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Ch 11: Gravity For the 1st three sections of this chapter, I’ve interjected most of the notes covered in 11th grade. You’ll see they are in small (12-pt) font. Read them if you like to help remind you of the conceptual feel of the material as you work to put the relevant math behind the theory. 11-1 Kepler’s Laws As an intro to Kepler’s Laws and Gravitational PE, let’s take a look at the notes we saw from 11th grade: Kepler’s Laws pf Planetary Motion- Notes The chain of events… From Brahe Kepler Newton Einstein* Brahe – measured positions of planets Took 20 years So accurate, still valid today Entrusted his data to Kepler Q: What is an ellipse? How are a circle and an ellipse related? A: The set of all points for which r1 + r2 = constant. A circle is a special case ellipse where the two foci coincide. Kepler’s Laws of Planetary Motion Law 1: Each planet moves in an elliptical orbit with the sun at one of the foci. The speed varies – faster the closer they are to the sun / slower the farther away they are from the sun. Sweeps out equal areas in “spoke-like” fashion The triangular area swept out when a planet is far from the sun is equal to the area swept out when it is much closer to the sun. NOTE: On p 340, there are two sentences that read: “The earth’s orbit is nearly circular, with the distance to the sun at perihelion (closest point) being 1.48 x 10 11 m and at aphehelion (farthest point) being 1.52. x 1011 m. The semi-major axis equals the average of these two, which is 1.50 x 1011 m (93 million miles) for the earth’s orbit. The mean earth-sun distance defines the astronomical unit, 1 AU where 1 AU = 1.50 x 1011 m. You may want to dig up your notes on parsecs and AU’s! You will need to use the concept of AVERAGING your two distances when you do problem 14. Also, problem 14 is NOT in the SSM, as denoted! Law 2: The line from the sun to any planet sweeps out equal areas of space in equal time intervals. http://astro.unl.edu/naap/pos/animations/kepler.swf Kepler was first to coin the term “satellite” though he didn’t understand that satellites were simply projectiles moving under the force of gravity. A satellite moves slower when working against gravity, and faster when working with it. Because of this, he spent 10 “needless” years crankin’ out useless geometrical formulas. Then, he came up with his third law. Law 3: The squares of the times of revolutions (periods) of the planets are proportional to the cubes of their average distances from the sun. T2 ~ R3 or T2 = C R3 where C is the same for all planets. Table 11-1, p 340, gives the mean orbital radii and orbital periods for all the planets. (Yep, Pluto is still on the list!) Also, one last note: the mean sun-earth distance defines the following astronomical unit, AU where 1 AU = 1.50 x 10 11 m = 93.0 x 106 miles. This unit, AU, is frequently used in problems dealing with our solar system/ NOTE: When working the problems, be mindful of the following things: Convert km to m when necessary Make SURE you square and / or cube, when necessary Use parentheses (when needed) around your numbers to avoid unnecessary errors!! See Ex 11-1, p 341 The mean distance from the sun to Jupiter is 5.20 AU. What is the period of Jupiter’s orbit around the sun? (NOTE: This means that Jupiter is about 5 times farther from the sun than the earth!) Use T2 = C R3 where TE = 1 year and RE = 1 AU = the period and mean distance for the Earth. Then, TJ and RJ = 5.20 AU = the period and mean distance for Jupiter. If you set up a ratio, and knowing that the C’s will cancel out, you get: T2Jupiter = R3Jupiter T2Earth R3Earth So, TJupiter = TEarth (RJupiter / REarth)3/2 = 1 y(5.20 AU/1 AU)3/2 = 11.9 y As a last note to this problem, if you look at Fig 11-5, part (a) depicts the periods versus the mean distance from the sun, while in (b), the squares of the periods are plotted versus the cubes of their mean distances from the sun. Q: What does this do to the data? What would you get if you plotted the logs of the periods versus the logs of their mean distances from the sun? A: Linearizes it. You’d also get a straight line. Q: What is the slope of the line in graph b? A: C, of course! Kepler’s Laws apply not only to planets, but to any satellite. He was familiar with all of Brahe’s calculations, which initially led him to think the orbits were circular. He was also familiar with Galileo’s ideas about inertia and accelerated motion, but did not use them in his work. He did not, however, appreciate the concept of inertia, and really, not until Newton, did inertia figure prominently in the discussion of planetary motion. 