Universal Gravitation

Transcription

1 J - The Foce of Gavity Chapte J Univesal Gavitation Blinn College - Physics 45 - Tey Honan Intoduction If Isaac Newton had meely witten down his thee laws of motion he would pobably still be known as the most impotant physicist of all time Instead, he went fa beyond that; he solved all elementay poblems involving mechanics and he found the foce law descibing gavity His theoy of gavity is the topic of this chapte The histoy of physics is lagely a histoy of the unification of the fundamental foces Befoe Newton thee wee sepaate notions of gavity What was known as gavity befoe Newton was the foce holding things to the suface of the Eath; we will call this teestial gavity It was not clea that this foce was the same as what was esponsible fo the motion of planets, moons and comets; this we will efe to as astonomical gavity Newton, with his theoy of univesal gavitation showed that these two foces wee actually the same; he unified teestial and astonomical gavity Oigin of the Invese Squae Law Newton was tying to undestand the foce law between point masses m and m sepaated by a distance The weight of a body is the gavitational foce on it Since the weight is popotional to the mass, the foce on m must be popotional to m F m Because of Newton's thid law the magnitude of the foce on m is the same Thus the foce on m must also be popotional to m F m How does the foce vay with distance? It should also decease with the distance and should go to zeo as appoaches infinity Some possibilities ae Newton settled on an invese squae law F, F, F F He deduced the invese squae law by compaing the atio of the acceleation of the moon towad the eath to the acceleation due to gavity at the suface of the eath to the atio of the adius of the moon's obit to the adius of the eath Suppose the foce has the fom F p fo some powe p, whee an invese squae law gives p - Since the acceleation of a body is popotional to the foce F a we can conclude a p Fom this popotionality we can wite the atio We can find the powe p using logs a a 3 p ln(a /a ) p ln( / ) Let a and efe to the moon's obit At Newton's time the acceleation of the moon towad the eath could be calculated The peiod of the moon's obit is T 73 days The eath-moon distance, which was measued by paallax, is R EM m This gives a π T π s o F e -α m 0007 m s The acceleation at the suface of the eath is a g 980 m s at a distance given by the eath's adius R E m Combining these expessions gives

2 Chapte J - Univesal Gavitation which veifies the invese squae law ln(a /a ) p ln( / ) ln ln , Newton ealized that thee was a cucial flaw in his logic hee He wanted a foce law between paticles; by a paticle it is meant that the distance to an object is lage compaed to the size of the object Clealy this is not coect fo an object on the suface of the eath Newton felt that the gavitational foce fo a spheical body, as long as one is outside of the body, is the same as if all of the mass is at the cente To pove this Newton had to fist invent integal calculus and then do a difficult integal This we will call the shell theoem Newton's Law of Univesal Gavitation Newton's law of gavity is the attactive invese squae law between point masses If m and m ae point masses sepaated by distance the magnitude of the foce between them is F G m m This can be witten as a vecto expession Let be the vecto fom mass to mass and let F be the foce on mass due to mass ṙ is the unit vecto in the diection of the vecto F -G m m ṙ ṙ / 0 m m ṙ F The constant G is known as Newton's univesal gavitational constant Gavitational foces between eveyday objects ae incedibly small This is eflected in the small value of G in ou SI system of units Newton neve knew the value of this constant It was eventually measued to be N m - G kg The Foce on a Mass Due to a Distibution of Masses Suppose we want to find the foce on a mass m due to a discete distibution of masses, m, m, Define the vecto i to be fom m i to m The foce of m i to m is F i -G m m i whee ṙ i i / i The total foce is the sum of all these pats F i F i This gives m i F -G m i ṙ i i This is just using the idea that foce is a vecto and that foces add as vectos In ou discussion of moments of inetia we dealt with continuous distibutions Beak up a continuous distibution into an infinite numbe of infinitesimal pieces; take d m to be one of these infinitesimal pieces The vecto is fom d m to m i ṙ i ṙ F -G m d m Example J - Net Foce on Moon What is the magnitude net foce on the moon, due to the eath and sun, at a (i) full moon, (ii) new moon and (iii) half moon?

