Variation of g with Height and Depth

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Go up from the Earth's surface, and small g, the acceleration due to gravity, gets smaller. Go down into a deep mine, and small g gets smaller again! Strange, right? Now, climb thirty two kilometres, and g drops by about one percent. So how deep must you dig for the same one percent drop? Thirty two kilometres again? Hold on to your answer. And look, this topic comes up often in JEE gravitation questions. So, let's begin.

0:36Going up: g at height h

So, look at this graph first. Along the bottom is small r, the distance from the centre of the Earth, and going up is small g, or the field capital E, the same thing. The dashed line at capital R is the surface: inside on its left, outside on its right. With me? Start at the surface. From the last concept, small g equals capital G M over R squared. Capital G is the gravitational constant, capital M is the Earth's mass, and R is its radius. This treats the Earth as a spherically symmetric body, okay? Now go up, to a height h above the surface. Tell me: how far are you from the centre of the Earth now? R plus h, correct! In every gravitation formula, height h means a distance R plus h from the centre. Students often forget this one. Clear? So, at this point outside the Earth, g h, with a small h written below for height, equals capital G M, over R plus h, the whole squared. Divide the top and bottom by R squared. Capital G M over R squared is just g, so g h equals g, over one plus h over R, the whole squared. This is exact, for any height: the falling curve, marked g proportional to one over r squared; the sign like an open alpha means proportional to. Got it?

2:00Small heights: a shortcut, with a condition

Now, many heights in questions are small, like a hill. When h is much less than R, there is a shortcut; the double less-than sign means much less than. Then g h is approximately g times, one minus two h over R, okay? The wavy equals sign means approximately. But tell me: can we use this shortcut at a height equal to the radius? Put h equals R, and see what you get. One minus two is minus one. So g h comes out as minus g! Negative gravity, pushing you away? Oh no, careful! Impossible. This is the classic trap: the shortcut works only when h is much less than R, right? So when h is comparable with R, always use the exact formula, the one on screen. Clear?

2:50Going down: g at depth d

Now let's go down, into a deep mine. You are closer to the centre now. So small g should increase, right? No, no, look at the graph! Left of the dashed line, inside the Earth, g is the straight line rising from zero, marked g proportional to r. From chapter one: inside a uniform solid sphere, g equals capital G M r, over R cubed. So deeper means smaller r, and smaller g. Here we treat the Earth as uniform, okay? Now, at a depth d below the surface, your distance from the centre is R minus d. Simple, right? Put it in. g d, with a small d written below for depth, equals capital G M times, R minus d, over R cubed. Take out capital G M over R squared, which is g, and we get g times, one minus d over R, okay? No approximation here: it holds at every depth, for a uniform Earth.

3:47The whole picture

Now, the whole graph. Going up, g falls. Going down, g falls. So small g is greatest right at the surface, at r equals capital R: the peak, marked g nought; the small zero written below is just a label. Clear? At the centre of the Earth, the depth d equals the full radius. One minus one is zero, so g d is zero at the centre. Right? And far away, as h goes to infinity, g h goes to zero. The arrow means tends to, and the sideways eight is infinity. Clear? Now, back to our puzzle. For small distances, which drops g more: going up a small height, or going down the same depth? Going up, yes! At a small height h, g drops by a fraction two h over R; at a depth d, only by d over R, okay? So for small distances, g falls twice as fast going up. With me? So here is a topper's trick. For the same small drop in g, the depth is twice the height: d equals two h. Our puzzle: thirty two kilometres up gives a one percent drop, so you must dig sixty four kilometres down, not thirty two! Mystery solved.

5:06Example 1: g at height R and at 3200 km

Example one, from your course. Find small g at a height of, first, R, and second, three thousand two hundred kilometres above the Earth's surface. Pause the video, and try it first. Done? Okay, let's check, with g as nine point eight, and R as six thousand four hundred kilometres. On screen, m s to the power minus two means metres per second squared. Okay? Part a: h equals R, so the exact formula. One plus one is two, and two squared is four. So g h is g over four: nine point eight over four, two point four five metres per second squared. Just one quarter, at one Earth radius up! Right? Now, part b: three thousand two hundred is exactly half of six thousand four hundred, so h equals R over two. Now, is this height small compared with the radius? No! Half the radius is not small at all. So, no shortcut; use the exact formula. Clear? One plus one half is one point five, and one point five squared is two point two five. So g h is g over two point two five, which is four ninths of g. Four ninths of nine point eight is about four point three six metres per second squared. Okay?

6:29Example 2: height where g equals g at depth R/2

Now Example two, a little harder, and it uses both formulas. At what height is small g equal to its value at a depth of R over two? Pause the video, and try it. Done? Okay, let's check. Depth first. g d equals g times, one minus one half, which is g over two. Half the surface value. Got it? Now the height. Quick way: set one minus two h over R equal to one half. That gives h equals R over four. Done, right? No, no, careful! g has fallen to half, so this height is not small; the shortcut fails. Use the exact formula: g, over one plus h over R, the whole squared, equals g over two. The double arrow means implies: cancel g and flip, and one plus h over R, the whole squared, equals two. With me? Take the square root: one plus h over R equals root two. So h equals, root two minus one, times R, right? Root two minus one is about zero point four one four. Times six thousand four hundred kilometres: about two thousand six hundred and fifty kilometres. Much bigger than R over four! See why the condition matters?

7:51Quick revision: g at height and depth, the d = 2h rule

So, quick revision. Height: g h equals g over one plus h over R, the whole squared, and the shortcut, one minus two h over R, only when h is much smaller than the radius. Depth: g d equals g times, one minus d over R, and it is zero at the centre. And g is greatest at the surface; for the same small drop, d equals two h. Now go and try the practice problems on this page. All the best!