CHAPTER 5 GRAVITATIONAL FIELD AND POTENTIAL


 Winfred Murphy
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1 CHATER 5 GRAVITATIONAL FIELD AND OTENTIAL 5. Intoduction. This chapte deals with the calculation of gavitational fields and potentials in the vicinity of vaious shapes and sizes of massive bodies. The eade who has studied electostatics will ecognize that this is all just a epeat of what he o she aleady knows. Afte all, the foce of epulsion between two electic chages q and q a distance apat in vacuo is q q 4πε, whee ε is the pemittivity of fee space, and the attactive foce between two masses M and M a distance apat is GM M, whee G is the gavitational constant, o, phased anothe way, the epulsive foce is GM M. Thus all the equations fo the fields and potentials in gavitational poblems ae the same as the coesponding equations in electostatics poblems, povided that the chages ae eplaced by masses and 4πε is eplaced by /G. I can, howeve, think of two diffeences. In the electostatics case, we have the possibility of both positive and negative chages. As fa as I know, only positive masses exist. This means, among othe things, that we do not have gavitational dipoles and all the phenomena associated with polaization that we have in electostatics. The second diffeence is this. If a paticle of mass m and chage q is placed in an electic field E, it will expeience a foce qe, and it will acceleate at a ate and in a diection given by qe/m. If the same paticle is placed in a gavitational field g, it will expeience a foce mg and an acceleation mg/m g, iespective of its mass o of its chage. All masses and all chages in the same gavitational field acceleate at the same ate. This is not so in the case of an electic field. I have some sympathy fo the idea of intoducing a ationalized gavitational constant Γ, given by Γ /(4πG), in which case the gavitational fomulas would look even moe like the SI (ationalized MKSA) electostatics fomulas, with 4π appeaing in poblems with spheical symmety, π in poblems with cylindical symmety, and no π in
2 poblems involving unifom fields. This is unlikely to happen, so I do not pusue the idea futhe hee. 5. Gavitational Field. The egion aound a gavitating body (by which I meely mean a mass, which will attact othe masses in its vicinity) is a gavitational field. Although I have used the wods aound and in its vicinity, the field in fact extents to infinity. All massive bodies (and by massive I mean any body having the popety of mass, howeve little) ae suounded by a gavitational field, and all of us ae immesed in a gavitational field. If a test paticle of mass m is placed in a gavitational field, it will expeience a foce (and, if eleased and subjected to no additional foces, it will acceleate). This enables us to define quantitatively what we mean by the stength of a gavitational field, which is meely the foce expeience by unit mass placed in the field. I shall use the symbol g fo the gavitational field, so that the foce F on a mass m situated in a gavitational field g is F mg. 5.. It can be expessed in newtons pe kilogam, N kg . If you wok out the dimensions of g, you will see that it has dimensions LT, so that it can be expessed equivalently in m s. Indeed, as pointed out in section 5., the mass m (o indeed any othe mass) will acceleate at a ate g in the field, and the stength of a gavitational field is simply equal to the ate at which bodies placed in it will acceleate. Vey often, instead of using the expession stength of the gavitational field I shall use just the gavitational field o pehaps the field stength o even just the field. Stictly speaking, the gavitational field means the egion of space suounding a gavitating mass athe than the field stength, but I hope that, when I am not speaking stictly, the context will make it clea. 5. Newton s Law of Gavitation. Newton noted that the atio of the centipetal acceleation of the Moon in its obit aound the Eath to the acceleation of an apple falling to the suface of the Eath was invesely as the squaes of the distances of Moon and apple fom the cente of the Eath. Togethe with othe lines of evidence, this led Newton to popose his univesal law of gavitation: Evey paticle in the Univese attacts evey othe paticle with a foce that is popotional to the poduct of thei masses and invesely popotional to the squae of the distance between them. In symbols: GMM F. N 5..
3 Hee, G is the Univesal Gavitational Constant. The wod univesal implies an assumption that its value is the same anywhee in the Univese, and the wod constant implies that it does not vay with time. We shall hee accept and adopt these assumptions, while noting that it is a legitimate cosmological question to conside what implications thee may be if eithe of them is not so. Of all the fundamental physical constants, G is among those whose numeical value has been detemined with least pecision. Its cuently accepted value is % N m kg. It is woth noting that, while the poduct GM fo the Sun is known with vey geat pecision, the mass of the Sun is not known to any highe degee of pecision than that of the gavitational constant. Execise. Detemine the dimensions (in tems of M, L and T) of the gavitational constant. Assume that the peiod of pulsation of a vaiable sta depends on its mass, its aveage adius and on the value of the gavitational constant, and show that the peiod of pulsation must be invesely popotional to the squae oot of its aveage density. The gavitational field is often held to be the weakest of the fou foces of natue, but to ave this is to compae incompaables. While it is tue that the electostatic foce between two electons is fa, fa geate than the gavitational foce between them, it is equally tue that the gavitational foce between Sun and Eath is fa, fa geate than the electostatic foce between them. This example shows that it makes no sense meely to state that electical foces ae stonge than gavitational foces. Thus any statement about the elative stengths of the fou foces of natue has to be phased with cae and pecision. 5.4 The Gavitational Fields of Vaious Bodies. In this section we calculate the fields nea vaious shapes and sizes of bodies, much as one does in an intoductoy electicity couse. Some of this will not have much diect application to celestial mechanics, but it will seve as good intoductoy pactice in calculating fields and, late, potentials Field of a oint Mass. Equation 5.., togethe with the definition of field stength as the foce expeienced by unit mass, means that the field at a distance fom a point mass M is In vecto fom, this can be witten as GM g N kg o m s 5.4.
