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Orbits Chapter 13.5-8 Today – Will not cover section 13.9. Newton’s Shell Theorem A uniform spherical shell of matter attracts a particle that is outside the shell as if all of the shell’s mass were concentrated at its center. It is also true that a uniform shell of matter exerts no net gravitational force on a particle located inside it. Let’s pretend there is a hole all the way thru the earth and we could drop an object into it – What would happen? 1. Mass outside the object we can ignore 2. Only mass inside radius of object matters 3. Via the shell theorem can compute the force assuming all of the inside mass is at the center € € Mass inside the Earth Gmobj M ins F= r2 3 4 πr M ins = ρ 3 4 πGmobj ρ F= r 3 F = −kr Mins mobj The r vector points out from the center, so we have and forth. and object would just oscillate back Clicker question1 Set frequency to BA A satellite in circular orbit about the Rock Earth is directly over Denver, 300 miles Satellite above the city, and traveling eastward at 16,000 mph. At the same time, a Denver rock is released from rest 300 miles above the city, right next to the satellite. True or false: The acceleration of satellite and rock at this point in time is the same in both magnitude and direction. A. True B. False Clicker question1 Set frequency to BA A satellite in circular orbit about the Rock Earth is directly over Denver, 300 miles Satellite above the city, and traveling eastward at 16,000 mph. At the same time, a Denver rock is released from rest 300 miles above the city, right next to the satellite. True or false: The acceleration of satellite and rock at this point in time is the same in both magnitude and direction. A. True If the satellite and rock are a distance r from the center B. False of the Earth, the acceleration of either one is independent of whether it is a satellite or rock Direction is toward the center of the Earth in each case. Clicker question 2 Set frequency to BA An astronaut is floating around in the space shuttle's cabin 200km up. Her acceleration, as measured from the earth's surface, A: B: C: D: zero - she's floating very small, in some random direction quite large, nearly g, directed towards the center of the earth quite large, nearly g, directed along the line of travel of the shuttle Clicker question 2 Set frequency to BA An astronaut is floating around in the space shuttle's cabin 200km up. Her acceleration, as measured from the earth's surface, A: B: C: D: zero - she's floating very small, in some random direction quite large, nearly g, directed towards the center of the earth quite large, nearly g, directed along the line of travel of the shuttle She's in orbit, so she's in uniform circular motion and has a large acceleration! F(grav) is almost the same up there, so her acceleration is almost the same as on earth in freefall, namely, nearly g, directed towards the center of the earth. She's just like a ball that's been tossed up in the air, accelerating straight down. Gravitational potential energy Gravitational force: & potential energy: Potential energy due to Earth’s gravity is RE = radius of Earth = 6380 km Total energy is potential plus kinetic and is constant (since Distance from center of the Earth (km) gravity is a conservative force) For total energy < 0, object is bound by the gravitational field (and orbits are ellipses). Examples are planets around the sun. For total energy = 0, object is barely unbound (parabolic orbit). For total energy > 0, object is unbound with a hyperbolic orbit. Escape velocity We define escape velocity as the minimum speed needed to escape a gravitational field (usually from the surface). Escaping means being able to reach r = ∞. On the previous slide we found this is possible if we have a total energy ≥ 0. Total energy of 0 means , that is For object m1 on the surface of a planet with mass MP and radius RP this translates to Solving for the speed gives This equation actually works for any radius outside the planets radius. Just replace RP with the distance from the planet’s center. Clicker question 3 Set frequency to BA Does escape velocity depend on launch angle? That is, if a projectile is given an initial speed vo, is it more or less likely to escape an airless, non-rotating planet, if fired straight up than if fired at an angle? A. Yes B. No Clicker question 3 Set frequency to BA Does escape velocity depend on launch angle? That is, if a projectile is given an initial speed vo, is it more or less likely to escape an airless, non-rotating planet, if fired straight up than if fired at an angle? A. Yes We derived the escape velocity using conservation of B. No energy. All that is needed is for the projectile to have enough kinetic energy such that the total energy is 0. Kinetic energy only depends on the magnitude of the velocity (squared), not