3 Chapte J - Univesal Gavitation 3 Sun (iii) half moon Moon (ii) new moon Eath (i) full moon R ES (not to scale) The masses of the eath, the moon and sun, and the eath-moon and the eath-sun distances ae: R EM M E kg, M S kg, M M kg, R EM m and R ES 50 0 m Hee assume a cicula obit Looking at thei numeical values we can also see that R EM << R ES, we will assume this, as well The magnitude of the foce of the eath on the moon is F E G M E M M N R EM The sun-moon distance vaies, but with the R EM << R ES assumption we can set the moon-sun distance to the eath-sun distance, R MS R ES The magnitude of the sun s foce on the eath becomes F S G M S M M N R ES (i) Fo the case of the full moon both foces F S and F E ae in the same diection, to the left in the diagam The magnitude of the net foce is the sum of the two magnitudes F net F S + F E N (ii) With the new moon, the foces F S and F E ae in opposite diections, in the diagam: to the left fo the sun and to the ight fo the eath The magnitude of the net foce is now the diffeence F net F S - F E N (ii) At the half moon phase, the foces F S and F E ae pependicula, since in the limit R EM << R ES the hypotenuse of the eathmoon-sun tiangle becomes paallel to the long side The magnitude of the net foce found by the Pythagoean theoem F net F S + F E N Note that since F S > F E the moon s tajectoy must always cuve towad the sun The Shell Theoem and Spheical Distibutions Conside a unifom spheical shell of mass M and adius R If we conside a point mass m a distance fom the cente then the shell theoem states that: fo < R, fo > R, F 0 and F -G M m ṙ To state this in wods: When m is outside the shell ( > R) the shell behaves as if all the mass is at the cente When m is inside the shell ( < R) the total gavitational foce on the mass is zeo Clealy, by symmety, at the exact cente of the shell the foce should be zeo It is a emakable popety of the invese squae law that it is zeo eveywhee inside We will not pove the shell theoem this semeste It will be poven in the second semeste couse in the analogous case of electostatics As a consequence of the shell theoem we can find the foce on a mass m a distance fom the cente of any spheical distibution Beak up the sphee into thin concentic shells If the shell is inside the adius the shell behaves as if all its mass is at the oigin If it is outside it doesn't contibute to the foce This gives the esult: F -G m M enclosed ṙ

4 4 Chapte J - Univesal Gavitation whee M enclosed is the total mass enclosed by a sphee of adius The Gavitational Field We will efe to the acceleation due to gavity at some position as the gavitational field g; typically this no longe will be unifom Since the foce on a mass m is F m g we can define the field at some position by the following pocedue Add a test mass m 0 at some position Define the field at that position to be the foce divided by the test mass The esult is then independent of the test mass g F m 0 Now conside a spheical planet of mass M and adius R A test mass at the suface of the planet expeiences a foce of Dividing m 0 into this gives F -G m 0 M R ṙ g -G M R ṙ o g G M R Example J - Weight on a Diffeent Planet An astonaut has an eath weight of 0lb What is he weight, in pounds, on a planet with twice the eath s adius and five times its mass The gavitational field (o acceleation due to gavity) g on the suface of a spheical planet of mass M and adius R is: g G M R The weight W of an astonaut of mass m is elated to g by W mg Both fomulas apply to both cases, on the eath and the othe planet The astonaut s mass m is the same on both planets The two fomulas apply to each case Fo the weights we see that the atio of the weights equals the atio of the g values Similaly fo g we see: W m g and W m g Dividing W g W g g G M R and g G M R Take case to be the new planet and case the eath Dividing g g M /M (R / R ) We can solve fo the esult W W M /M W (R / R ) W E M /M E 5 (R / R E ) 5 W W 5 E 0lb 50lb J - Enegy Consideations Potential Enegy of Two Masses by Recall that a consevative foce is any foce whose wok is independent of path We can define a potential enegy fo any consevative foce

5 Chapte J - Univesal Gavitation 5 Δ U -W - F d We want to find the potential enegy fo two masses m and M sepaated by a distance To get this we will conside M to be at a fixed position, which we will take as the oigin, and move m in its pesence Since the foce on m due to M is F -G M m ṙ, we get Δ U - F d - i f -G M m Multiplying a vecto by a unit vecto gives the vecto component in the unit vecto's diection ṙ d becomes the adial component of d which is just d, the change in the adial vaiable ṙ d d The limits of the integal then only depend only on the adial distances of the endpoints ṙ d f d ΔU G M m i -G M m - f We ae looking fo some function U(), descibing the potential enegy as a function of position Since it must satisfy Δ U U f - U( i ), it is unique up to an abitay constant The simplest choice is U() -G M m whee the abitay constant is chosen to make the potential enegy zeo at infinity Potential Enegy of a Configuation U( ) 0 o moe pecisely lim U() 0 Now we conside the case of seveal point masses, m, m, Take the distance between m and m to be and geneally ij as the distance between the i th and j th masses Fo two masses we have U -G m m / Fo thee masses we have this tem plus enegy tems between m and m 3, and between m and m 3 We get The geneal expession is U -G m m - G m m G m m 3 3 U -G i<j m i m j ij The sum is ove all indices i and j whee we insist that i < j to avoid double counting i Enegy and Escape Speed The total enegy is the sum of the kinetic enegies of all masses and the total potential enegy above Fo a single mass m moving in the pesence of a lage mass M the total mechanical enegy is E K + U m v - G M m The potential enegy is negative, going to zeo at infinity To escape the gavitational pull of M, m must have enough enegy to each infinity, o zeo enegy The escape speed is the citical speed needed to escape the gavity of a planet fom the suface of the planet of mass M and adius R This becomes Solving fo the escape speed we get 0 E m v esc - G M m R v esc G M R

