[MUSIC] Orbits, the operations of the orbs from whom we do exist and cease to be, as Shakespeare put it. To explain orbits without mathematics, Newton asks us to imagine a cannon on a mountain. It fires the ball horizontally, but as we know, the ball curves away, below the straight line because of gravity. Fire it faster and it travels further with less curvature. Fire it fast enough and earth also curves away below the line. Fast enough, in fact about eight kilometers per second and the curvature of the trajectory will equal the curvature of the Earth. Actually there was a lot of empirical knowledge about orbits before Newton. Johannes Kepler used the beautiful observational data of Tycho Brahe to induce empirical laws about the orbits of the planets. Approximating the orbits is circles for the purposes of this course, Kepler's Law says that the square of the period of the orbit is proportional to the cube of the distance from the sun. No explanation, just a very precise generalization. [LAUGH] I know what you're thinking. Show me how that works. It's truly wonderful and it's not too tricky, so, let's go. Remember circular motion. Travelling in a circle, an object is accelerating towards the center. Here, the horizontal component of the tension in the string produces the centrifugal force without which, it wouldn't curve in a circle. So let's consider a small object mass m in circular orbit around the large mass, capital, big M. For little m, Newton's second law of motion is F equals ma equals mr omega squared. Omega, the angular speed, is 2pi over the period T so we have F equals MR times 2pi over T, all squared. For a planet or satellite the force involved is gravity, so we substitute using Newton's law of gravity, F equals GMm over r squared. Cancel the little m and rearrange. We do indeed get T squared proportional to r cubed. Further, we now know that the constant proportionality is 4pi squared on GM, an amazing result. Amaze yourself by answering this question. It's hard to explain how stunning Newton's discoveries were. Before Newton, how many learned men and women had wondered, what mechanics governs the heavens? In Marlowe's play, this is what Doctor Faustus sells his soul for! Yet here, Newton's laws and a few lines of Algebra, show that the force that causes the apple to fall also explains the orbit of the moon around the earth, and that of the earth and the other planets around the sun. Newton was and is an intellectual hero, and we're still using his legacy. We can do similar calculations for other bodies. For instance, the Earth-moon distance is 380,000 kilometers. Its period is 27 days. What would be the orbital radius of a satellite with a period of one day? Let's recap, because this is an important result. We write the law of periods and take the cube root. The cube root of 27 is 3, 3 squared is 9, so our satellite is 9 times closer than the moon at 42,000 kilometers. At this radius, a satellite orbits in about one day, so it stays in the same place above the equator. This, the geosynchronous orbit is where we put most of our communications satellites. [LAUGH] It's a busy orbit. Another busy region is the low orbit, just above the atmosphere. Here, a similar calculation gives a period of about 90 minutes. For the International Space Station, about 400 kilometers above the Earth, the period is 92 minutes. [LAUGH] The astronauts don't get bored with the scenery. Now this brings us to an awkward problem in language. We say that these astronauts are weightless as they drift around inside the space station. But what does weightless mean? In fact, their weight, their gravitational attraction to the Earth, goes as 1 on r squared. So it's only reduced by 12%. They are in free fall. They are accelerating towards the Earth at 88% of 9.8 meters per second per second. However, the space station is also accelerating towards the Earth at the same rate and that's why they float around inside it. Well, we're stuck with the word weightless for the experience of free fall. If you are in a falling elevator, you would also feel weightless. Even though it's your weight that's accelerating you downwards towards a nasty collision with the ground. So we're left with this awkward paradoxical statement, weightlessness, is the name for what you feel when the only force acting on you is your weight. Back to orbits. So far, we've discussed them in terms of forces. Let's now discuss them in terms of energy. Again, a small mass, N, in circular orbit, radius R, about a large mass, capital M. For that circular orbit, radius R, the centripetal force is gravity, GMm on R squared. This time, let's write write centripetal acceleration as v squared on r. Now we can easily turn this term into kinetic energy. Just multiply both sides by r on two. This gives half GMm on r equals half mv squared. Now we remember that the potential energy of M and m at separation r is minus GMm on r. So the left hand side of the previous equation is just minus a half times the potential energy, so minus a half times the potential energy equals the kinetic energy. And, since the potential energy is never positive, the kinetic energy is never negative. Save negative kinetic energies and imaginary speeds for when we get to quantum mechanics. So, the total mechanical energy, u plus K is, let's substitute E equals minus GMm on 2r or one-half the potential energy. So this means that a close orbit has low total energy. You have to do work to get from low orbit to high. Alternatively, if you fall from high orbit to low, your potential energy goes more negative, but your kinetic energy more positive. Here's a toy we can use as a model. In this gadget, the height, measured from the top, goes roughly as minus one on R. So here, close to the earth, we have U equals MGH proportional to minus one on r. So here we have orbits in a minus one on r potential. Small r gives very negative potential energy, but high kinetic energy. Large r gives less negative U and smaller K. Somewhat complicated here by friction and rolling, neither which occur in space. We can also think about the orbits of comets. Fast, when close to the sun, but slow when far away. So keep that in mind. A low orbit, small r, has very negative U and E, but large K. Low orbits are fast. A high orbit, large r, has less negative U and less negative E, but smaller K. High orbits are slow. Here's some simple physics that will probably surprise you. Imagine you're in orbit, following another spacecraft, and you want to dock with it. How do you catch up? Think about it. Well, it's simple if we go step by step. To catch up, you have to go faster. So you want to fall down into a lower, faster orbit. So you point your rocket in the opposite direction of your motion. This does negative work on you, so you fall down into a lower orbit, lower orbits are faster, so you catch up. When you have caught up, you want to slow down. This time you point your engine behind you and fire. This does positive work on you, which takes you to a higher, slower orbit, so you slow down. Well if that seems counter intuitive, you're in good company. It was counter intuitive to the first cosmonauts and astronauts who did rendezvous maneuvers. Finally, here is a seriously weighty problem. Our sun is orbiting around the center of the Milky Way galaxy. It's at a distance of 27,000 light years and with a period of roughly 240 million years. So, making a few assumptions, how many solar masses do we need to provide the centripetal acceleration of our sun around the galactic center? [LAUGH] Let's put that one in the quiz.