the direction so the angle is irrelevant. On rotating planets the escape velocity is the same but the initial velocity is not zero so it does make sense to take off at an angle. History of solar system understanding 1543: Nicholas Copernicus theorizes that the Earth orbits the sun. 1609 & 1619: Johannes Kepler uses 30 years of data collected by Tycho Brahe to derive 3 laws for planetary motion 1. Planets move in elliptical orbits with sun at one focus planet Sun 2. A line drawn between sun and planet sweeps out equal areas during equal intervals of time 3. The square of a planet’s orbital period is proportional to the cube of the semimajor-axis length These laws can be derived from Newton’s 1687 gravity theory. Circular orbits from a force perspective All closed orbits are ellipses but we will analyze the simpler case of circular orbits. Very good approximation for planets and moons. Not so good for comets. A small object m in orbit around a large object M is called a satellite. Newton’s 2nd law is M The only force is gravity: The acceleration is only radial: Therefore: Solving for v gives Note that orbital velocity only depends on M/r. m Need for Dark Matter Solving for v gives It is observed that disk galaxies have a rotation curve which is flat from centre to edge (line B in illustration), i.e. stars are observed to revolve around the centre at constant speed over a large range of distances. It was expected that these galaxies would have a rotation curve that slopes down from the centre to the edge (dotted line A in illustration), in the same way as other systems with most of their mass in the centre, such as the Solar System of planets or the Jovian System of moons. Clearly, something else is happening to these galaxies besides a simple application of the laws of gravity to the observed matter. Hence Dark Matter! Law of Areas A line drawn between sun and planet sweeps out equal areas during equal intervals of time or dA/dt is constant 1 2 ΔA ≈ r Δθ 2 r dA 1 2 dθ 1 2 ≈ r = rω dt 2 dt 2 rΔθ L = r × p = rp⊥ = r(mv ⊥ ) = r(mωr) € 2 L = mr ω dA L = dt 2m € L must be a constant – Kepler’s 2nd law is just conservation of angular momentum Law of Periods: Orbital period Orbital period T is the time it takes to complete one revolution. Orbital speed can be determined by distance covered in one revolution (circumference) divided by the period. We already know that So Can also write as proving Kepler’s 3rd law (for circular orbits) m M What is the moon’s orbital period? The moon is a satellite of the Earth and is at a distance of 384,000 km. m We can directly solve for the period M Could also calculate speed: Geosynchronous Orbits Consider a satellite whose orbital period is exactly 24 hours. Think about this - the earth rotates once in 24 hours, and in that time the satellite has also run around the earth once. From our perspective on the ground, that satellite is at rest above us! This is convenient for TV or satellite communication - the satellite is always above the same spot. You aim your dish at it and don’t have to keep tracking the satellite. We call this a geosynchronous orbit. How high up is such a satellite? GM E 2 We have R = T 2 4π 3 Plugging in all the constants, with T=24 hrs, and taking the cube root , I get R=42,000 km. This is a very high orbit. Since R_earth = 6,000 km, it is 36,000 km above the ground. (Compare to the shuttle, which is only about 250 km up) Geosychronous Orbits cont. Geosynchronous satellites have v = GM E /RSAT = 3 km/s , slower than the shuttle. The farther out you are in orbit, the slower you go. (Bigger distance, but still bigger T.) € Clicker question 4 Set frequency to BA Two communications satellites are in orbit at the same height, but one weighs twice as much as the other. The speed of the heavier satellite is A: Less than B: Equal to D: (need more information) the speed of the lighter one.. C: Greater than Clicker question 4 Set frequency to BA Two communications satellites are in orbit at the same height, but one weighs twice as much as the other. The speed of the heavier satellite is A: Less than B: Equal to D: (need more information) C: Greater than the speed of the lighter one.. Mass cancels out. Speed depends only on radius, for spherical orbits. Clicker question 5 The international space station altitude gradually decreases due to drag and is periodically boosted back up. Assuming perfectly circular orbits, is the station velocity higher before or after the boosts. A. Before B. After C. Same D. Can’t tell Set frequency to BA Clicker question 5 The international space station altitude gradually decreases due to drag and is periodically boosted back up. Assuming perfectly circular orbits, is the station velocity higher before or after the boosts. A. Before B. After C. Same D. Can’t tell Set frequency to BA for circular orbits. Smaller r results in larger v.