Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a puzzle to start. Picture a satellite skimming just above the Earth's surface. One full round takes about eighty four point six minutes. Now, dig a smooth, straight tunnel through the Earth, not through the centre, just along a chord. Let a ball go at one end. It slides to the other end, and comes back. And its round trip also takes eighty four point six minutes! So how does a ball in a tunnel keep time with a satellite in the sky? Strange, right? Today's mixed problems are about periods in gravitation, in the short JEE Main style, and questions like these come up often. So, let's begin.

0:41Toolkit 1: periods of circular orbits

So, the first tool. Take a satellite on a circular orbit of radius small r, measured from the centre of the planet, not from its surface. Its period, capital T, equals two pi, times the square root of r cubed over capital G M; the tick-shaped sign is the square root. Capital G is the gravitational constant, and capital M is the planet's mass. Okay? Now, take two satellites around the same planet. Two pi, capital G and capital M are all the same for both. So tell me: if the orbit radius doubles, does the period just double? No! Square the formula: T squared is proportional to r cubed; the curly sign means proportional to. That is Kepler's third law. So T two over T one equals, r two over r one, to the power three over two; the small one and two below just label the two satellites. Double the radius, and the period more than doubles. Clear? Now, a satellite skimming just above the surface. Here small r is just capital R, the Earth's radius. And swap capital G M for small g R squared, which holds for a spherically symmetric Earth. Then T equals two pi, root of R over g; the wavy equals sign means approximately, so about eighty four point six minutes. That is the number from our puzzle. Got it? And one more famous orbit: the geostationary orbit. Its period is twenty four hours, and its radius is about forty two thousand three hundred kilometres from the centre. Right?

2:10Toolkit 2: inside the Earth, and SHM

Now, the second tool. Look at this graph. 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, and the peak is g nought, with a small zero below: g at the surface. Outside, on the right, the field falls as one over r squared. With me? But what happens inside the Earth? Read the left part of the graph. As you go down towards the centre, does the field grow, or shrink? It shrinks, yes, exactly! For a uniform Earth, look at the straight line on the left: inside, the field is proportional to r. The field equals g over R, times r, pointing towards the centre. At the surface that gives g, and at the centre, zero. Okay? Now, recall simple harmonic motion, or SHM for short. When the acceleration, small a, equals minus omega squared x, the motion is simple harmonic. Here x is the displacement from the middle, and omega, the Greek letter, is the angular frequency. The minus sign means the pull always points back to the middle. Then the double arrow, meaning so: the period T is two pi over omega. Right? And the greatest speed comes at the middle: v max, with max written small below, equals omega times capital A, the amplitude, which is the farthest distance from the middle. Clear?

3:35Example 1: period at half the geostationary radius

Example one, a warm-up. Here is a satellite orbiting at half the radius of the geostationary orbit. Find its period. Pause the video, and try it first. Done? Okay, let's check. Half the radius, so half the period: half of twenty four hours is twelve hours. Easy, isn't it? No, no, careful! That is the trap. Around the same planet, on circular orbits, T goes as r to the power three over two, not as r. Got it? So, T over twenty four hours equals one half, to the power three over two. That is one over, two root two: about zero point three five four. With me? So T is about twenty four times zero point three five four hours: about eight point five hours. Not twelve, eight and a half! Okay? Now, a quick sanity check. Half of forty two thousand three hundred is about twenty one thousand one hundred and fifty kilometres, between the Earth's radius and the geostationary radius. So the period must lie between eighty four point six minutes and twenty four hours; the angled signs mean less than. Eight point five hours fits. Perfect!

4:45Example 2: SHM in a tunnel along a chord of the Earth

Now a harder one, and this is our puzzle. There is a smooth, straight tunnel along a chord of the Earth, at a distance R over two from the centre. Release a body at one end, and it moves to and fro. Show the motion is simple harmonic, and find its period and greatest speed. Pause and try it. Hint: which part of the field acts along the tunnel? Done? Okay, let's check. Take a uniform Earth. At a distance small r from the centre, the field is g over R, times r, towards the centre, just like our graph. Right? But the body cannot fall towards the centre; the tunnel guides it. So which part of this field actually moves it? The part along the tunnel, exactly! Measure x from the tunnel's midpoint. The part along the tunnel is the field times x over r; the dot means times. The r cancels, leaving g over R, times x, towards the midpoint. Clear? So the acceleration is minus g over R, times x. Compare with minus omega squared x: it is SHM, with omega equal to root g over R! So T equals two pi, root R over g: about eighty four point six minutes. Okay? And look: the distance R over two never entered the period. Both periods, the skimming satellite's and this one, are two pi, root R over g. Puzzle solved! That is why the ball in the tunnel keeps time with the satellite. Now, the greatest speed. For that we need the amplitude. The tunnel is R over two from the centre, so the amplitude is R over two, right? No, no, wait! R over two is the tunnel's distance from the centre. The body starts from rest at one end, so the amplitude, capital A, is half the chord. By Pythagoras: the root of, R squared minus, R over two the whole squared. That is R times root three over two. With me? So, v max equals omega times capital A: root g over R, times root three over two, times R, okay? The R's combine, leaving root g R, times root three over two. Put in the numbers: about six point nine kilometres per second, reached at the midpoint. On screen, k m s to the minus one is just kilometres per second. That's it!

7:06Quick revision: orbit periods and SHM in an Earth tunnel

So, quick revision. One: for circular orbits around the same planet, T goes as r to the power three over two, not as r. Two: inside a uniform Earth, the field is g over R, times r, and along a chord tunnel the acceleration is minus g over R, times x, so the motion is simple harmonic. Three: its period is two pi, root R over g, about eighty four point six minutes, and v max is omega times the amplitude, which is half the chord. You handled two classic traps today, so well done. Now go and try the practice problems on this page. All the best!