4 4 g GM ˆ N kg o m s 5.4. Hee ˆ is a dimensionless unit vecto in the adial diection. It can also be witten as GM N kg o m s 5.4. g Hee is a vecto of magnitude hence the in the denominato Field on the Axis of a Ring. Befoe stating, one can obtain a qualitative idea of how the field on the axis of a ing vaies with distance fom the cente of the ing. Thus, the field at the cente of the ing will be zeo, by symmety. It will also be zeo at an infinite distance along the axis. At othe places it will not be zeo; in othe wods, the field will fist incease, then decease, as we move along the axis. Thee will be some distance along the axis at which the field is geatest. We ll want to know whee this is, and what is its maximum value. θ FIGURE V. z a z a δm
5 5 Figue V. shows a ing of mass M, adius a. The poblem is to calculate the stength of the gavitational field at. We stat by consideing a small element of the ing of mass δm. The contibution of this element to the field is G δm a z diected fom towads δm. This can be esolved into a component along the axis (diected to the cente of the ing) and a component at ight angles to this. When the contibutions to all elements aound the cicumfeence of the ing ae added, the latte component will, by symmety, be zeo. The component along the axis of the ing is G δm a z G δm cos. z G δm z θ. / a z a z, ( a z ) On adding up the contibutions of all elements aound the cicumfeence of the ing, we find, fo the gavitational field at g GMz ( a z ) / diected towads the cente of the ing. This has the popety, as expected, of being zeo at the cente of the ing and at an infinite distance along the axis. If we expess z in units of a, and g in units of GM/a, this becomes g z ( z ) / This is illustated in figue V.. GM.85GM Execise: Show that the field eaches its geatest value of whee 9a a z a/.77a. Show that the field has half this maximum value whee z.47a and z.896a.
6 6.4 FIGURE V Field Distance fom cente 5.4. lane discs. α θ z FIGURE V.A δ
7 7 Conside a disc of suface density (mass pe unit aea) σ, adius a, and a point on its axis at a distance z fom the disc. The contibution to the field fom an elemental annulus, adii, δ, mass πσ δ is (fom equation 5.4.) z δ δ g πgσ / ( z ) To find the field fom the entie disc, just integate fom to a, and, if the disc is of unifom suface density, σ will be outside the integal sign. It will be easie to integate with espect to θ (fom to α), whee z tan θ. You should get g πgσ( cos α), o, with M πa σ, g ( cos α). GM a Now π( cosα) is the solid angle ω subtended by the disc at. (Convince youself of this don t just take my wod fo it.) Theefoe g G σω This expession is also the same fo a unifom plane lamina of any shape, fo the downwad component of the gavitational field. Fo, conside figue V.. GσδAcos θ The downwad component of the field due to the element δa is Gσδω. Thus, if you integate ove the whole lamina, you aive at Gσω. θ δω δa FIGURE V.
8 8 Retuning to equation 5.4.8, we can wite the equation in tems of z athe than α. If we expess g in units of GM/a and z in units of a, the equation becomes. z g 5.4. z This is illustated in figue V.4. FIGURE V Field Distance fom cente The field is geatest immediately above the disc. On the opposite side of the disc, the field changes diection. In the plane of the disc, at the cente of the disc, the field is zeo. Fo moe on this, see Subsection If you ae calculating the field on the axis of a disc that is not of unifom suface density, but whose suface density vaies as σ(), you will have to calculate M a π σ( ) d 5.4. a σ( ) d and g πgz. / z 5.4. You could ty, fo example, some of the following foms fo σ():
9 9 k k k,,, k σ. σ σ σ a a a a If you ae inteested in galaxies, you might want to ty modelling a galaxy as a cental spheical bulge of density ρ and adius a, plus a disc of suface density σ() and adius a, and fom thee you can wok you way up to moe sophisticated models. In this section we have calculated the field on the axis of a disc. As soon as you move off axis, it becomes much moe difficult. Execise. Stating fom equations 5.4. and 5.4., show that at vay lage distances along the axis, the fields fo a ing and fo a disc each become GM/z. All you have to do is to expand the expessions binomially in a/z. The field at a lage distance fom any finite object will appoach GM/ Infinite lane Laminas. Fo the gavitational field due to a unifom infinite plane lamina, all one has to do is to put α π/ in equation o ω π in equation to find that the gavitational field is g πgσ This is, as might be expected, independent of distance fom the infinite plane. The lines of gavitational field ae unifom and paallel all the way fom the suface of the lamina to infinity. Suppose that the suface density of the infinite plane is not unifom, but vaies with distance in the plane fom some point in the plane as σ(), we have to calculate σ( ) d g πgz / z Ty it, fo example, with σ() being one of the following: σ e k k σ, σ,. e k
10 5.4.5 Hollow Hemisphee. Execise. Find the field at the cente of the base of a hollow hemispheical shell of mass M and adius a Rods. θ h O A x θ α FIGURE V.5 δx B x θ β Conside the od shown in figue V.5, of mass pe unit length λ. The field at due to the element δx is Gλ δx/. But x h tan θ, δx hsec θδθ, hsecθ, so the field at is Gλ δθ / h. This is diected fom to the element δx. The xcomponent of the field due to the whole od is Gλ β sin θdθ h α Gλ ( cos α cosβ). h The ycomponent of the field due to the whole od is Gλ β cos θ dθ h α Gλ ( sinβ sin α). h The total field is the othogonal sum of these, which, afte use of some tigonometic identities (do it!), becomes at an angle ( α β) Gλ g sin ( β α) h  i.e. bisecting the angle AB.
