Chapte 5 Gavitation In this Chapte we will eview the popeties of the gavitational foce. The gavitational foce has been discussed in geat detail in you intoductoy physics couses, and we will pimaily focus on specifying the popeties of the foce and its associated foce field, using the vecto notions we have intoduced in the pevious Chaptes. The Gavitational Foce The gavitational foce between two point-like paticles is popotional to the poduct of thei masses, and invesely popotional to the squae of the distance between them. The foce is always attactive, and diected along the line connection the two paticles. F = G mm ˆ The constant G is the gavitational constant, whose value is 6.673 x 10-11 Nm 2 /kg 2. This elation is known as Newton s law of univesal gavitation. The pinciple of supeposition allows us to calculate the foce on mass m due to multiple othe point-like masses M 1, M 2, : F = Gm n i=1 M i i 2 ˆ i whee i is the distance between mass M i and mass m. The pevious elation is coect only if both masses ae point-like objects. If one of the masses is a continuous, we must eplace the sum with an integal: F = Gm ( ) ˆ ' ' 2 ρ ' dv' The Gavitational Field The gavitational field geneated by the gavitational foce is defined in much the same way as we defined the electostatic field geneated by the electostatic foce: g = F m = G ( ) ˆ ' ' 2 ρ ' dv' - 1 -
The units of the gavitational field ae N/kg = m/s 2. The Gavitational Potential The gavitational potential is defined as the scala function Φ whose gadient is equal to the opposite of the gavitational field: g = Φ One way to detemine if we can find such a scala function is to calculate the cul of the gavitational field: g = Φ = ˆx ŷ ẑ x Φ x y Φ y z Φ z = 0 In the case of the gavitational field we find g = GM ˆ = GM 1 sinθ φ 1 ˆθ 1 θ 1 ˆφ = 0 and we conclude that thee is a scala function that can geneate the gavitational field. Since the gavitational field depends just on, we expect that the gavitational potential is also just a function of. The following scala function can geneate the gavitational field: Φ = G M This is the gavitational potential due to a point mass M. If we have a continuous mass distibution, the gavitational potential will be equal to Φ = G In this equation we have assumed that the constant of integation is equal to 0 (o that the gavitational potential is 0 at infinity). ( ) ' ρ ' dv' - 2 -
One of the easons that the gavitational potential is intoduced is that even fo extended mass distibutions, it is in geneal easie to calculate the gavitational potential (which is a scala) instead of the gavitational potential enegy. Once we have detemined the gavitational potential, we can detemine the gavitational foce by calculating the gadient of the gavitational potential. The Gavitational Potential Enegy The gavitational potential can be used to detemine the gavitational potential enegy U of a paticle of mass m. Recall fom Physics 121 o Physics 141 that the change in the potential enegy of an object, when it moves fom one position to anothe position, is the opposite of the wok done by the foces acting on the object. Conside an object that moves in the gavitational field of a point mass M located at the oigin of a coodinate system. The wok done on the object of unit mass by the gavitational foce is equal to dw = g d = ( Φ) d = i Φ dx i = dφ x i If we move the object of unit mass fom infinity to a specific position the wok done is equal to W ( ) = dφ = Φ( ) The wok done to move an object of mass m to this position is thus equal to W ( ) = mφ( ) The potential enegy of the object at this position is thus equal to U ( ) = mφ( ) Infomation about the foce on the object can be obtained by taking the gadient of the potential enegy: F( ) = U ( ) = G mm ˆ = G mm ˆ which is of couse equal to the gavitational foce we used as ou stating point. - 3 -