11-2 Newton’s Law of Gravity You might also remember from 11th grade that Newton said there was a force of attraction between any pair of particles that was jointly proportional to their masses, and inversely proportional to the square of the distance between them. where G = 6.67 x 10-11 N m2 / kg2. Q: Is it necessary to memorize the units for G? A: Nope. Just use the units that would commonly cancel out any mass units on top, m units on the bottom, and leave you with kg m/s2, which is the same as a Newton. After all, the SI unit of Force is the Newton. This opens up some possibilities for calculating and / or manipulating acceleration, as F ~ ma. See Ex 11-2, p 344. Ex 11-2, p 344: What is the free-fall acceleration of an object at the altitude of the space shuttle orbit, about 400 km above the Earth’s surface? a = F = GmMEarth/r2 = GMEarth m m r2 To find r, use the radius of the Earth, plus the altitude, so r = REarth + h = 6370 km + 400 km = 6770 km. Convert this to m and you get 6.77 x 106 m. Then, a = (6.67 x 10-11 N m2 / kg2)(5.98 x 1024 kg) (6.77 x 106 m)2 = 8.70 m/s2 One last note: a = F/m = GME / r2 also can be rewritten as gR2Earth / r2. In other words: GME = gRE2 TRY THIS!! At what distance above the surface of the earth is the acceleration of gravity half of its value at sea level? (Don’t forget that “r” is the TOTAL of RE + rabove the earth)!!!! Make sure you read through the equation derivations on p 345 – 347 (equations 11-10 – 11-16). Many are just simple algebraic manipulations of the same equations, but contain useful shortcuts and explanations with regard to the measurement, manipulation, and accuracy of G. Derivation of Kepler’s Laws Newton showed that when an object moves around a force center such as the sun, the object’s path is a conic section (parabola, hyperbola, or ellipse). The two former types are not closed paths; only the ellipse is closed, thus, Kepler’s 1st Law is really a consequence of Newton’s Law of Gravity. His second law of equal areas follows from the understanding that the force exerted on an orbiting object is centripetal. See Fig 11-8 (a) and (b) below. (a) In a given time, dt, the planet moves a distance v dt which is │r x v dt│. So, the area, dA, swept out by the radius r in time dt is given by dA = ½ │r x v dt│= 1/2m │r x mv│dt or NOTE: The ½ is there because you are assuming, for very small values of time (dt), that your shapes are triangles; hence, the formula A = ½ b h. Furthermorr, they are putting m under the 1 (in the fraction ½) so that they can also out m into the cross product. That way, they can make it look like angular momentum! Lastly, though the cross product typically has a sin component, when dt is very small, sin is basically sin 900 which = 1, so that part is not shown. dA = _1_ L where L = │r x mv│= the magnitude of the orbital angular dt 2m momentum of the planet about the sun. What this really means is that the area swept out in a given time is proportional to the magnitude of the angular momentum. Since the force on a planet is along the line from the planet to the sun, there is no torque about the sun. Thus the magnitude of the angular momentum is conserved. Another way of saying that is that L is constant!! Further, what this really means is that the rate at which the area is swept out is the same for all parts of the orbit. This is Kepler’s 2nd Law. Now, Check out the rest of the derivations in your text that bring you to: So, C 4 2 GM s where Ms = the mass of the sun. Try Ex 11-3, p 347-348 now. (It’s an easy one. ) You can either solve it like they do in the text, or by using the information in Table 11.1. 11-3 Gravitational Potential Energy Energy Conservation and Satellite Motion Q: What is KE? PE? For a planet moving in a perfectly circular orbit, PE and KE would have to be constant. Gravity is completely centripetal, so there is no component of force acting in the direction of the motion. Thus, speed, and consequently KE, don’t change. In elliptical orbits, PE is greatest when the satellite is farthest away (apogee), and least when closest (perigee). Obviously, the opposite is true for KE. So, in the case of elliptical orbits, KE and PE vary, but their sum is constant. In an elliptical orbit there is a component of force in the direction of motion, which will change the speed in most every place except at apogee and perigee. KE + PE KE + PE KE + PE KE + PE Near or at the surface of the earth, g is basically uniform because r = R E + h is essentially RE, for h <<RE. That’s why when you have a coconut on top of a tree, and it is subject to gravity, it’s essentially the same amount of g even if you double the height of the tree. Unless that tree starts at the core of the earth the change in height is irrelevant, compared to RE. Q: If an object is far from the earth, by how much does the force of g decrease? A: 1 / r2, according to Newton’s Law of Gravity. In Chapter 6, we learned that PE can be defined as dU = - F∙ ds where F is a conservative force on a particle, and ds is its general displacement. ( W = F∙ ds which is related to KE.) To find the radial gravitational force, we can substitute dU = - F∙ ds = - Fr∙ ds = - Fr dr = - (- GMEm ) dr = GMEm dr r2 r2 Q: What do you get in you integrate both sides of the equation? A: U = - GMEm + U0 where U0 is the constant of integration r Really, all that matters is the change in PE, so, we can choose U0 = 0 most of the time. Q: What does U equal if r → ∞? A: U = 0, of course! Figure 11-9 below shows a plot of U(r) versus r for the choice U = 0 at r = ∞ for an object of mass m, and an earth mass of ME. As you can see, the function begins at the negative value U = - GMEm / RE = - mgRE at the earth’s surface and increases as r increases. Ultimately, it approaches zero as r → ∞. The slope of the curve at r = RE is GMEm / R2E = mg. The equation of the tangent line is shown in blue, and is given by f(h) = U(RE) + mgh, where h = r - RE is the distance above the earth’s surface. From the Figure, you can see that if h is very small, U(RE) + mgh ~ U(r). Figure 11-9 Escape Speed / Velocity Q: What would happen to a rock launched horizontally at 8 km / s? Why? Q: What would happen to a rock launched vertically at 8 km / s? Why? Q: Is there a critical vertical speed / velocity that lets a projectile “outrun” gravity? DEF: Escape Speed / Velocity = The critical speed / velocity required to outrun gravity = 11.2 km / s. 2GM E D Q: What is the KE required for an object to escape gravity? The formula is v = A: mgR, of course!!!! The reason for this value is as you launch something up with some initial KE, as the object rises, its KE decreases and its PE increases where the maximum increase in PE is GMEm / RE. So this is also the maximum amount by which the KE can decrease. If the initial KE is greater than GMEm / RE, then the total energy will be greater than zero, and the object will still have some KE even when r is very large. What this really means is that the object will be able to escape the earth’s gravity. Further, since the PE at the earth’s surface is - GMEm / RE, the total energy, E = KE + PE or K + U must be ≥ 0, and the object escapes. See Figure 11-10. Figure 11-10 The KE of an object at a distance r from the center of the earth is E – U(r). When the total energy is less than zero (E1 in the figure) the KE is zero at r = rmax and the object is bound to the earth. When the total energy is greater than zero (E2 in the figure), the object has enough speed to escape the earth’s gravity. This speed is called escape speed, and is given by: When you substitute the value for g and the radius of the earth, you get ve = 11.2 km/s. If superman flies to the top of a mountain that’s high enough to escape air drag, and hurls a rock horizontally at 8 km / s, what will happen to the rock? How much time will go by before he can turn around and catch it? What if he hurls it faster, but less than 11.2 km / s—what will be the shape of its orbit? Will it take: shorter – same – longer for the rock to return? Why? What if he hurls it at a speed greater than 11.2 km / s? Guess what happens if he hurls it at a speed > 42.5 km / s? (He is, after all, Superman .) Q: How much work would be required to lift a payload against gravity to a distance far, far away (like infinity)? Most of the work done in lifting a payload occurs within the first 10,000 km of the Earth. The change of PE for an object moved to an infinite distance is 62 MJ, which mathematically corresponds to 11.2 km / s. If given morr than 62 MJ, it will escape and never return. Q: What happens to its PE and KE as it continues outward? Q: Will its velocity ever be zero? Q: What if a payload is launched such that 11.2 km / s < payload speed < 42.2 km / s ? Classification of Orbits by Energy Going back to figure 11-10, a negative total energy means KE < + GMEm / RE so that KE + PE (or K + U) is never greater than zero. Also, if the total energy is negative, the total energy line intersects the PE curve at some maximum separation, rmax, and the system is “bound.” If the total energy is either zero, or positive, there is no such intersection, and the system is “unbound.” Hmmm…sounds “calculusy.” Further, when E is negative, its absolute value, │E│ is called the binding energy, which is how much energy has to be added to the system to bring its total energy up to zero. The PE of an object, such as a planet or comet, having mass m at a distance r from the sun is given by U(r) = - GMsm / r. Does that formula sound familiar? One last bit: KE is defined as ½ mv2. If KE + PE < 0, then the orbit is either an ellipse or a circle, and the object will be bound to the sun. If KE + PE > 0, the orbit is a hyperbola. The object will make one swing around the sun, and vamoose out of our solar system. If KE + PE = 0, the orbit will be a parabola, and the net effect will be the same as when KE + PE > 0. In essence, when TE ≥ 0, the object escapes. So far, no comet or asteroid has satisfied this criterion, and thus all are bound to our solar system! Pretty neat! Try Ex’s 11-4 – 11-6 now. Ex 11-4 will lead you into doing the other two.