11 If the od is of infinite length, we put α π/ and β π/, and we obtain fo the field at g Gλ h β α h B A A FIGURE V.6 B Conside an ac A B of a cicle of adius h, mass pe unit length λ, subtending an angle β α at the cente of the cicle. Gλ Execise: Show that the field at is g sin ( β α). This is the same as the h field due to the od AB subtending the same angle. If A B is a semicicle, the field at Gλ would be g, the same as fo an infinite od. h
12 An inteesting esult following fom this is as follows. FIGURE V.7 Thee massive ods fom a tiangle. is the incente of the tiangle (i.e. it is equidistant fom all thee sides.) The field at is the same as that which would be obtained if the mass wee distibuted aound the incicle. I.e., it is zeo. The same esult would hold fo any quadilateal that can be inscibed with a cicle such as a cyclic quadilateal Solid Cylinde. We do this not because it has any paticula elevance to celestial mechanics, but because it is easy to do. We imagine a solid cylinde, density ρ, adius a, length l. We seek to calculate the field at a point on the axis, at a distance h fom one end of the cylinde (figue V.8). h z δz l FIGURE V.8
13 The field at fom an elemental disc of thickness δz a distance z below is (fom equation 5.4.9) δg Gρδz ω Hee ω is the solid angle subtended at by the disc, which is the field at fom the entie cylinde is z π / ( z a ). Thus l h z g πgρ dz, h / ( z a ) 5.4. o g πgρ ( l ( l h) a h a ), 5.4. o g πgρ( l ) It might also be of inteest to expess g in tems of the height y l h) of the point above the midpoint of the cylinde. Instead of equation 5.4., we then have g πgρ ( l ( y l) a ( y l) a ) If the point is inside the cylinde,at a distance h below the uppe end of the cylinde, the limits of integation in equation 5.4. ae h and l h, and the distance y is l h. In tems of y the gavitational field at is then g πgρ ( y ( y l) a ( y l) a ) In the gaph below I have assumed, by way of example, that l and a ae both, and I have plotted g in units of πgρ (counting g as positive when it is diected downwads) fom y to y. The potion inside the cylinde y l ), epesented by ( equation 5.4.4, is almost, but not quite, linea. The field at the cente of the cylinde is, of couse, zeo. (
14 4.5 Above the cylinde.4... Inside the cylinde g Below the cylinde y Below, I daw the same gaph, but fo a thin disc, with a and l.. We see how it is that the field eaches a maximum immediately above o below the disc, but is zeo at the cente of the disc.
15 g y Hollow Spheical Shell. a ξ O δx x θ FIGURE V.9
16 6 We imagine a hollow spheical shell of adius a, suface density σ, and a point at a distance fom the cente of the sphee. Conside an elemental zone of thickness δx. The mass of this element is πaσ δx. (In case you doubt this, o you didn t know, the aea of a zone on the suface of a sphee is equal to the coesponding aea pojected on to the cicumscibing cylinde.) The field due to this zone, in the diection O is πaσg cos θδx. ξ Let s expess this all in tems of a single vaiable, ξ. We ae going to have to expess x and θ in tems of ξ. We have a ξ ξcos θ ξ x, fom which a ξ cosθ and ξ δx ξδξ. πagσ a Theefoe the field at due to the zone is δξ.. ξ If is an extenal point, in ode to find the field due to the entie spheical shell, we integate fom ξ a to a. This esults in GM g But if is an intenal point, in ode to find the field due to the entie spheical shell, we integate fom ξ a to a, which esults in g. Thus we have the impotant esult that the field at an extenal point due to a hollow spheical shell is exactly the same as if all the mass wee concentated at a point at the cente of the sphee, wheeas the field inside the sphee is zeo. Caution. The field inside the sphee is zeo only if thee ae no othe masses pesent. The hollow sphee will not shield you fom the gavitational field of any othe masses that might be pesent. Thus in figue V., the field at is the sum of the field due to the hollow sphee (which is indeed zeo) and the field of the mass M, which is not zeo. Antigav is a useful device in science fiction, but does not occu in science fact.
17 7 M. FIGURE V Solid Sphee. A solid sphee is just lots of hollow sphees nested togethe. Theefoe, the field at an extenal point is just the same as if all the mass wee concentated at the cente, and the field at an intenal point is the same is if all the mass inteio to, namely M, wee concentated at the cente, the mass exteio to not contibuting at all to the field at. This is tue not only fo a sphee of unifom density, but of any sphee in which the density depends only of the distance fom the cente i.e., any spheically symmetic distibution of matte. M If the sphee is unifom, we have, so the field inside is M a GM GM g a Thus, inside a unifom solid sphee, the field inceases linealy fom zeo at the cente to GM / a at the suface, and theeafte it falls off as GM /. If a unifom hollow sphee has a naow hole boed though it, and a small paticle of mass m is allowed to dop though the hole, the paticle will expeience a foce towads the cente of GMm / a, and will consequently oscillate with peiod given by 4 π a GM
18 Bubble Inside a Unifom Solid Sphee. c FIGURE V. is a point inside the bubble. The field at is equal to the field due to the entie sphee minus the field due to the missing mass of the bubble. That is, it is g 4 ( ) π. 4 ( 4 ) 4 πg ρ c πgρ πgρ Gρ That is, the field at is unifom (i.e. is independent of the position of ) and is paallel to the line joining the centes of the two sphees.
