Circular Orbits for m << M; a planet and Star

Transcription

1 Circular Orbits for m << M; a planet and Star In the Gravity PhET lab, you established that for circular orbits, the speed is proportional to the inverse of the square root of the radius. We can show that: Free Body Diagram of m: For Uniform Circular Motion: Therefore, we have the circular orbit speed: This is what you demonstrated in the Gravity PhET lab.

2 Circular Orbits for m << M; a planet and Star Remember Kepler s Third Law from the text: the square of the orbital period is proportional to the cube of the semimajor axis of the orbit? We can calculate this as an equality: So, we have Newton s Form of Kepler s Third Law: This equation can also be used for an elliptical orbit if you replace the radius, r, with the semimajor axis a. The above equations work when m << M. If that is not the case, both masses orbit about their common center of mass see next WB.

3 Whiteboard Problem 13-4: Binary Stars Consider two stars with masses M 1 and M 2 separated by a distance d and orbiting their center of mass in circular orbits. a) Find an expression for the orbital speed of M 1 in terms of M 1, M 2, r 1, and r 2. (LC) b) Without doing any work, what is the speed of M 2?

4 Elliptical Orbits for m << M For an elliptical orbit, the force varies, and the planet speeds up and slows down that s Kepler s Second Law. PhET But, we know that for any orbit, some things are conserved. Since there are no nonconservative forces acting on m, the mechanical energy is conserved. This can be used to connect any two points in the orbit:

5 perihelion Elliptical Orbits for m << M Note that, since the force of gravity is always directed toward the center, the torque on m about M is always zero, i.e. aphelion So, the angular momentum is conserved. This can be cumbersome to use at an arbitrary point in the orbit, but if we look at just the perihelion point and the aphelion points where the angle is 90 o :

6 Whiteboard Problem 13-5 The dwarf planet Pluto moves in a fairly elliptical orbit. At its closest approach to the Sun of 4.43 X 10 9 km (perihelion), Pluto s speed is 6.12 km/s. What is Pluto s speed at its most distant point in its orbit (aphelion), 7.30 X 10 9 km? (LC) Pluto from the New Horizons Spacecraft in 2015 Hint: Draw the orbit and the points mentioned. There are two ways to do this problem. They both work, but one is a lot easier.

7 Orbits in a Cloud of Mass (not in the text) We have been concentrating on the problem of one object orbiting another where each object can be represented as a point mass: r > R R m Mass, M r < R m What about an object orbiting inside or outside a spherical gravitating cloud of matter of mass M and radius R? We can do this if we use two important facts: For a spherically symmetric distribution of mass M: 1. For r < R, mass m only feels the gravity of that part of M that is interior to it, M int for r < R. And M int acts as a point mass at r = For r > R, mass m feels the gravity of all of M as if all of M was a point mass at r = 0.

8 Whiteboard Problem 13-6: Orbits in a Cloud Consider a mass m orbiting inside or outside a spherical, uniformly distributed, mass M of radius R. a) Find an expression for the circular R orbit speed of m for r > R. (M acts as point mass at r = 0.) A slightly different way, for mass m: and for UCM: So: Mass, M b) Find an expression for the circular orbit speed for r < R. (m sees only M int that acts as point mass at r = 0.) Now: r < R m r > R m So:

9 Dark Matter Lab : Background (remember the video: The Missing Universe?) We have seen that the orbital speeds of the planets in the Solar System decrease with distance from the Sun, where: The stars in the Milky Way orbit the center of the galaxy in circular orbits; however, it s a little more complicated because here, all of the ~50 billion stars provide the gravity. However, most of these stars are closer to the center of the galaxy than the Sun. So, near the Sun and beyond, the rotation curve should begin to fall off inversely with the square root of r. But it stays relatively flat! Why? (this is true for almost all spiral galaxies as well)

10 Dark Matter Lab : Background The standard explanation for this discrepancy is that there is matter out there that we cannot see that provides the extra gravity Non Luminous Dark Matter. Thus, the Milky Way galaxy and other spiral galaxies are embedded in large halos of this mysterious Dark Matter. It is estimated that there must be more than 10X as much dark matter as luminous matter. To date, no one has been able to discover what this Non-Baryonic Dark Matter is. An Alternative Explanation: Modified Newtonian Dynamics (MOND) Another possibility is that our understanding of gravity is lacking. In 1983, Mordechai Milgrom proposed a modification for Newtonian gravity for very low accelerations near a characteristic value of a 0 = 1.2 X m/s 2. A gravitational force law of 1/r will produce flat rotation curves, but we know that gravity goes as 1/r 2 where it has been tested for high accelerations like in the Solar System, so Milgrom s modification has to give Newtonian gravity for high accelerations, but something different at low accelerations. One form of MOND gives : Where a n is the Newtonian acceleration and a m is the modified acceleration. Thus, for large accelerations compared to a 0, the Newtonian acceleration is unmodified, but for accelerations near a 0 and below, there is a modification to the acceleration.

11 Dark Matter Lab Complete the steps in the handout. You may want to refer to the slide about calculating circular orbital speeds in a uniform cloud of matter; We put copies of these with the lab handout. Do the work on your whiteboards, and fill in the table on the back and plot your results in the grid. You can use Excel or another spreadsheet to do the calculations and plots if you know how and want to. We strongly suggest that you use the units described in the handout; it really does make the calculations a lot easier. Remember, in these units masses are in M s (solar masses); lengths in parsecs (pc); and time in millions of years (My) so that: G = pc 3 /(My 2 M s ) a 0 = 3.88 pc/my 2 1 pc/my ~ 1 km/s Make sure to put your Group number on the lab and all of your names. Turn it in to a TA before you leave.