Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, a puzzle to start. Release a body from rest, three Earth radii above the ground. How fast does it land? Quick way: v squared equals two g h. That gives about nineteen point four kilometres per second. But hold on. Eleven point two kilometres per second is enough to escape from the Earth forever! So a body that simply falls from rest lands faster than the escape speed? Something is wrong here, right? Today's problems mix ideas from the gravitation unit, in the short JEE Main style, and questions of this kind do appear in the exam. So, let's begin.

0:43Toolkit 1: g above and below the surface

So, look at this graph first. Along the bottom is small r, the distance from the Earth's centre; going up is small g, or capital E, the field. The dashed line at capital R is the surface: inside on the left, outside on the right. The peak is g nought, with a small zero below: g at the surface. Inside, g is proportional to r; the curly sign means proportional to. Outside, it falls as one over r squared. With me? Now, this chapter's numbers. Unless a question says otherwise, small g is nine point eight metres per second squared, the m s to the power minus two on screen. And capital R, the Earth's radius, is six thousand four hundred kilometres. Okay? Now, the first question in most problems here. At a height h above the surface, how far are you from the centre of the Earth? R plus h, yes, exactly! Small r equals R plus h. Gravitation formulas use this distance from the centre, never the height alone. Right? So, small g at a height h. g h, with a small h written below, equals g over, one plus h over R, the whole squared. This is exact for any height, outside a spherically symmetric Earth. It is the falling green curve on the graph. Clear? Now, the shortcut. When h is much less than R, the double less-than sign, g h is approximately g times, one minus two h over R; the wavy equals sign means approximately. But only then! For a height comparable with the radius, it goes badly wrong. Our first example tests exactly this. Got it? And going down, to a depth d: g d, with a small d written below, equals g times, one minus d over R, for a uniform Earth. That is the straight line on the left of the graph. Okay?

2:34Toolkit 2: energy, when g changes on the way

Now, the second tool: energy, with a spherically symmetric Earth. Masses capital M and small m, a distance r apart, have potential energy capital U, minus capital G M m over r, zero at infinity; capital G is the gravitational constant. And the swap you will use again and again: capital G M equals small g R squared. With me? Now, a body falls from thousands of kilometres up. Can we still use v squared equals two g h, the formula from kinematics? No! Up there, small g is much weaker, as we just saw. So we use energy. Ignoring air, the loss in potential energy equals the gain in kinetic energy. Okay? So, v squared equals two g h assumes a constant g, and that is fine only when h is much less than R, the same condition as before, right? Now, a topper's trick for checking answers. Per kilogram, the energy needed to go from the surface to infinity is capital G M over R, which is just g times R, clear? So, half v e squared equals g R; v e, with a small e below, is the escape velocity. The double arrow means so: v e is the root of two g R, about eleven point two kilometres per second; the tick-shaped sign is the square root, and the k m s to the minus one is kilometres per second. This number will settle our puzzle. Got it?

4:01Example 1: g at 1600 km, and why the shortcut fails

Example one, an easy one to warm up. Find small g at a height of one thousand six hundred kilometres. Pause the video, and try it first. Done? Okay, let's check. First, compare the height with the radius. h over R is one thousand six hundred over six thousand four hundred: one quarter. So one plus h over R is one point two five. Okay? Now, the quick way: the shortcut. One minus two times a quarter is one half, so g h is half of nine point eight: four point nine metres per second squared. Done, right? No, no, careful! h is a quarter of the radius; that is not small at all. The shortcut needs h much less than R, so this four point nine is wrong. Use the exact formula. Clear? One point two five squared is one point five six two five. And g over one point five six two five is exactly zero point six four times g. With me? Zero point six four times nine point eight: about six point two seven metres per second squared. The shortcut said four point nine. Quite a difference! Always check the condition first.

5:13Example 2: landing speed from 3R by energy conservation

Now, remember our puzzle from the start? Here it is, as a JEE Main style question: a body is released from rest at a height three R above the Earth's surface. Find its speed on reaching the surface. Pause and try it. Hint: does small g stay the same all the way down? Done? Okay, let's check. Quick way, like in the puzzle: v squared equals two g, times three R, and that gives about nineteen point four kilometres per second. Correct? No, no, careful! That formula needs a constant g, and three R is not small compared with the radius at all. Up there, g is far weaker, so this answer is wrong. Right? So, energy. Distances from the centre first. It starts at R plus three R, that is four R, and it ends at the surface, at a distance R from the centre. Okay? Potential energy lost becomes kinetic energy; the mass cancels, so we work per kilogram. Half v squared equals capital G M times, one over R, minus one over four R; one minus one quarter is three quarters, so we get three quarters of capital G M over R, that is, three quarters of g R, using the swap. With me? Multiply by two: v squared is three halves of g R, right? So v is the root of one point five g R, the answer in symbols. Clear? Now the number: one point five, times nine point eight, times six point four times ten to the power six, that is capital R in metres, then the root; m s to the minus one is metres per second. That comes to about nine point seven kilometres per second. Got it? Now the check, with our topper's trick. Half v e squared is g R, and half v squared is three quarters of g R, so v squared is three quarters of v e squared. Nine point seven is less than eleven point two, the escape velocity. Puzzle solved: nineteen point four came from pretending g stays nine point eight all the way down!

7:15Quick revision: exact g at height, energy for long falls

So, quick revision. One: measure every distance from the centre, r equals R plus h, and swap capital G M for g R squared. Two: g at height h is g over, one plus h over R, the whole squared; use the shortcut only when h is much less than the radius. Three: for long falls, use energy, not v squared equals two g h, and check against v e, root of two g R, the escape velocity. Now go and try the practice problems on this page. All the best!