19 9 5.5 Gauss s Theoem. Much of the above may have been good integation pactice, but we shall now see that many of the esults ae immediately obvious fom Gauss s Theoem itself a tivially obvious law. (O shall we say that, like many things, it is tivially obvious in hindsight, though it needed Cal Fiedich Gauss to point it out!) Fist let us define gavitational flux Φ as an extensive quantity, being the poduct of gavitational field and aea: FIGURE V. g δa δφ g δa If g and δa ae not paallel, the flux is a scala quantity, being the scala o dot poduct of g and δa: δa FIGURE V. g δφ g δa If the gavitational field is theading though a lage finite aea, we have to calculate g δa fo each element of aea of the suface, the magnitude and diection of g possibly vaying fom point to point ove the suface, and then we have to integate this all ove the suface. In othe wods, we have to calculate a suface integal. We ll give some examples as we poceed, but fist let s move towad Gauss s theoem. In figue V.4, I have dawn a mass M and seveal of the gavitational field lines conveging on it. I have also dawn a sphee of adius aound the mass. At a distance fom the mass, the field is GM/. The suface aea of the sphee is 4π. Theefoe the total inwad flux, the poduct of these two tems, is 4πGM, and is independent of the size of the sphee. (It is independent of the size of the sphee because the field falls off invesely as the squae of the distance. Thus Gauss s theoem is a theoem that applies to invese squae fields.) Nothing changes if the mass is not at the cente of the sphee. No does it change if (figue V.5) the suface is not a sphee. If thee wee seveal
20 masses inside the suface, each would contibute 4πG times its mass to the total nomal inwads flux. Thus the total nomal inwad flux though any closed suface is equal to 4πG times the total mass enclosed by the suface. O, expessed anothe way: M FIGURE V.4 The total nomal outwad gavitational flux though a closed suface is equal to 4πG times the total mass enclosed by the suface. This is Gauss s theoem. Mathematically, the flux though the suface is expessed by the suface integal g d If thee is a continuous distibution of matte inside the suface, of density ρ A. which vaies fom point to point and is a function of the coodinates, the total mass inside the suface is expessed by ρ dv. Thus Gauss s theoem is expessed mathematically by
21 da You should check the dimensions of this equation. g 4πG ρ dv M FIGURE V.5 V. In figue V.6 I have dawn gaussian spheical sufaces of adius outside and inside hollow and solid sphees. In a and c, the outwad flux though the suface is just 4πG times the enclosed mass M; the suface aea of the gaussian suface is 4π. This the outwad field at the gaussian suface (i.e. at a distance fom the cente of the sphee is GM/. In b, no mass is inside the gaussian suface, and theefoe the field is zeo. In d, the mass inside the gaussian suface is M, and so the outwad field is GM /.
22 a b d c FIGURE V.6
23 In figue V.7 I daw (pat of an) infinite od of mass λ pe unit length, and a cylindical gaussian suface of adius h and length l aound it. h l FIGURE V.7 The suface aea of the cuved suface of the cylinde is πhl, and the mass enclosed within it is λl. Thus the outwad field at the suface of the gaussian cylinde (i.e. at a distance h fom the od) is 4πG λl πhl Gλ/h, in ageement with equation In figue V.8 I have dawn (pat of) an infinite plane lamina of suface density σ, and a cylindical gaussian suface o cosssectional aea A and height h. A h FIGURE V.8
24 4 The mass enclosed by the cylinde is σa and the aea of the two ends of the cylinde is A. The outwad field at the ends of the cylinde (i.e. at a distance h fom the plane lamina) is theefoe 4πG σa A πgσ, in ageement with equation Calculating Suface Integals. While the concept of a suface integal sounds easy enough, how do we actually calculate one in pactice? In this section I do two examples. Example. δz m θ g δa FIGURE V.9 In figue V.9 I show a small mass m, and I have suounded it with a cylinde o adius a and height h. The poblem is to calculate the suface integal g.da though the entie
25 5 suface of the cylinde. Of couse we aleady know, fom Gauss s theoem, that the answe is 4πGm, but we would like to see a suface integal actually caied out. I have dawn a small element of the suface. Its aea δa is dz times aδφ, whee φ is the usual azimuthal angle of cylindical codinates. That is, δa a δz δφ. The magnitude g of the field thee is Gm/, and the angle between g and da is 9 o θ. The outwad flux o Gma cos( θ 9 ) δz δφ though the small element is g δa. (This is negative i.e. it is actually an inwad flux because cos (θ 9 o ) sin θ.) When integated aound the πgma sin θδz elemental stip δz, this is. To find the flux ove the total cuved suface, let s integate this fom z to h and double it, o, easie, fom θ π/ to α and double it, whee tan α a/h. We ll need to expess z and in tems of θ (that s easy: z a cot θ and a csc θ),and the integal becomes α π/ 4πGm sin θdθ 4πGm cos α Let us now find the flux though one of the flat ends of the cylinde. δa ρ g h θ m FIGURE V.
