Working with Gravity: Potential Energy

Transcription

1 pevious index next Woking with Gavity: Potential negy Michael Fowle 31/1/07 Gavitational Potential negy nea the ath We fist biefly eview the familia subject of gavitational potential enegy nea the ath s suface, such as in a oom The gavitational foce is of couse F = mg vetically downwads To aise a mass m, we must apply an upwad foce F, balancing gavity, so the net foce on the body is zeo and it can move upwads at a steady speed (ignoing ai esistance, of couse, and assuming we gave it a tiny exta push to get it going) Applying the steady foce F as the mass moves a small distance Δ takes wok F Δ, and to aise the mass m though a height h takes wok mgh This enegy is stoed and then, when the object falls, eleased as kinetic enegy Fo this eason it is called potential enegy, being potential kinetic enegy, and witten U = U( h) = mgh Note one obvious ambiguity in the definition of potential enegy: do we measue h fom the floo, fom the top of ou wokbench, o what? That depends on how fa we will allow the aised object to fall and convet its potential enegy to kinetic enegy but the main point is it doesn t matte whee the zeo is set, the quantity of physical inteest is always a diffeence of potential enegies between two heights that s how much kinetic enegy is eleased when it falls fom one height to the othe (Pehaps we should mention that some of this potential enegy may go to anothe fom of enegy when the object falls if thee is substantial ai esistance, fo example, some could end up eventually as heat We shall ignoe that possibility fo now) U(h) h

2 Onwad and Upwad Let s now conside the wok involved in lifting something so high that the ath s gavitational pull becomes noticeably weake It will still be tue that lifting though F( ) = GMm/, downwads So Δ takes wok F Δ, but now GMm du = F d = d and to find the total wok needed to lift a mass m fom the ath s suface ( fom the cente of the ath) to a point distance fom the cente we need to do an integal: GMm 1 1 U( ) U( ) = d = GMm Fist check that this makes sense close to the ath s suface, that is, in a oom Fo this case, Theefoe = + h, whee h 1 1 U( ) U( ) = GMm + h = GMm ( h ) + h GMm = mgh whee the only appoximation is to eplace + h by in the denominato, giving an eo of ode h/, pats pe million fo an odinay oom To see what this potential function looks like on a lage scale, going fa fom the ath, it is necessay fist to decide whee it is most natual to set it equal to zeo The standad convention is to set the potential enegy equal to zeo at = infinity! The eason is that if two bodies ae vey fa fom each othe, they have no influence on each othe s movements, so it is pointless to include a tem in thei total enegy which depends on thei mutual inteaction Taking the potential enegy zeo at infinity gives the simple fom

3 3 GMm ( ) =, U we plot it below with in units of ath adii The enegy units ae GMm/, the 1 at the fa left being at the ath s suface ( = 1), and the fist steep almost linea pat coesponds to mgh The above is a map of the potential enegy hill to be climbed in going away fom the ath vetically upwads fom any point To gain something close to a theedimensional pespective, the ath can be visualized as being at the bottom of a potential well with flaed sides, like this: O, fom a diffeent pespective:

4 4 Of couse, this is still only in two dimensions, but that s fine fo most gavitational poblems: planetay obits ae only two-dimensional A satellite in a cicula obit aound the ath can be imagined as a fictionless paticle sliding aound inside this cone at a fixed height, fo an elliptic obit the paticle would slide between diffeent heights Gavitational Potential negy in a Two Body System By this, I mean how do we extend the above pictue of gavitational potential as a well going down out of a flat plane to, fo example, the combined potential enegies of a mass in the gavitational fields of both the ath and the Moon, as would occu on a flight to the Moon Fom the beginning of the pevious section, the potential enegy diffeence between any two points fom the gavitational foce of a single body is the wok done against that foce in going fom one point to the othe, U U = F d ( ) ( ) 1 1 It doesn t matte how the path gets fom 1 to : if it took diffeent amounts of wok depending on the path, we could gain enegy by having a mass go up one path and down the othe, a pepetual motion machine The fact that this is not tue means the gavitational field is consevative: gavitational potential enegy can b a tem in a consevation of enegy equation Recall fom the pevious lectue that the gavitational field obeys the Law of Supeposition: to find the total gavitational foce on a mass fom the gavitational field of both the ath and the Moon, we just add the vectos epesenting the sepaate foces It follows immediately fom this that, putting F = Fath + FMoon, the gavitational potential enegy diffeence between two points is simply the sum of the two tems

