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Gravity.Notes Chapter 6 Gravity 6.1 Newtonian Gravity Let’s review the original theory of gravity, and see where and how it might be updated. Newton’s theory of gravity begins with a specified source of mass, ⇢(r), the mass density, as a function of position. Then the gravitational field g is related to the source ⇢ by g = 4 ⇡G⇢, (6.1) r· − where G is the gravitational constant (G 6.67 10 11 Nm2/kg2). Particles ⇠ ⇥ − respond to the gravitational field through the force F = m g used in Newton’s second law (or its relativistic form): m x¨(t)=m g, (6.2) for a particle of mass m at location x(t). The field g is a force-per-unit-mass (similar to E, a force-per-unit-charge), and it has g =0, so it is natural to introduce a potential φ with g = φ, r⇥ r just as we do in E&M. Then the field equation and equation of motion read (again, given ⇢) 2φ =4⇡G⇢ r (6.3) m x¨(t)= m φ. − r 6.1.1 Solutions From the field equation, we can immediately obtain (by analogy if all else fails) the field g for a variety of symmetric source configurations. Start with the point 187 188 CHAPTER 6. GRAVITY source solution – if ⇢ = mδ3(r) for a point source located at the origin, we have Gm φ = (6.4) − r similar to the Coulomb potential V = q ,butwith 1 G (and 4 ⇡✏0 r 4 ⇡✏0 ! q m, of course). ! We can generate, from (6.1), the “Gauss’s law” for gravity – pick a domain ⌦, and integrate both sides of (6.1) over that domain, using the divergence theorem, to get g da = 4 ⇡G ⇢(r0) d⌧ 0 . (6.5) · − I@⌦ Z⌦ Menc ⌘ As usual, this form is always true, but not always| {z useful} for finding g. Suppose we have an infinite line of mass with constant mass-per-unit-length λ.Thensymmetrydemandsthatg = g(s) ˆs, and using a Gaussian cylinder of height h and radius s centered on the wire, we have g da = g(s)2⇡hs, (6.6) · I@⌦ with Menc = λh,sothat 2 Gλ g(s)= (6.7) − s which mimics the electrostatic result E(s)= λ , and the two are again 2 ⇡✏0 s related under 1 G. 4 ⇡✏0 ! Problem 6.1 Find the gravitational field above and below an infinite sheet of mass with constant mass-per-unit-area σ. Problem 6.2 What is the potential φ for the infinite line of mass and for the infinite sheet? Problem 6.3 What is the potential inside and outside a sphere of radius R with constant mass density ⇢0? 6.2. OPTIONS 189 6.2 Options 6.2.1 Three types of mass There are a few natural questions we might ask at this point. The first – is it clear that the mass appearing on the left and right of (6.2) are the same? In E&M, for example, Newton’s second law reads: m x¨(t)=q E and we don’t associate m and q (units aside), why do we have the same m appearing on both sides of (6.2)? We could imagine that the inertial mass m on the left is di↵erent from the “passive” gravitational mass, mp, on the right: m x¨(t)=mp g. (6.8) While experiment suggests m m , it is impossible (within the confines of ⇡ p Newtonian gravity and dynamics) to rule out the possibility that m = m . 6 p Einstein’s theory of general relativity requires that m = mp,soitscorrectness suggests that m = mp exactly, i.e. that the inertial and passive gravitational masses are identical. There is a third type of mass built into Newtonian gravity – the “active” gravitational mass that generates the gravitational field – in theory, then, there are three di↵erent masses that could be present in the equations of Newtonian gravity: g = 4 ⇡G⇢a r· − (6.9) m x¨(t)=mp g, where ⇢a is the active mass density. Again, up to experimental error, the active and passive masses are the same, so we don’t worry about these distinctions (people worry about them all the time – if you could find, say, a di↵erence between m and mp, then general relativity would automatically be unavailable as the theory of gravity). 