26 6 This time, δa ρ δρ δφ, g Gm/ and the angle between g and δa is 8 o θ. The o Gmρ ( ) outwads flux though the small element is cos 8 θ δρ δφ and when integated πgm cos θρδρ aound the annulus this becomes. We now have to integate this fom ρ to a, o, bette, fom θ to α. We have h sec θ and ρ h tan θ, and the integal becomes πgm α sin θdθ πgm( cos α) Thee ae two ends, so the total flux though the entie cylinde is twice this plus equation 5.6. to give as expected fom Gauss s theoem. Φ 4πGm, 5.6. Example. g δa FIGURE V. θ
27 7 In figue V. I have dawn (pat of) an infinite od whose mass pe unit length is λ. I have dawn aound it a sphee of adius a. The poblem will be to detemine the total nomal flux though the sphee. Fom Gauss s theoem, we know that the answe must be 8πGαλ. The vecto δa epesenting the element of aea is diected away fom the cente of the sphee, and the vecto g is diected towads the neaest point of the od. The angle between them is θ 9º. The magnitude of δa in spheical coodinates is Gλ a sin θδθδφ, and the magnitude of g is (see equation 5.4.5). The dot poduct g δa is a sin θ Gλ. a a sin θ o sin θδθδφ.cos( θ 9 ) Gλa sin θδθδφ To find the total flux, this must be integated fom φ to π and fom θ to π. The esult, as expected, is 8πGαλ. 5.7 otential. If wok is equied to move a mass fom point A to point B, thee is said to be a gavitational potential diffeence between A and B, with B being at the highe potential. The wok equied to move unit mass fom A to B is called the potential diffeence between A and B. In SI units it is expessed in J kg. We have defined only the potential diffeence between two points. If we wish to define the potential at a point, it is necessay abitaily to define the potential at a paticula point to be zeo. We might, fo example define the potential at floo level to be zeo, in which case the potential at a height h above the floo is gh; equally we may elect to define the potential at the level of the laboatoy bench top to be zeo, in which case the potential at a height z above the bench top is gz. Because the value of the potential at a point depends on whee we define the zeo of potential, one often sees that the potential at some point is equal to some mathematical expession plus an abitay constant. The value of the constant will be detemined once we have decided whee we wish to define zeo potential. In celestial mechanics it is usual to assign zeo potential to all points at an infinite distance fom any bodies of inteest. Suppose we decide to define the potential at point A to be zeo, and that the potential at B is then ψ J kg. If we move a point mass m fom A to B, we shall have to do an amount of wok equal to mψ J. The potential enegy of the mass m when it is at B is then mψ. In these notes, I shall usually use the symbol ψ fo the potential at a point, and the symbol V fo the potential enegy of a mass at a point.
28 8 In moving a point mass fom A to B, it does not matte what oute is taken. All that mattes is the potential diffeence between A and B. Foces that have the popety that the wok equied to move fom one point to anothe is outeindependent ae called consevative foces; gavitational foces ae consevative. The potential at a point is a scala quantity; it has no paticula diection associated with it. If it equies wok to move a body fom point A to point B (i.e. if thee is a potential diffeence between A and B, and B is at a highe potential than A), this implies that thee must be a gavitational field diected fom B to A. g A. B δx. FIGURE V. Figue V. shows two points, A and B, a distance δx apat, in a egion of space whee the gavitational field is g diected in the negative x diection. We ll suppose that the potential diffeence between A and B is δψ. By definition, the wok equied to move unit mass fom A to B is δψ. Also by definition, the foce on unit mass is g, so that the wok done on unit mass is gδx. Thus we have g dψ dx The minus sign indicates that, while the potential inceases fom left to ight, the gavitational field is diected to the left. In wods, the gavitational field is minus the potential gadient. This was a onedimensional example. In a late section, when we discuss the vecto opeato, we shall wite equation 5.7. in its theedimensional fom g gadψ ψ While ψ itself is a scala quantity, having no diectional popeties, its gadient is, of couse, a vecto.
29 9 5.8 The Gavitational otentials Nea Vaious Bodies. Because potential is a scala athe than a vecto, potentials ae usually easie to calculate than field stengths. Indeed, in ode to calculate the gavitational field, it is sometimes easie fist to calculate the potential and then to calculate the gadient of the potential otential Nea a oint Mass. We shall define the potential to be zeo at infinity. If we ae in the vicinity of a point mass, we shall always have to do wok in moving a test paticle away fom the mass. We shan t each zeo potential until we ae an infinite distance away. It follows that the potential at any finite distance fom a point mass is negative. The potential at a point is the wok equied to move unit mass fom infinity to the point; i.e., it is negative. M x x δx FIGURE V. The magnitude of the field at a distance x fom a point mass M (figue V.) is GM/x, and the foce on a mass m placed thee would be GMm/x. The wok equied to move m fom x to x δx is GMmδx/x. The wok equied to move it fom to infinity dx GMm is GMm. The wok equied to move unit mass fom to, which is the x potential at is GM ψ The mutual potential enegy of two point masses a distance apat, which is the wok equied to bing them to a distance fom an infinite initial sepaation, is V GMm I hee summaize a numbe of similalooking fomulas, although thee is, of couse, not the slightest possibility of confusing them. Hee goes:
30 Foce between two masses: GMm F. N 5.8. Field nea a point mass: GM g, N kg o m s which can be witten in vecto fom as: g GM ˆ N kg o m s GM o as: g. N kg o m s Mutual potential enegy of two masses: V GMm. J otential nea a point mass: GM ψ. J kg I hope that s cystal clea otential on the Axis of a Ring. We can efe to figue V.. The potential at fom the element δm is G δm. This ( a z ) is the same fo all such elements aound the cicumfeence of the ing, and the total potential is just the scala sum of the contibutions fom all the elements. Theefoe the total potential on the axis of the ing is: ( a z ) / GM ψ /
31 The zcomponent of the field (its only component) is d/dz of this, which esults in GMz g. This is the same as equation 5.4. except fo sign. When we deived / ( a z ) equation 5.4. we wee concened only with the magnitude of the field. Hee dψ/dz gives the zcomponent of the field, and the minus sign coectly indicates that the field is diected in the negative zdiection. Indeed, since potential, being a scala quantity, is easie to wok out than field, the easiest way to calculate a field is fist to calculate the potential and then diffeentiate it. On the othe hand, sometimes it is easy to calculate a field fom Gauss s theoem, and then calculate the potential by integation. It is nice to have so many easy ways of doing physics! 5.8. lane Discs. Refe to figue V.A. The potential at fom the elemental disc is GδM πg σ δ dψ ( z ) / ( z ) The potential fom the whole disc is theefoe / ψ a d πg σ ( z ) / The integal is tivial afte a billiant substitution such as and we aive at X z o z tan θ, ψ πgσ( z a z) This inceases to zeo as z. We can also wite this as ψ πgm. a z πa z / z, 5.8. and, if you expand this binomially, you see that fo lage z it becomes, as expected, Gm/z.