isualization of the Gavitational Potential We can visualize the gavitational potential in a numbe of diffeent ways. The most common ways ae the contou plot, showing equipotential sufaces, and 3D plots showing the gavitational potential as function of the two-dimensional position (x, y). A simple pogam that can be used to geneate such plots is gavitationalpotential which can be found in the Mathematica folde unde Computing Tools on ou website: (* Make a 3D plot of the gavitational potential due to two point masses of mass 1 and mass 4, located at (-2, -2) and (2, 2), espectively. *) Plot3D[(1/Sqt[(x + 2)^2 + (y + 2)^2]) + (4/Sqt[(x - 2)^2 + (y - 2)^2]), {x, -5, 5}, {y, -5, 5}, PlotRange -> {0, 7.5}, PlotPoints -> 50, Ticks -> {Automatic, Automatic, Automatic}] (* Make a 2D contou plot of the gavitational potential due to two point masses of mass 1 and mass 4, located at (-2, -2) and (2, 2), espectively. *) ContouPlot[(1/Sqt[(x + 2)^2 + (y + 2)^2]) + (4/ Sqt[(x - 2)^2 + (y - 2)^2]), {x, -5, 5}, {y, -5, 5}]; The plots that ae geneated using this pogam ae shown in Figue 1. Figue 1. Diffeent ways to visualize the gavitational potential due to two point masses of mass 1 and mass 3, located at (-2,-2) and (2,2), espectively. - 4 -
The Shell Theoem When we calculate the gavitational foce o the gavitational potential geneated by a mass distibution, we can always use the most geneal expession fo these quantities in tems of the volume integal ove the mass distibution. Howeve, if the mass distibution has spheical symmety, we can use the shell theoem to calculate the gavitation foce and the gavitational potential. The shell theoem states that: The gavitational potential at any point outside a spheically symmetic mass distibution is independent of the size of the distibution, and we can conside all of its mass to be concentated at the cente of the mass distibution. The gavitational potential is zeo at any point inside a spheically symmetic mass distibution. Using the shell theoem it is easy to calculate the gavitational potential and the gavitational foce due to a spheical shell, which is shown in Figue 2. Figue 2. The gavitational potential and the gavitational foce due to a spheical shell. The shell theoem can be used to make impotant pedictions about obital motion of planets aound the cental sta and sola systems aound the cente of a galaxy. The obsevation of the otational motion of the planets aound the sun let to the fist detemination of its mass. Since the sun is much moe massive than any of the planets in the sola system, the motion of the - 5 -
planets could be descibed in tems of the gavitational foce due to just the mass of the sun (see Figue 3. Figue 3. The obital velocity of the planets in ou sola system as function of thei distance fom the sun. The theoetical dependence, which depends on the mass of the sun, is shown by the solid cuve, and does an excellent job descibing the tend in the data. Figue 4. Measued obital velocity of stas as function of distance fom the cente of the galaxy. The otational cuve can only be explained if we assume that thee is halo of dak matte in univese, distibuted thoughout the galaxy. The sola systems in most galaxies cay out an obit aound the cente of the galaxy. Since it is assumed that a massive black hole is located in the cente of most galaxies, we expect to see a tend in the obital velocity vesus distance, simila to the tend seen in ou sola system (see - 6 -
Figue 3). In eality, we see a distibution that deceases at a much smalle ate as function of the distance fom the cente of the galaxy (see Figue 4). This implies that thee is moe mass in the galaxy than was assumed, but also that this exta mass is NOT located in the cente of the galaxy, but thoughout the galaxy. This exta matte, that we can not see diectly, is called dak matte and you teache is looking fo it in a mine in England. The Poisson Equation In Electicity and Magnetism we geatly benefited fom the Gauss law, which elated the electic flux though a closed suface with the total chage enclosed by that suface. Given the fact that the natue of the gavitational foce (its 1/ dependence) is simila to the natue of the electic foce, we expect that simila laws apply to the gavitational foce. Based on the appoach we took in Electicity and Magnetism, we define the gavitational flux due a point mass m in the following manne: Φ gav = S ( ˆn g)da whee S is an abitay suface suounding the mass m (see Figue 5). Using the definition of the gavitational field, we can ewite this equation as Φ gav = Gm S cosθ da π 2π cosθ = Gm sinθdθdφ = 2πGm cosθ sinθdθ = 4πGm θ=0 φ=0 π θ=0 Figue 5. Suface used to calculate gavitational flux associated with a point mass m. - 7 -
