Energy in Planetary Orbits

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Eunice Haynes
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1 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 1 Energy in Planetary Orbits Consequences of Kepler s First and Third Laws According to Kepler s First Law of Planetary Motion, all planets move around the Sun in elliptical orbits with the Sun at one focus. In fact, all orbits except that of Pluto are practically circles with the Sun at the center. Consider a planet of mass m with an orbital radius R and a speed v. It will have a period of revolution T given by T = 2πR = v = 2πR v T However, the centripetal force can be expressed in terms of the usual formula mv 2 /R and also Newton s Law of Universal Gravity F c = mv2 R = GmM S = F R 2 G where M S is the mass of the Sun. We substitute now for v = 2πR/T mv 2 R = m4π2 R 2 T 2 R 3 = GM S 4π 2 T 2 = G mm S R 2 For planetary orbits the cube of the Radius Is proportional to the square of the Period All planets have negative mechanical energy? Believe it or not, all planets have negative total mechanical energy! How can this be? It surely can t be because of the Kinetic Energy: K = 1 2 mv2 which has to be positive. This it must the the fault of the potential energy: U G (R) = G mm S R Recall that the absolute value, or zero reference, of potential energy is arbitrary. Conventionally, we say the gravity potential energy is zero when two masses are infinitely far apart.

2 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 2 Energy in Planetary Orbits The Escape Speed For a given initial speed v i we have the total energy conserved at any distance r: E = 1 2 mv2 i G mm E R E = 1 2 mv2 G mm E r To get the maximum distance r max we set the final speed equal to 0. This then gives us the expression for r max 1 r max = 1 GmM E ( 1 2 mv2 i G mm E R E ) Now suppose we want the mass to travel infinitely far away for the Earth, so then 1/r max = 0. Then the expression above in parenthesis must also be zero: ( 1 2 mv2 i G mm E R E ) = 0 = v i (r max = ) v esc = 2GM E R E Objects with a speed v esc at the Earth s surface will travel to infinity and never return. Obviously, you can substitute for any other planet or even the Sun to get their escape speed. The concept of escape speed is critical to understand why the moon has no atmosphere, why the Earth has no hydrogen or helium in its atmosphere, and why the outer heavy planets all contain hydrogen and helium as their most abundant elements.

3 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 3 The Ultimate in Escape Speed: The BLACK HOLE According to relativity, the fastest speed anything can have is the speed of light c. Nothing, not even light, can go faster then c. So what happens when the escape speed becomes bigger than the speed of light? What you have then is something called a Black Hole! This is an object so massive, and so compact, that nothing can escape from it not even light itself. So you can t see it, but it s there. And if you get to close to it, then you will never escape yourself. Do Black Holes exists? Most physicists think so. It is believe that Stars with a mass several times that of the Sun will eventually explode into a Super-nova and leave behind an incredibly dense core. For example, the core might be the mass of the Earth and the size of a dime. The only evidence for a Black Hole would be if another object is too close, like a binary star companion, and one sees the companion gradually consumed by the Black Hole.

4 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 4 Nagging discrepancies The Bohr Atom In the early 1900s there were two nagging discrepancies in the physics of the time. The first had to do with the concept of the ether and the speed of light. We have seen that this small problem eventually led to the collapse of Newton s classical mechanics and the introduction of Einstein s Special Relativity. The second problem had to do with atomic spectra. In the 1900s, physicists had learned about atoms, and that atoms had negatively charged electrons inside. Moreover, the experiments of Rutherford had shown that there was a positively charged nucleus deep inside the atoms, about which the electrons orbited. For all practical purposes, the atom appeared to be a miniature solar system. Light was emitted from the atoms when the electrons re-arranged themselves into different orbits. Each element had its own characteristic frequencies or wavelengths of light and these were called the atomic spectra. Essentially these atomic spectra were like fingerprints for the atom. The problem was that nobody was able to calculate these spectra from the know principles of Electricity and Magnetism. And there was an even worse worry. If the electrons were actually in orbit around the nucleus, then the electron was being accelerated. And the theory of electricity predicted that a centripetally accelerated electron would gradually lose all its kinetic energy and fall down into the nucleus in less than microseconds. But this did not happen. Why? Simple Experimental Data needed a Simple Answer In the late 1800s, atomic physicists studying atomic spectra began to find very simple relationships among the frequencies or wavelengths of light emitted. One of these physicists was named Johann Balmer and he studied hydrogen, the simplest atom. He found that some of the most prominent wavelengths of the light from hydrogen obeyed a remarkably simple formula 1 λ n = R H ( n 2) where R H = m 1, and n = 3, 4, 5,.... Nobody could figure out how to derive this simple formula!