32 5.8.4 Infinite lane Lamina. The field above an infinite unifom plane lamina of suface density σ is πgσ. Let A be a point at a distance a fom the lamina and B be a point at a distance b fom the lamina (with b > a), the potential diffeence between B and A is ψ ψ πg σ( b a) B A If we elect to call the potential zeo at the suface of the lamina, then, at a distance h fom the lamina, the potential will be πgσh Hollow Hemisphee. Any element of mass, δm on the suface of a hemisphee of adius a is at a distance a fom the cente of the hemisphee, and theefoe the potential due to this element is meely G δm/a. Since potential is a scala quantity, the potential of the entie hemisphee is just GM/a Rods. Refe to figue V.5. The potential at due to the element δx is Gλ δx Gλsecθδθ. The total potential at is theefoe ψ sec sec β β β G λ secθ dθ Gλ ln. α sec sec α α α β A l FIGURE V.4 B
33 Refe now to figue V.4, in which A 9 o α and B 9 o β. secβ tanβ secα secα cos α cosβ ( sinβ) ( sin α) sin A sin B ( cos B) sin Acos A.cos B ( cos A) sin B cos B.sin A cot Acot B s( s ). ( s )( s l) s( s ) ( s l)( s ), whee s ( l ). (You may want to efe hee to the fomulas on pp. 7 and 8 of Chapte.) l Hence ψ G λ ln l If and ae vey lage compaed with l, they ae nealy equal, so let s put and wite equation as ψ Gm ln l l l Gm ln l l ln l. Maclauin expand the logaithms, and you will see that, at lage distances fom the od, the potential is, expected, Gm/. Let us etun to the nea vicinity of the od and to equation We see that if we move aound the od in such a manne that we keep constant and equal to a, say that is to say if we move aound the od in an ellipse (see ou definition of an ellipse in Chapte, Section.) the potential is constant. In othe wods the equipotentials ae confocal ellipses, with the foci at the ends of the od. Equation can be witten ψ a l ln G λ a l Fo a given potential ψ, the equipotential is an ellipse of majo axis ψ /( Gλ) e a l, /( 5.8. ψ Gλ e ) whee l is the length of the od. This knowledge is useful if you ae exploing space and you encounte an alien spacecaft o an asteoid in the fom of a unifom od of length l.
34 Solid Cylinde. Refe to figue V.8. The potential fom the elemental disc is dψ [( z a ) z] / and theefoe the potential fom the entie cylinde is πgρδz 5.8. h l ψ. π ρ ( ) / h l G z a dz z dz. const 5.8. h h I leave it to the eade to cay out this integation and obtain a final expession. One way to deal with the fist integal might be to ty z a tan θ. This may lead to sec θd θ. Fom thee, you could ty something like sec θ secθ tan θ secθd tan θ secθ tan θ sec θ secθdθ,andsoon. tan θd sec θ secθ tan θ sec θ tan θdθ Hollow Spheical Shell. Outside the sphee, the field and the potential ae just as if all the mass wee concentated at a point in the cente. The potential, then, outside the sphee, is just GM/. Inside the sphee, the field is zeo and theefoe the potential is unifom and is equal to the potential at the suface, which is GM/a. The eade should daw a gaph of the potential as a function of distance fom cente of the sphee. Thee is a discontinuity in the slope of the potential (and hence in the field) at the suface Solid Sphee. a x δx FIGURE V.4A M δμ
35 5 The potential outside a solid sphee is just the same as if all the mass wee concentated at a point in the cente. This is so, even if the density is not unifom, and long as it is spheically distibuted. We ae going to find the potential at a point inside a unifom sphee of adius a, mass M, density ρ, at a distance fom the cente ( < a). We can do this in two pats. Fist, thee is the potential fom that pat of the sphee below. This M is GM /, whee M is the mass within adius. Now we need to deal with a the mateial above. Conside a spheical shell of adii x, x δx. Its mass is 4π x δx. Mx δx δ M M. The potential fom this shell is 4 πa a GMx δx G δm/ x. This is to be integated fom x to a, and we must then add a the contibution fom the mateial below. The final esult is GM ψ ( a ) a Figue V.5 shows the potential both inside and outside a unifom solid sphee. The potential is in units of GM/, and distance is in units of a, the adius of the sphee. FIGURE V otential GM a GM (hypebola) . GM a ( a ) (paabola) .4 GM a Distance fom cente
36 6 5.9 Wok Requied to Assemble a Unifom Sphee. Let us imagine a unifom solid sphee of mass M, density ρ and adius a. In this section we ask ouselves, how much wok was done in ode to assemble togethe all the atoms that make up the sphee if the atoms wee initially all sepaated fom each othe by an infinite distance? Well, since massive bodies (such as atoms) attact each othe by gavitational foces, they will natually eventually congegate togethe, so in fact you would have to do wok in disassembling the sphee and emoving all the atoms to an infinite sepaation. To bing the atoms togethe fom an infinite sepaation, the amount of wok that you do is negative. Let us suppose that we ae pat way though the pocess of building ou sphee and that, 4 at pesent, it is of adius and of mass M π ρ. The potential