 2R to escape

Now a harder one, example two, the only medium problem in this set. Read it with me: find the energy needed for a satellite of mass small m, orbiting at radius two R, to escape. Pause and try. Hint: start from the total energy in the orbit. Done? Okay, let's check. Quick thought: to escape from radius two R, just cancel the potential energy there. So the answer is G M m over two R, right? No, no, careful! That is for a body at rest there. Our satellite is already moving, so it already has kinetic energy. For a circular orbit, its total energy, capital E, is minus G M m over two small r. And here two R is given as the radius, not a height, so no R plus h this time. So E is minus G M m over four R, okay? Now, to escape, the satellite must reach infinity, where the potential energy, capital U, is zero. The least energy gets it there with zero speed, so its total energy must reach zero. Right? So the energy needed is final minus initial: zero, minus, minus G M m over four R, which gives G M m over four R, okay? Exactly half of the quick thought, because the kinetic energy already pays the other half. Got it? Finally, use G M equals g R squared once again. G M m over four R becomes m g R over four. That's the answer. Perfect!

6:42Quick revision: r = R + h, exact g, orbit speed and escape

So, quick revision. Height h means small r equals R plus h, and G M equals g R squared. The short formula for g is only for h much less than R; otherwise use g over, one plus h over R, the whole squared. In a circular orbit, the orbital speed is root G M over r, the period goes as r to the power three over two, and to escape from the orbit, supply minus E, which is G M m over two r. Now go and try the practice problems on this page, like a real test. All the best!