Effect of the Earth's Rotation

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Right now, as you watch this video, the Earth is spinning, and you are spinning with it, once every day. Now imagine the Earth spinning faster, and faster, and faster. At some point, things resting on the ground at the equator would start to float! So tell me: how short would the day have to be for that? Twelve hours? Six hours? Hmm, keep your guess in mind. By the end of this video, we will work it out. And look, the effect of the Earth's rotation on small g comes up often in JEE gravitation questions. So, let's begin.

0:38The spin makes g a little smaller

So, first the set-up. The Earth spins about its axis, once a day. How fast it turns is its angular speed, written omega, the Greek letter omega, okay? And everything on the Earth goes round with it: you, me, this classroom, everything. Now, tell me. Does this spin make small g, the acceleration due to gravity, bigger or smaller? Smaller, yes! See, to move in a circle, a body needs a pull towards the centre. And who gives that pull? Gravity itself. So part of gravity's pull is used up just to keep us moving in a circle. Right? So what is left over is a little less than g. We call it the effective value, and write it g prime: g with a small dash, okay? Small g is gravity alone; g prime is what is left after the spin takes its share. And here is the key result of this concept. At latitude lambda, the Greek letter lambda, g prime equals g, minus omega squared R cos squared lambda. Here capital R is the radius of the Earth, and cos squared lambda means the cosine of lambda, squared. It is an approximation, but a very good one. Write it down. Clear?

1:54At the poles and at the equator

Now, what is latitude? It tells you how far north or south of the equator you are. Lambda is zero degrees at the equator, and ninety degrees at the poles; the small circle means degrees. With me? So, tell me. If you stand exactly at the North Pole, how much does the Earth's spin reduce small g? Not at all! Absolutely zero. Cos ninety degrees is zero, so the whole term disappears. The double arrow means this gives: g prime equals g. At the pole you just turn on the spot; you do not go round any circle. Okay? And at the equator? Cos zero degrees is one, so g prime equals g, minus omega squared R, the biggest drop. Why the biggest? Because cos squared lambda can never be more than one. Got it? Now, how big is this drop for the Earth? Omega squared R is only about zero point zero three four metres per second squared. The wavy equals sign means about, and m s to the minus two is read metres per second squared. Compare it with g, nine point eight. Tiny, right? We will calculate it ourselves in Example one.

3:05The trap: cos squared, not cos

Now, a small trap. Look at the drop again: omega squared R, times cos squared lambda. Not cos, cos squared, okay? So, quick question. At latitude forty five degrees, the drop is what fraction of the drop at the equator? Cos forty five degrees is one over root two; the long tick is the square root sign. Put it in brackets and square it, and you get one half. So the drop is exactly half of its value at the equator. Did you get one half? Very good! But many students forget the square, and say one over root two of it. Oh no, careful! That is the trap. Always square the cosine. Clear?

3:46Example 1: effective g at the equator

Alright, Example one. Read it with me. Find omega squared R for the Earth, and the effective g at the equator. Pause the video, and try it first. Done? Okay, let's check. First we need omega. So, how long does the Earth take for one full turn? One day, yes! Twenty four hours, that is twenty four times three thousand six hundred seconds. And one full turn is two pi radians; rad is short for radians. So omega is two pi, over twenty four times three thousand six hundred seconds. Okay? Work it out: omega is seven point two seven times ten to the minus five radians per second. Rad s to the minus one is read radians per second. With me? Now put it in brackets, square it, and multiply by capital R, the Earth's radius, six point four times ten to the six metres. Omega squared R comes to zero point zero three three eight metres per second squared. That is about zero point zero three four, the value we saw earlier. Perfect! Now the effective g at the equator. Which value of lambda do we put in? Zero degrees, exactly! Cos zero is one, so g prime is just g minus omega squared R, nothing more. Nine point eight minus zero point zero three four: nine point seven six six metres per second squared. Smaller than g, but only just. That's it!

5:18Example 2: day length for weightlessness at the equator

Now, remember our puzzle from the start? Let's finish it. How long would the day have to be, for bodies at the equator to feel weightless? Pause the video, and try it. Done? Okay, let's check. First, what does weightless mean here? What must g prime be? Zero, exactly! All of gravity's pull goes into keeping the body on its circle, and nothing is left to press it on the ground. So g prime, which is g minus omega squared R, equals zero. Right? Solve it for omega: omega equals the square root of g over R, okay? Put in nine point eight, over six point four times ten to the six: omega is about one point two four times ten to the minus three radians per second. Got it? Now compare it with today's omega, from Example one. Is this new omega bigger, or smaller? Much bigger! Faster spin, shorter day. Capital T, the period, is two pi over omega. That comes to about five thousand and eighty seconds, which is about eighty four point six minutes. Careful: the answer comes out in seconds first, so change it to minutes. Clear? So, not twelve hours, not six hours. Under an hour and a half! How close was your guess? The Earth would have to spin that fast for things at the equator to float. Very good!

6:44Quick revision: latitude formula, poles and equator

So, quick revision. The effective g at latitude lambda is g, minus omega squared R cos squared lambda, and do not forget the square. At the poles, g prime equals g; at the equator, g prime equals g minus omega squared R, and omega squared R is about zero point zero three four metres per second squared. And bodies at the equator would float if omega were the square root of g over R: a day of about eighty four minutes. Now go and try the practice problems on this page. All the best!