at its suface is GM G 4π ρ. 4 π G. The amount of wok equied to add a laye of thickness δ and mass 4π ρ δ to this is πg 4π ρδ π Gρ δ. The wok done in assembling the entie sphee is the integal of this fom to a, which is 5 6π Gρ a GM a 5. Nabla, Gadient and Divegence. We ae going to meet, in this section, the symbol. In Noth Ameica it is geneally ponounced del, although in the United Kingdom and elsewhee one sometimes heas the altenative ponunciation nabla, called afte an ancient Assyian haplike instument of appoximately that shape. In section 5.7, paticulaly equation 5.7., we intoduced the idea that the gavitational field g is minus the gadient of the potential, and we wote g dψ/dx. This equation efes to an essentially onedimensional situation. In eal life, the gavitational potential is a thee dimensional scala function ψ(x, y, z), which vaies fom point to point, and its gadient is ψ ψ ψ gad ψ i j k, 5.. x y x
37 7 which is a vecto field whose magnitude and diection vay fom point to point. The gavitational field, then, is given by g gad ψ. 5.. Hee, i, j and k ae the unit vectos in the x, y and zdiections. The opeato is i j k, so that equation 5.. can be witten x y x g ψ. 5.. I suppose one could wite a long book about, but I am going to ty to estict myself in this section to some bae essentials. Let us suppose that we have some vecto field, which we might as well suppose to be a gavitational field, so I ll call it g. (If you don t want to be esticted to a gavitational field, just call the field A as some sot of undefined o geneal vecto field.) We can calculate the quantity.g i j k ig g g. x x x x j y k z 5..4 When this is multiplied out, we obtain a scala field called the divegence of g: g g x y g z.g div g. x y z 5..5 Is this of any use? Hee s an example of a possible useful application. Let us imagine that we have some field g which vaies in magnitude and diection though some volume of space. Each of the components, g x, g y, g z can be witten as functions of the coodinates. Now suppose that we want to calculate the suface integal of g though the closed bounday of the volume of space in question. Can you just imagine what a headache that might be? Fo example, suppose that g x i xyj xzk, and I wee to ask you to calculate the suface x y z integal ove the suface of the ellipsoid. It would be had to know a b c whee to begin. Well, thee is a theoem, which I am not going to deive hee, but which can be found in many books on mathematical physics, and is not paticulaly difficult, which says: The suface integal of a vecto field ove a closed suface is equal to the volume integal of its divegence.
38 . 8 In symbols: g da div g dv If we know g x, g y and g z as functions of the coodinates, then it is often vey simple and staightfowad to calculate the divegence of g, which is a scala function, and it is then often equally staightfowad to calculate the volume integal. The example I gave in the pevious paagaph is tivially simple (it is a athe atificial example, designed to be idiculously simple) and you will eadily find that div g is eveywhee zeo, and so the suface integal ove the ellipsoid is zeo. If we combine this vey geneal theoem with Gauss s theoem (which applies to an invese squae field), which is that the suface integal of the field ove a closed volume is equal to 4πG times the enclosed mass (equation 5.5.) we undestand immediately that the divegence of g at any point is elated to the density at that point and indeed that div g.g 4πGρ This may help to give a bit moe physical meaning to the divegence. At a point in space whee the local density is zeo, div g, of couse, is also zeo. Now equation 5.. tells us that g ψ, so that we also have.( ψ).(ψ) 4πGρ If you wite out the expessions fo and fo ψ in full and calculate the dot poduct, you will find that.(ψ), which is also witten ψ ψ ψ ψ, is. Thus we x y z obtain ψ ψ ψ ψ x y z 4πGρ This is oisson s equation. At any point in space whee the local density is zeo, it becomes ψ 5.. which is Laplace s equation. Thus, no matte how complicated the distibution of mass, the potential as a function of the coodinates must satisfy these equations. We leave this topic hee. Futhe details ae to be found in books on mathematical physics; ou aim hee was just to obtain some feeling fo the physical meaning. I add just a few small comments. One is, yes, it is cetainly possible to opeate on a vecto field with the opeato %. Thus, if A is a vecto field, %A is called the cul of A. The cul of a gavitational field is zeo, and so thee is no need fo much discussion of it in a chapte on gavitational fields. If, howeve, you have occasion to study fluid dynamics o
