Orbital Speed and Period

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Some satellites seem to hang still above one point of the Earth; dish antennas on rooftops point at them and never turn. So tell me: is such a satellite floating there, with no gravity? Or is it moving? Hold on to your answer. Questions on orbital speed and period come up often in JEE papers. So, let's begin.

0:27Gravity supplies the centripetal force

So, here is our Earth: a uniform sphere, mass capital M, radius capital R, and it pulls as if all its mass sat at its centre, okay? Now a satellite goes round it in a circle of radius small r, where small r equals capital R plus small h, the height above the surface. See, small r is measured from the centre of the Earth, not from the ground, right? Now think. Anything moving in a circle needs a force towards the centre, the centripetal force. So what supplies it here? Gravity, exactly! So capital G M small m over r squared equals small m v o squared over r; the little two up top means squared. Capital G is the gravitational constant, small m the satellite's mass, and v o, v with a small o at the bottom, the orbital speed. Clear? Cancel one small m and one r. v o equals the square root of G M over r; that tick-shaped sign is the square root. And from chapter two, G M equals small g R squared, with small g the gravity at the surface. So v o is also the root of g R squared over r. With me? Now tell me. Take a satellite twice as heavy, in the same orbit. Does it need to move faster? No! There is no small m left; it cancelled. The orbital speed does not depend on the satellite's mass. Heavy or light, same orbit, same speed. Got it?

1:56Period and angular momentum

Next, the period, capital T, the time for one round. One round is two pi r, with pi the Greek letter, and time is distance over speed. So T equals two pi r over v o, which becomes two pi root of r cubed over G M, okay? The little three means cubed. Now think. Move the satellite to a higher orbit. Faster or slower? Longer period, or shorter? Slower, and longer! This sign, like an open alpha, means proportional to. v o goes as r to the power minus one half, and T as r to the power three over two. Bigger circle, slower satellite: the period grows fast. Right? And the angular momentum, capital L, equals m v o r, which is m root of G M r. This one does carry the satellite's mass; the speed and the period do not. Clear?

2:49Just above the surface: r≈R

Now, satellites just above the surface. The height is small compared with capital R, so small r is approximately R; the wavy equals sign means approximately. Put r equal to R in our formula. What is left? Root of g R, yes! Small g, nine point eight, times capital R, six point four times ten to the six metres; the cross means times, and the little six is the power of ten. That gives about seven point nine kilometres per second; k m s with a little minus one is read kilometres per second. Right? And the period: two pi root of R over g; the slash means over. About five thousand and eighty seconds, which is eighty four point six minutes. Under an hour and a half! Got it? One link with chapter three. The escape speed, v e, with a small e at the bottom, is root two times v o at the same point: about eleven point two kilometres per second here. Oh no, careful: seven point nine to orbit, eleven point two to escape! With me?

3:53Geostationary and polar satellites

Now, back to our puzzle. The Earth is spinning, right? So what period lets a satellite stay above one point? Twenty four hours, exactly! Capital T equals twenty four h, h for hours. It goes round once while the Earth turns once, in the same direction, west to east. Okay? And it must move in the equatorial plane, the plane of the equator. Then it stays above one point. Twenty four hours, west to east, equatorial plane: all three together make it geostationary. Its height? Example two will tell us. Polar satellites are different: low orbits, passing over the poles. As the Earth turns beneath them, they scan the whole Earth. Clear?

4:40Example 1: speed and period ratios for orbits r and 4r

Alright, example one, an easy one. Two satellites orbit at radii r and four r, okay? Find the ratios of their speeds and of their periods. Pause the video and try it. Done? Okay, let's check. Speeds first. Four times farther out, so four times slower, right? No, no, careful! Speed goes as r to the power minus one half. So v one over v two, the small one and two just label the satellites, is the root of four r over r: root four, which is two. Two is to one; the colon is read as is to. Twice as fast, not four times. Clear? Now the periods. T goes as r to the power three over two, so T one over T two is r over four r, in brackets, which is one quarter, to the power three over two. Root of one quarter is one half, and one half cubed is one eighth. So one is to eight: four times farther, eight times longer! With me?

5:42Example 2: height of a geostationary satellite

Now example two, a little harder, and it finishes our puzzle. Find the height of a geostationary satellite. Take capital G as six point six seven times ten to the minus eleven newton metre squared per kilogram squared, capital M as six times ten to the twenty four kilograms, and capital R as six point four times ten to the six metres. Pause the video and try it. Done? Okay, let's check. Capital T is twenty four hours, but capital G is in S I units, so time must be in seconds: twenty four times three thousand six hundred, eighty six thousand four hundred seconds. Put in twenty four, and everything goes wrong. Okay? Now square the period formula: T squared equals four pi squared r cubed over G M; the double arrow means so. So r is GMT squared over four pi squared, all in big brackets, to the power one third: the cube root. With me? Put in the numbers: capital G times capital M, times eighty six thousand four hundred squared, all over four pi squared. Then the cube root. Clear? It comes to about four point two three times ten to the seven metres, or forty two thousand three hundred kilometres. So that is the height, right? No, no, wait! Small r is measured from the centre of the Earth, remember? Height is from the surface. So h equals r minus capital R, the Earth's radius. Four point two three minus zero point six four, times ten to the seven: about three point five nine times ten to the seven metres. Got it? About thirty five thousand nine hundred kilometres up. So, our puzzle: floating with no gravity? No! It is always moving, going round once in twenty four hours, in step with the turning Earth, and gravity keeps it on its circle. Mystery solved!

7:43Quick revision: orbital speed, period, geostationary

So, quick revision. v o equals root G M over r, and T equals two pi root r cubed over G M, with no satellite mass in either. Small r is from the centre; near the surface, seven point nine kilometres per second and eighty four point six minutes. Geostationary: twenty four hours, west to east, equatorial plane, about thirty five thousand nine hundred kilometres up. Now go and try the practice problems on this page. All the best!