Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take two planets made of the same rock, so they have the same density. One is small; the other has ten times its radius. Now a satellite skims just above each surface. Which satellite takes less time to go round once? The small planet, surely? Hold that thought; the answer will surprise you. Today is Set two of our mixed gravitation problems: three quick ideas, then two solved problems. Short questions like these come up often in JEE Main. So, let's begin.

0:37Idea 1: double the orbital speed

Idea one. To double its orbital speed, how must a satellite's orbital radius change? First, just the direction. Move it farther out, or closer in? Closer in, yes! Smaller radius, bigger speed. For a circular orbit, v o, that is v with a small o below for orbital, equals the square root of capital G M over r. Capital G is the gravitational constant, capital M the planet's mass, and r is measured from the planet's centre. The double arrow means so, and this sign, like a stretched alpha, means proportional to: v o is proportional to r to the power minus half. Okay? So v o goes as one over root r. To double v o, root r must become half. So the new radius is r over two, right? No, no, careful! Root r becomes half, so r itself becomes one half, squared: one quarter. So r new, with the small word new written below, is r over four. Move it in, to a quarter of the radius. Clear?

1:41Idea 2: escape speed and orbital speed

Idea two. Find the ratio of the escape velocity to the orbital velocity, close to the Earth's surface. Close to the surface means the height is small compared with capital R, the Earth's radius. So r is about R; the wavy equals sign means approximately. Then v o is root of G M over R, which is root of small g times R, since G M equals small g R squared. Got it? And the escape velocity from the surface, v e, with a small e below, is root of two g R, which we met in Chapter three. Right? Now, you tell me. What is v e over v o? Root two, exactly! The g R cancels, and only the two survives under the root. Okay? Root two is about one point four one. So near the Earth, seven point nine kilometres per second to orbit, and eleven point two to escape. Remember, both speeds here are taken near the surface; that was our starting condition. With me?

2:40Idea 3: the energy to free a satellite

Idea three. Find the energy needed to remove a satellite of small mass m, from an orbit of radius r, to infinity. Its potential energy, capital U, is minus G M m over r. So we must supply G M m over r, right? No, no, wait! The satellite is moving, so it already has kinetic energy. In a circular orbit, its total energy, capital E, is minus G M m over two r: only half as negative. Total means kinetic plus potential, with zero at infinity. Clear? Now, the least energy we can give is just enough to reach infinity, with nothing left over. So E final, with the small word final below, is zero. Okay? Energy needed is final minus initial: zero minus, in brackets, minus G M m over two r. That is G M m over two r, half of what the trap gave. That's it!

3:35Example 1: comet speed at aphelion by Kepler's second law

Now, Example one. Here, a comet's speed at perihelion, point five A U from the Sun, is sixty kilometres per second. The letters A U mean the astronomical unit, and k m s to the minus one means kilometres per second. Its aphelion is thirty A U away. Find its speed at aphelion. Pause the video, and try it first. Done? Okay, let's check. Look at the figure. The orange dot, labelled Sun, is at one focus of the ellipse. Perihelion, the closest point, is on the right; aphelion, the farthest, is on the left. The shaded slices, A one and A two, remind us: equal areas in equal times. The two a below, and T squared proportional to a cubed in the corner, we won't need today. Right? At these two points, and only there, the velocity v is at right angles to r, the line from the Sun. The upside down T sign means perpendicular. Okay? Gravity is a central force, so capital L, the angular momentum about the Sun, is conserved. At the two ends it is just m v r, so v p r p equals v a r a; small p for perihelion, small a for aphelion. Clear? Now, before any numbers, think. At aphelion, will the comet be faster or slower? Slower, of course! It is sixty times farther away. So v a equals v p times r p over r a: sixty times point five over thirty; the cross means times. If you flip that ratio, you get a huge speed at the far end. Wrong way round! Got it? Sixty times point five is thirty, and thirty over thirty is one. So v a is just one kilometre per second. With me?

5:20Example 2: skimming satellite period from density alone

Now the harder one, and it is our puzzle from the start. Show that the period of a satellite skimming the surface of a planet depends only on the planet's density. Then evaluate it for rho equal to five thousand five hundred kilograms per cubic metre. Rho, this Greek letter, is the density, and k g m to the minus three is its unit. Pause and try. Hint: write the mass using the density. Done? Okay, let's check. Skimming means a circular orbit just above the surface, so r is about capital R, the planet's radius. Then capital T, the period, is two pi, times the square root of R cubed over G M, from Chapter four. Okay? But the question wants density, not mass. So what can you write in place of capital M? Yes, exactly! Mass is volume times density. The volume of a sphere is four thirds pi R cubed, so M equals four thirds pi R cubed rho. Right? Now substitute. R cubed over G M becomes R cubed, over G times four thirds pi R cubed rho; the dot means times. R cubed cancels from top and bottom, leaving three over four pi G rho. With me? So T is two pi, times root of three over four pi G rho. Take the two pi inside the root as four pi squared, and four pi cancels: T equals root of three pi over G rho. Look, no R left! Only the density. So, our puzzle: same rock, same density, same period, whether the planet is small or huge! Clear? Now the numbers. Capital G is six point six seven times ten to the minus eleven newton metre squared per kilogram squared. Three pi, over G times five thousand five hundred, then the square root: about five thousand and seventy seconds. Got it? Divide by sixty: about eighty four point five minutes. And five thousand five hundred is about the Earth's mean density, which we found in Chapter two. So this is close to the period of a satellite just above the Earth itself. Beautiful, isn't it?

7:27Quick revision: orbital vs escape speed, satellite energy

So, quick revision. In a circular orbit, v o goes as r to the power minus half, and near the surface, v e is root two times v o. To free a satellite, supply G M m over two r, and for a comet, v p r p equals v a r a, at the two ends only. And a satellite skimming a planet has a period of root of three pi over G rho: density only. Now go and try the practice problems on this page. All the best!