5 5 Fom this, then, the potential enegy of a mass somewhee between the ath and the Moon is U total ( ) GM m GM Mm = C taking as usual U ( ) = 0, and C, CM ae the coodinates of the centes of the ath and the Moon espectively It s woth visualizing this combined potential: it would look like two of these cone-like wells, one much smalle than the othe, in what is almost a plain Going in a staight line fom inside one well to the inside of the othe would be uphill then downhill, and at the high point of the jouney the potential enegy would be flat, meaning that the gavitational pull of the ath just cancels that of the Moon, so no wok is being done in moving along the line at that point The total potential enegy thee is still of couse negative, that is, below the value (zeo) fa away in the plain Gavitational Potential The gavitational potential is defined as the gavitational potential enegy pe unit mass, and is often witten ϕ ( ) We shall aely use it the poblems we encounte involve the potential enegy of a given mass m (But ϕ ( ) is a valuable concept in moe advanced teatments It is analogous to the electostatic potential, and away fom masses ϕ = 0 ) obeys the same patial diffeential equation, ( ) scape! How fast must a ocket be moving as it escapes the atmosphee fo it to escape entiely fom the ath s gavitational field? This is the famous escape velocity, and, neglecting the depth of the atmosphee, it clealy needs sufficient initial kinetic enegy to climb all the way up the hill, CM 1 GMm GM mvescape =, vescape = This woks out to be about 11 km pe sec Fo the Moon, escape velocity is only 3 km pe second, and this is the eason the Moon has no atmosphee: if it had one initially, the Sun s heat would have been sufficient to give the molecules enough themal kinetic enegy to escape In an atmosphee in themal equilibium, all the molecules have on aveage the same kinetic enegy This means lighte molecules on aveage move faste On ath, any hydogen o helium in the atmosphee would eventually escape fo the same eason

6 6 xecise: Satun s moon Titan is the same size as ou Moon, but Titan has a thick atmosphee Why? xecise: Imagine a tunnel boed staight though the ath emeging at the opposite side of the globe The gavitational foce in the tunnel is F = mg /, as deived above (a) Find an expession fo the gavitational potential enegy in the tunnel Take it to be zeo at the cente of the ath (b) Now sketch a gaph of the potential enegy as a function of distance fom the ath s cente, beginning at the cente but continuing beyond the ath s adius to a point fa away This cuve must be continuous Conventionally, the potential enegy is defined by equiing it to be zeo at infinity How would you adjust you answe to give this esult? Potential and Kinetic negy in a Cicula Obit The equation of motion fo a satellite in a cicula obit is mv GMm = It follows immediately that the kinetic enegy ( ) K = mv = GMm / = U, that is, the Kinetic negy = 1/ (Potential negy) so the total enegy in a cicula obit is half the potential enegy The satellite s motion can be visualized as cicling aound tapped in the cicula potential well pictued above How fast does move? It is easy to check that fo this cicula obit v obit = GM obit Recalling that the escape velocity fom this obit is v = GM /, we have escape obit v escape = v obit elating speed in a cicula planetay obit to the speed necessay, stating at that obit, to escape completely fom the sun s gavitational field This esult isn t supising: inceasing the speed by doubles the kinetic enegy, which would then exactly equal the potential enegy: that means just enough kinetic enegy fo the satellite to climb the hill completely out of the well

7 7 Bottom line: the total enegy of a planet of mass m in a cicula obit of adius about a Sun of mass M is tot GMm = pevious index next