5 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 5 The Bohr Atom Making a virtue out of a necessity After Balmer, other atomic physicists (Lymann, Paschen, others) began to play numerical games with the wavelengths of light from hydrogen. They came up with a general formula that explained even more sets of light wavelengths: 1 = R H ( 1 λ nfi n 2 1 f n 2 ) i where n f and n i (n f < n i so as to get a positive difference) are any sets of integers. It all was kind of fun, but so far just numerology. It shouldn t be happening because the electrons should be falling into the center of the atom. The resolution of this dilemma came first from Neils Bohr, a Danish physicist. Essentially, he took the same approach as Einstein. Make a virtue out of a necessity. For Einstein, if it was necessary that the speed of light c be the same for everybody, then build a mechanics based on that assumption. For Bohr, if it was necessary that the hydrogen atom survive, then simply state that to be the case. Bohr s Model of the Atom Bohr stated four simple rules for the hydrogen atom: 1) The electron does move around the proton just like a planet around a Sun. 2) The electron is allowed in only certain stable orbits 3) The radii r of stable orbits obey an angular momentum quantization rule: mvr = n h 2π where n is an integer and h is a constant which another physicist called Planck had found in seemingly unrelated studies of thermal matter. 4) An electron was allowed to jump from one orbit to another and radiate energy (light) according to the formula E i E f = hf where E i is the orbital energy in the initial state, E f is the orbital energy in the final state, and f is the frequency of the light. With these four simple rules, Bohr was able to explain precisely all the features of the hydrogen atomic spectra.

6 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 6 The Energy of Orbits Making Electrons and Planets the Same We saw that the energy of a planetary orbit is given by E G = 1 2 mv2 G mm s r We will have the same thing for an electron orbiting a proton E e = 1 2 mv2 k e e 2 where r is the radius of the electron s orbit. As before the centripetal force comes from the center of the orbit, the electric force here mv 2 e 2 = k e r r 2 So this gives for the kinetic energy mv 2 /2 r K = 1 2 mv2 = k ee 2 So now the total energy of the orbit becomes E = k ee 2 2r Just as in the planetary case, this total energy is negative. The final step comes in applying Bohr s third condition 2r mvr = n h 2π = r n = h2 4π 2 1 mk e e 2 to get rid of the r in terms of n. This produces the famous Bohr energy formula for the hydrogen atom E n = k ee 2 ( electron-volts 2a 0 n2) = n 2 where a 0 is a constant called the Bohr radius. So now we have the energies of all the orbits, and their differences produce the atomic spectra light for hydrogen, including the 13.6 ev ionization energy for hydrogen.

7 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 7 CHAPTER 12: Oscillatory Motion Consider a spring lying in a horizontal position. A mass is attached to the spring, and the spring is stretched and then released. In the absence of friction the mass will oscillate along the horizontal axis. That means the position x of the mass will go from a maximum positive displacement, to zero displacement, and then to a maximum negative displacement, and then repeat the cycle. This position function x(t) obeys the Simple Harmonic Motion equation (SHM): Simple Harmonic Motion x(t) = x m cos (ωt + φ) (12.6) The position function x(t) = x m cos (ωt + φ) has three parameters: x m is the amplitude of the motion ω(= 2πf) is the angular frequency of the motion φ is the phase constant of the motion The time for one complete cycle of the oscillation is called the period T. The number of cycles per second is called the frequency f. The frequency f is the inverse of the period T frequency f = 1/T There are two simple systems, pendulum and spring for which you should now the equation for their periods in terms of the physical parameters: m Period for a Spring of a given mass and force constant: T = 2π k Period for a Pendulum of a given length: T = 2π L g (12.25) The velocity and the acceleration of the mass in oscillatory motion can be calculated directly from the position function x(t) = x m cos (ωt + φ) by taking the first and the second time derivatives respectively: v(t) = dx dt = d(x m cos (ωt + φ)) dt a(t) = dv dt = d( ωx m sin (ωt + φ)) dt = ωx m sin (ωt + φ) (12.15) = ω 2 x m cos (ωt + φ) (12.16)

8 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 8 Oscillatory or Simple Harmonic Motion The general equation for simple harmonic motion is given by x(t) (or y(t) ) = x m cos (ωt + φ) (12.6) This equation describes the motion of a mass attached to a spring (either horizontally or vertically), or the motion of a pendulum. The purpose of this chapter is to study this type of motion, and in particular to become familiar with the concepts of Amplitude, frequency, and phase. Take a specific case of a spring which is stretched to an initial distance x 0, and an attached mass is also given an initial speed v 0. We will see that the constants x m and φ can be expressed in terms of the two initial conditions x 0 and v 0. We first substitute in the position and the velocity equations at time t = 0 x(t = 0) = x m cos (ω 0 + φ) = x m cos φ = x 0 v(t = 0) = ωx m sin (ω 0 + φ) = ωx m sin φ = v 0 Now divide the second equation by the first in order to get an equation for the phase angle by itself in terms of x 0 and v 0 tan φ = v 0 ωx 0 With a little more work, you can substitute this expression for tan φ into one of the two other equations and obtain an expression for the amplitude x m just in terms of x 0, v 0, and ω x m = x ( v0 ω ) 2 Special case where v 0 = 0 (no initial velocity) tan φ = 0 = φ = 0 x m = x = x m = x 0 So this special case gives a very simple Simple Harmonic Motion equation x(t) = x 0 cos ωt (only when v 0 = 0)