When we have a mass distibution inside the suface S we need to eplace m by a volume integal ove the mass distibution: Φ gav = 4πG ρdv The left-hand side of this equation can be ewitten using Gauss s divegence theoem: We thus conclude that fo any volume and thus Φ gav = ( ˆn g)da = ( g)dv S ( g)dv + 4πG ρdv = {( g) + 4πGρ}dv = 0 ( g) = 4πGρ This equation can also be expessed in tems of the gavitational potential and becomes o This equation is known as Poisson s equation. ( Φ) = 2 Φ = 4πGρ 2 Φ = 4πGρ The Tides We all know that the tides ae caused by the motion of the moon aound the eath, but most of us have a moe difficult time to explain why we have two high tides a day. Ou simple pictue would suggest that you have high tide on that side of the eath closest to the moon, but this would only explain one high tide a day. The calculation of the effect of the moon on the wate on eath is complicated by the fact that the eath is not a good inetial system, so we have to assume that we can find a good inetial fame in which we can descibe the foces acting on the wate (see Figue 6). Conside the foces on a volume of wate of mass m, located on the suface of the eath (see Figue 6), due to the moon and the eath: - 8 -
F m = m m = GmM E ˆ GmM m R 2 ˆR Figue 6. Geomety used to detemine the foces on a volume of wate of mass m, located on the suface of the eath. The foce exeted by the moon on the cente of the eath is equal to F E = M E E = GM m M E D 2 ˆD When we view the motion of the wate on eath, we view its motion with espect to the eath. We have used a coect inetial fame to detemine the acceleation of the individual components of ou system, and we can now tansfom to a efeence fame in which the eath is at est and centeed aound the oigin (note: this is a non-inetial efeence fame). In this efeence fame, the acceleation of mass m is equal to = m E = GM E ˆ GM m R 2 ˆR + GM m D 2 GM ˆD = E ˆR ˆ GM m R 2 ˆD D 2-9 -
The fist tem on the ight-hand side is just the gavitational foce on mass m due to the eath. It will be the same anywhee on the suface of the eath and is not esponsible fo the tides. The second pat of the ight-hand side is equal to the acceleation associated with the tidal foce. It is elated to the diffeence between the gavitational pull of the moon on the cente of the eath and on the suface of the eath. Let us conside what this equation tells us about the magnitude and the diection of the tidal foce at 4 diffeent positions on the suface of the eath (see Figue 6). Point a: At point a, the unit vectos associated with R and D ae pointing in the same diection. Since R > D, the (1/R 2 ) tems will be smalle than the (1/D 2 ) tem, and the tidal acceleation will be diected towads the ight: 1 T = GM m D 1 2 R 2 ˆD 1 = GM m D 1 2 D + ( ) 2 ˆD = GM m 1 1 D 2 1+ D 2 ˆD = 2GM m D 3 ˆD Point b: At point b, the unit vecto associated with R and D ae pointing in the same diection. But, since R < D, the (1/R 2 ) tems will be lage than the (1/D 2 ) tem, and the tidal acceleation will be diected towads the left but with the same magnitude as the acceleation at point a: 1 T = GM m D 1 2 R 2 ˆD 1 = GM m D 1 2 D ( ) 2 ˆD = GM m 1 1 D 2 1 D 2 ˆD = 2GM m D 3 ˆD Point c: The x components of the vectos associated with R and D ae simila, and the x components of the acceleation will cancel. The vecto associated with D will have no y component, and the acceleation will be due to the y component associated with R (which points towads the cente of the eath): ˆR T = GM m R 2 ˆD = GM m D 2 ˆR y R 2 = GM m 1 D 2 D ŷ = GM m D 3 ŷ Point d: The calculation fo the foce on point c is simila to the calculation of the foce on point a. The foce will be diected towads the cente of the eath and will be equal to T = GM m D 3 ŷ - 10 -
We thus conclude that the acceleation of the wate on the suface of the eath is diected away fom the suface at two diffeent locations. It will thus be high tide at these two locations and since the moon otates aound the eath in one day, each location will see a high tide twice a day (one time when the moon that location is facing the moon, and one time when the moon is located on the opposite side of the eath fom that location). A summay of the tidal foces on the suface of the eath is shown in Figue 7. Figue 7. Tidal foces at vaious places on the suface of the eath. - 11 -