57Lecture 7 Tuesday Feb 10, 2015 Gravitational Force and Gravitational PotentialUp until now our attention has been on particle trajectories subject to specified forces. Before treatingthe important case of planetary motion in the sun's gravitational field, we take a bit of time with somegeneral considerations on gravitational forces on long length scales like those across the solar systemsuch that g ≠ constant, a topic that may be called Newtonian gravitational field theory. Later we willturn to the question of planetary orbits in such fields and Kepler’s laws.We are perhaps familiar with Newton’s law of gravity, which says that the gravitational force betweentwo objects of masses m1 and m2 is (for point masses or for masses that are a large distance away fromeach other compared to the extent of their distributions) rF = −Gm1m2r2ˆr = −Gm1m2r3rrHere G is the universal gravitational constant: G = 6.67x10-11 N-m2/kg2. The ˆr indicates the direction ofF is along the line connecting the centers of the two masses (defined by the vector r.) The minus signindicates that the force is always attractive,( Aside: The force on m1 due to m2 is Gm1m2r2 in the direction towards m2 , where m1 is the passivegravitational mass of particle 1 and m2 is the active gravitational mass of particle number 2. Thedistinction is purely conceptual; all of us believe the two types of gravitational mass are indistinguishable.NB; in the HW you are to consider the case that one of the masses is negative (BTW: no negative massparticle has ever been observed); this will reverse the sign of the force. But watch out: if the inertia of oneof the masses is also negative, then its acceleration gains an additional negative sign.We note in passing, that the formula is consistent with Newton's third law, that the forces on the twomasses are equal and opposite.The relation is valid for masses with nophysical extent (point masses) and for caseswhere the distance between the masses islarge compared to their physical size. Thatthe result is also true for spheres where r isthe distance between their centers is notobvious ( and anecdote says that to prove thisis why Newton invented integral calculus.)That this is indeed the case is provedbelow…If the mass(es) are distributed in space, thenwe have to integrate the force over the massdistribution. (assuming that thegravitational field from a set of masses ismerely the sum of the fields from each massseparately) For example, consider a point test mass m2 at position r2 and a distributed mass m1. (Seesimilar to Coulomb's law ofelectostrostatics: F ∝ q1q2/ r258figure) Then the force (on m2) is rF = −Gm2ρ1(rr1)V1∫rr2−rr1|rr2−rr1|3d3rr1(†)where ρ1 is the mass density of m1 and we have used the vector identity ˆr / r2=rr / r3.Inasmuch as the integrand is a vector, doing the integral could be complicated.Also we note the total mass is m1=ρ1(rr1)V1∫d3rr1The total force on m1 is of course equal and opposite to this.The gravitational acceleration vector at the point r2 due to the distributed mass m1 is eqn † withoutthe factor of m2: rg(rr2) = −Gρ1(rr1)V1∫rr2−rr1|rr2−rr1|3d3rr1(*)The curl of this vector field vanishes (perhaps not obvious) so we may define a potential energyper unit mass m2 by rg(rr ) = −r∇Φ(rr ) analogous to rF(rr ) = −r∇V (rr )For a field source point mass m1 at the origin, this corresponds to the familiar ( with the potentialset to zero at ∞) Φ(rr ) =PE of m2m2= −Gm1rwhich may be integrated to obtain the case of adistributed mass m1: Φ(rr ) = −GV1∫ρ1(rr1)|rr −rr1|d3rr1This integral for Φ is easier than the corresponding integral for g, labeled (*), because it is theintegral of a scalar, and not a vector.Remark: If m2 were distributed also , we would have to integrateover its mass distribution as well. The force on the total m2 is: rF = −Gρ2(rr2)V2∫ρ1(rr1)V1∫rr2−rr1|rr2−rr1|3d3rr1d3rr2The total force on m1 is of course equal and opposite to this.59---Evaluation of the above integral is in generaldifficult, but for the common case of a sphericaldistribution and for a disk (and for the HW3B.1 ofa pencil distribution ) we can do it. It willtranspire that a spherical distribution of mass hasmuch the same gravitation as a concentrated pointmass. ….Consider a uniform shell of mass M, infinitesimalthickness h, and radius R, centered on the origin.It has mass density ρ = M/ (4π R2h). We place atest mass at a distance x from the center of theshell. Without loss of generality (wo/log) weplace it a distance x > 0 above the center. Thefigure illustrates the case x > R, but we will also beinterested in x < R.The net gravitational potential at x is thesum of contributions from many infinitesimal masselements dm of the shell, each giving a potential–G dm/s, with dm = ρ h Rdθ Rsinθ dφ. ThusΦ(x) = −GρhR2θ= 0θ=π∫φ= 0φ= 2π∫1s(θ)sinθdφdθwhere s(θ) is defined by the figure and is thedistance from dm to the test mass.The φ integration is easy, just giving a factor of 2π. This leaves us with only the θ integration stillto do:Φ(x) = −2πGρhR2θ= 0θ=π∫1s(θ)sinθdθTo do the remaining integral we need s as a function of θ: Law of cosines:s2= R2+ x2− 2Rx cosθThe differential of this is2s ds = 2 R x sinθdθwhich implies60 sinθdθs=dsRxSoΦ(x) = −2πGρhxR dss mins max∫= −2πGρhxR(smax− smin)• If the point x is outside the shell ( x > R ) then smax is x+R and smin is x-R. ThusΦ(x) = −4πGρhR2x( for x > R )But the total shell mass is M = 4π ρh R2, so this is Φ = -GM/x. Just as if all the mass wereconcentrated at the center!------------------• If the point x is inside the shell ( x < R ) then smin is R-x and smax is R+x. ThusΦ(x) = −4πGρhR = −GM / Rindependent of x!The gravitational field g outside a uniform shell isas if all the mass were at its center.The gravitational potential Φ inside the shell is thesame everywhere inside.So the gravitational field rg inside the shell is zero.( But the potential Φ is not zero! )61-----------It is straightforward to extend this from the case of a uniform shell to the case of a sphericalvolume like a planet (as long as its mass density does not vary with angle) by superposing theresult for many nested shells. Outside the planet, the g field is as if all the mass were at its center.g = -GM/r2. Inside the planet at a distance y from