6.2.2 Negative mass In E&M, there are two types of charge, positive and negative – yet we only (typically) allow one sign for mass. What would happen if you had a negative mass? Take a pair of masses m` and mr (left and right) separated a distance x as shown in Figure 6.1. The force on each mass is: Gm m Gm m F = ` r xFˆ = ` r xˆ, (6.10) ` x2 r − x2 but if we assume all forms of mass are the same (inertial, active and passive), 190 CHAPTER 6. GRAVITY m mr xˆ x Figure 6.1: Two masses separated by a distance x. then the acceleration of each mass is Gm Gm a = r xaˆ = ` xˆ. (6.11) ` x2 r − x2 So far, we have made no assumptions about the signs of the masses – take mr > 0 and m` < 0, what happens? The details of the dynamics are, as always, complicated, but of interest is the direction of the acceleration – both accelerations point to the right (in the xˆ direction) – the masses “chase” each other. That implies that both masses are speeding up and traveling in the same direction. What happens to energy and momentum conservation? The total energy of the system is 1 1 Gm m E = m v2 + m v2 ` r , (6.12) 2 ` ` 2 r r − x for v` and vr the velocity vectors. The energy can remain constant for a constant separation x while both v` and vr increase because m` < 0.Similarly, the momentum of the initial configuration is zero (assuming the masses start from rest) – if the two masses move in the same direction with increasing speed, it would appear that the linear momentum is no longer zero. Again, since m` < 0, we have total momentum p = m` v` + mr vr that is zero even though both masses are moving in the same direction. 6.3 Predictions: Non-relativistic First, we can go back to the original data that motivated Newton to introduce the 1/r2 force of gravity, associated with two massive bodies. The Kepler laws, obtained by Kepler through observation, are encoded in Newtonian gravity. Our first goal will be to identify the motion of a test mass induced by the gravitational field of a spherically symmetric central body of mass M. Then we’ll move on to cases in which the “test body” (that feels the e↵ect of the field) is not a point particle. 6.3. PREDICTIONS: NON-RELATIVISTIC 191 6.3.1 Elliptical Orbits For a particle of mass m that travels in the presence of the gravitational field of a spherical central body of mass M and radius R, the total energy is 1 H m r˙2 + r2 φ˙2 + U(r)=E, (6.13) ⌘ 2 ⇣ ⌘ where we are working in the x y plane in spherical coordinates and U(r)= − GMm. Angular momentum is conserved here, from p˙ = @H =0,sothat − r φ − @ J mr2 φ˙ is a conserved quantity. Using this, we can write the total energy z ⌘ as 1 J 2 H = m r˙2 + z + U(r)=E. (6.14) 2 m2 r2 ✓ ◆ Because the potential energy here goes like 1/r,itisnaturaltodefinea new coordinate ⇢ 1/r to make the potential energy linear in ⇢.Withthis ⌘ substitution in place, our energy expression becomes 1 ⇢˙2 J 2 H = m + z ⇢2 + U(⇢)=E. (6.15) 2 ⇢4 m2 ✓ ◆ Finally, we want a geometric description of the trajectory – we don’t particularly care how long it takes the particle to move from one place to another, we’re more interested in the shape of its motion. To obtain a geometric description, re-parametrize in terms of φ instead of t –wehave: d⇢ Jz 2 ⇢˙ = φ˙ = ⇢0 ⇢ (6.16) dφ m using φ˙ = Jz ⇢2 and defining ⇢ d⇢ .Now m 0 ⌘ dφ 2 1 Jz 2 2 H = ⇢0 + ⇢ + U(⇢)=E, (6.17) 2 m and we can take the φ derivative of both sides to get rid of the constant E, then solve for ⇢00: m GMm2 ⇢00 = ⇢ 2 U 0(⇢)= ⇢ + 2 , (6.18) − − Jz − Jz and GMm2 ⇢(φ)=A cos(φ)+B sin(φ)+ 2 . (6.19) Jz 192 CHAPTER 6. GRAVITY We don’t need to keep both the sine and cosine terms, either one suffices, so take GMm2 ⇢ = 2 + A cos φ. (6.20) Jz Then the radius, as a function of angle, is 1 J 2 1 p r = = z (6.21) ⇢ GMm2 1+e cos φ ⌘ 1+e cos φ 2 2 where e A Jz is a new constant and p Jz .Fore =0,this ⌘ GMm2 ⌘ GMm2 6 expression for r(φ) describes an ellipse with “eccentricity” e and “semi-latus rectum” p. The closest point to the central body (situated at the origin) occurs when φ =0, and is called the “perihelion” (technically, that’s the closest point in an orbit about the sun, as the name suggests), and the furthest point from the central body is at φ = ⇡, called the “aphelion”. The details of the orbit provide those values, and from them we can extract information about the central body. 6.3.2 Kepler’s Laws Kepler’s first law is that the motion of bodies around the sun is described by elliptical orbits, and we have demonstrated this directly with the solution (6.21).
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