39 9 electomagnetism, you will need to become vey familia with it. I paticulaly daw you attention to a theoem that says The line integal of a vecto field aound a closed plane cicuit is equal to the suface integal of its cul. This will enable you easily to calculate twodimensional line integals in a simila manne to that in which the divegence theoem enables you to calculate theedimensional suface integals. Anothe comment is that vey often calculations ae done in spheical athe than ectangula coodinates. The fomulas fo gad, div, cul and ae then athe moe complicated than thei simple foms in ectangula coodinates. Finally, thee ae dozens and dozens of fomulas elating to nabla in the books, such as cul cul gad div minus nablasquaed. While they should cetainly neve be memoized, they ae cetainly woth becoming familia with, even if we do not need them immediately hee. 5. Legende olynomials. In this section we cove just enough about Legende polynomials to be useful in the following section. Befoe stating, I want you to expand the following expession, by the binomial theoem, fo x <, up to x 4 : ( x cos θ x ) /. 5.. lease do go ahead and do it. Well, you pobably won t, so I d bette do it myself: I ll stat with 5 5 ( X ) / X X X X 4 K and theefoe [ x( cos θ x) ] / x( cos θ x) ( ) 8 x cos θ x 5 6 x ( cos θ x) x 4 ( cos θ x) 4 K
40 4 x cos θ x 5 8 ( 4cos θ 4x cos θ x ) ( 8cos θ x cos θ 6x cos θ x ) x x 4 4 x( 6cos θ x cos θ 4x cos θ 8x cos θ x )K 5..4 x cos θ x 5 ( cos θ) x ( cos θ cos θ) x ( cos θ cos θ)k The coefficients of the powes of x ae the Legende polynomials l (cos θ ), so that ( x cos θ x ) / x 4 ( cos θ) x ( cos θ) x ( cos θ) x ( cos θ) K The Legende polynomials with agument cos θ can be witten as seies of tems in powes of cos θ by substitution of cos θ fo x in equations..5 in Section. of Chapte. Note that x in Section is not the same as x in the pesent section. Altenatively they can be witten as seies of cosines of multiples of θ as follows cos θ 4 8 (5cos θ 64 (cos θ ) cos θ) (5cos 4θ cos θ 9) (6cos 5θ 5cos θ cos θ) (cos 6θ 6cos 4θ 5cos θ 5) (49cos7θ cos 5θ 89cosθ 75cos θ) (645cos8θ 4cos 6θ 77cos 4θ 5cos θ 5) / Fo example, 6 (cos θ) can be witten eithe as given by equation 5..7, o as given by equation, namely 6 4 (c 5c 5c 5), whee c cos θ
41 4 The fome may look neate, and the latte may look awkwad because of all the powes. Howeve, the latte is fa faste to compute, paticulaly when witten as nested paentheses: ( 5 C(5 C( 5 C ))) /6, whee C cos θ 5. Gavitational otential of any Massive Body. You might just want to look at Chapte of Classical Mechanics (Moments of Inetia) befoe poceeding futhe with this chapte. In figue VIII.6 I daw a massive body whose cente of mass is C, and an extenal point at a distance R fom C. I daw a set of Cxyz axes, such that is on the zaxis, the coodinates of being (,, z). I indicate an element δm of mass, distant fom C and l fom. I ll suppose that the density at δm is ρ and the volume of the mass element is δτ, so that δm ρδτ. z R l C θ δm * y x FIGURE V.6
42 4 The potential at is. τ ρ ψ l d G l dm G 5.. But, cos θ R R l so. ) (cos ) (cos cos 4 τ θ ρ τ θ ρ τ θ ρ τ ρ ψ K d R d R d R d R G 5.. The integal is to be taken ove the entie body, so that τ ρ M d, whee M is the mass of the body. Also τ θ ρ dm z d cos, which is zeo, since C is the cente of mass. The thid tem is ( ) ( ). sin cos τ θ ρ τ θ ρ d R d R 5.. Now ( ) ( ) ( ) [ ], C B A dm y x x z z y dm d τ ρ whee A, B and C ae the second moments of inetia with espect to the axes Cx, Cy, Cz espectively. But A B C is invaiant with espect to otation of axes, so it is also equal to A B C, whee A, B, C ae the pincipal moments of inetia. Lastly, τ θ ρ d sin is equal to C, the moment of inetia with espect to the axis Cz. Thus, if R is sufficiently lage than so that we can neglect tems of ode (/R) and highe, we obtain. ) ( R C C B A MR GM ψ 5..4 In the special case of an oblate symmetic top, in which A B < C, and the line C makes an angle γ with the pincipal axis, we have ( ) ( ), / cos R Z A C A A C A C γ 5..5 so that. ψ R Z R A C M R G 5..6
43 4 Now conside a unifom oblate spheoid of pola and equatoial diametes c and a espectively. It is easy to show that (Execise: Show it.) It is slightly less easy to show (Execise: Show it.) that C Ma A M ( a ) c Fo a symmetic top, the integals of the odd polynomials of equation 5.. ae zeo, and the potential is geneally witten in the fom ψ GM R 4 J J 4 4 a R a (cos γ) R (cos γ) L 5..9 Hee γ is the angle between C and the pincipal axis. Fo a unifom oblate spheoid, C A J. This esult will be useful in a late chapte when we discuss pecession. Mc 5. essue at the Cente of a Unifom Sphee What is the pessue at the cente of a sphee of adius a and of unifom density ρ? (eliminay thought: Show by dimensional analysis that it must be something times Gρ a.) δ δ Conside a potion of the sphee between adii and δ and cosssectional aea A. Its volume is Aδ and its mass is ρaδ. (Wee the density not unifom thoughout the sphee, we would hee have to wite ρ()aδ. ) Its weight is ρgaδ, whee 4 g GM πgρ. We suppose that the pessue at adius is and the pessue / FIGURE V.7 at adius δ is δ. (δ is negative.) Equating the downwad foces to the upwad foce, we have A ( δ) 4 πagρ δ A. 5..
44 44 That is: δ 4 πgρ δ. 5.. Integate fom the cente to the suface: a 4 d πgρ d. 5.. Thus: πgρ a. 5..4
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