9 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 9 Worked example of Simple Harmonic Motion A particle oscillates in simple harmonic motion according to the following position equation where t is in seconds and x is in meters: x(t) = 4.0 cos (πt + π 4 ) Here the phase angle φ is given in radians instead of the more common degrees. (Recall that π radians is equal to 180 o, so π/4 = 45 o.) Determine the amplitude, frequency (and angular frequency), and period of the motion. Determine the position, velocity, and acceleration of the particle at t = 1 second. The amplitude x m and the angular frequency ω can be determined by simply comparing this equation to the general equation x(t) = x m cos (ωt + φ). You will see right away that in this example x m = 4.0 meters, and ω = π radians/second. To get the (plain) frequency f, and then the period T requires that you recall the relation between the angular frequency ω and f ω = 2πf = f = ω 2π (12.11) So in this case f = π/(2π) = 0.5 cycles per second. The period is given just as the inverse of the (plain) frequency f T = 1 f = 1 = 2.0 s 0.5 s 1 To get the position and velocity at t = 1 second, just substitute in position and the velocity equations of motion: x(t = 1) = 4.0 cos (π 1 + π 4 ) = 2.83 m v(t = 1) = (4.0)(π)(sin (π 1 + π )) = 8.89 m/s 4 a(t = 1) = (4.0)(π) 2 (cos (π 1 + π )) = 27.9 m/s2 4

10 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 10 Newton s Second Law and Simple Harmonic Motion Up to now we have been working with the position equation x(t) = x m cos (ωt + φ) (12.6) for oscillatory motion, and have simply stated that this is the correct equation. Now we will prove that is the correct for the case of a mass attached to a spring. This we will do using Newton s second law of motion, the force equation When a spring is stretched a distance x where x > 0, then there will be a force in the negative x direction. Such a force is called a Restoring Force because it tends to restore the spring back to its original configuration. We have already seen in Chapter 7 that the magnitude of the force depends upon the spring constant k F spring = kx (12.1) This force is exerted on the attached mass m, and by Newton s second law we have F = ma = kx = a = k m x (12.3) Like the force itself, the acceleration is linearly proportional and opposite in sign to the displacement from the equilibrium position. This equation can be re written using the calculus definition of acceleration a = d2 x dt 2 = k m x We now symbolize the ratio k/m by the symbol ω 2, and then obtain: ω k m = d2 x dt = 2 ω2 x

11 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 11 Newton s Second Law and Simple Harmonic Motion We have derived an equation for the position function x(t): d 2 x dt 2 = ω2 x This equation states that we need a function x(t) such that when we take two derivatives with respect to t, then we will get back the negative of the original function x(t) multiplied by ω 2. Very simply, Eq is exactly that function, and the prove is just do it. Start with Eq. 12.6, and then take two time derivatives: x(t) = x m cos (ωt + φ) (12.6) d dt x(t) = x mω sin (ωt + φ) d 2 dt x(t) = x mω 2 cos (ωt + φ) = ω 2 x(t) In general, whenever one has a restoring force equation coming in Newton s second law, then one will always find the solution to be the position function is just simple harmonic motion. Note that the constants x m and φ are completely arbitrary. Any pair on numbers x m and φ will satisfy the restoring force equation. In order to fix x m and φ to particular values, on must specify the initial conditions, namely the initial position (x 0 ), and the initial velocity (v 0 ) as we have done previously.

12 Lecture 19: Energy in Orbits, Bohr Atom, Oscillatory Motion 12 The Mass Spring System in Simple Harmonic Motion The general solution for simple harmonic motion is x(t) = x m cos (ωt + φ) (12.6) where for the spring we remember that ω 2 = k m = ω = k m (12.4) Now in general ω = 2πf so ω = 2πf = k m = f = 1 k 2π m (12.12) Finally one can determine the period T of the motion in terms of k and m since T = 1/f T = 1 f = 2π m k (12.13) Worked Example A car of mass 1300 kg has four shock absorber springs each with a force constant of 20,000 N/m. If two people riding in the car have a combined mass of 160 kg, what is the frequency f of the vibration of the car when it is driven over a pothole? Assume that the total mass (=1460 kg) is equally distributed over the four shocks, so each shock absorber has a mass of 325 kg attached to it. The solution is just to use Eq to solve for f f = 1 k 2π m = 1 2π 20, = 1.18 Hz The abbreviation Hz (after Heinrich Hertz) means 1 cycle/second. The period of the vibration T is simply given as the inverse of the frequency T = 1 f = = 1.70 seconds

Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12) 1

Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12) 1 REVIEW: Rotational Equilibrium (Chapter 11) With the use of torques one can solve problems in rotational equilibrium. Rotational