6 6 Chapte J - Univesal Gavitation Example J3 - Rocket Launched off a Spheical Planet A ocket is launch off the suface of a spheical planet of mass M and adius R with a speed v 0 in a vetical (pependicula to the suface) diection (a) If the speed v 0 is less than the escape speed, the ocket will each some maximum distance fom the cente max and fall back What is max? E m v - G M m m is the ocket s mass It must scale out of the poblem Take the initial position to be at the suface and the final to be at max The final kinetic enegy is zeo E i E f m v 0 - G M m R 0 - G M m max max - v 0 R - G M (b) Fo ocket speeds geate than the escape speed, the ocket will continue to infinity and still have kinetic enegy What is v, the speed at infinity? (To be pecise, this should be discussed in tems of limits: as, v v ) The expession fo the enegy is the same as is the initial values The final potential enegy is now zeo E i E f Note that this can be witten as v m v 0 - G M m R v 0 - v esc m v + 0 v v 0 - G M R J3 - Obits Cicula Obits We will now conside cicula obits of a satellite of mass m obiting a much lage object of mass M, m M Take the lage object to be a sphee of adius R If the satellite obits at a height h above the suface, then the adius of the obit is R +h We now apply the second law to the satellite The only foce is gavity and the acceleation is centipetal We get: F net,c m a c G M m m a c In this expession the satellite mass cancels; this is the essence of weightlessness Suppose one studies the obit of the intenational space station m is the station's mass If an astonaut is floating in the station and not touching anything, then m is the astonaut's mass The cancellation of m implies that the astonaut and station have the same obit; he will float elative to the station Afte canceling the masses we can then use the two expessions fo the centipetal acceleation fo unifom cicula motion This gives a pai of expessions, one that elates the speed and adius and the othe elating the peiod and adius Solving these expessions fo the speed and peiod gives G M a c v π T v G M 4 π T G M 3

7 Chapte J - Univesal Gavitation 7 Example J4 - Low-eath Obit On a planet without an atmosphee, it is possible to obit just above the suface Fo the eath the obit must be above the atmosphee Although the atmosphee gets less dense with the height above the suface, thee is a point whee the dag is small enough fo an obit Conside a low eath obit to be 00 km (about 0 mi) above the suface What ae the peiod and speed of this low-eath obit? R eath m, M eath kg h 00 km m R eath + h m 4 π T 3 T 530 s 89 min G M eath Low-eath obits can be highe but the height should not be much lage These will have slightly longe peiods, but low-eath obits ae close to 90 minutes v G M eath 7780 m /s Example J5 - Geostationay Obit It is useful fo communication satellites, weathe satellites and spy satellites to maintain a fixed position in the sky elative to the otating eath This is called a geostationay obit This equies an obit diectly ove the equato with a peiod of 4 h (a) Take the peiod of a low-eath obit to be 5 hous and the obital adius to be the eath s adius, R E Estimate the adius of a geostationay obit as a multiple of R E 4 π T G M 3 We will use atios to solve this, as we did in Example-J We will use the peiod equation above fo two cases: R E with T 5 h and find the unknown with T 4 h M M eath fo both cases 4 π 4 π T 3 and T 3 G M eath G M eath T T 3 4 h 5 h Dividing 3 R E 6 3 R E T T 64 R E (b) Calculate this geostationay obital adius pecisely While we ae being pecise, the obital peiod is not one sola day but one sideeal day 393 h (Note that a sideeal day is the time it takes fo the eath to otate once elative to the distant stas, as opposed to elative to the sun Thee ae 3654 sola days in a yea but, with one exta otation elative to the distant stas, thee ae 3664 sideeal days in a yea) 3 4 π T 3 G M T eath G M eath 4 π 3 N m kg Compae this with the adius of the eath ( s) kg 4 π m R eath m 66 R eath

8 8 Chapte J - Univesal Gavitation Keple's Laws Keple came afte both Copenicus and Galileo, and came befoe Newton Keple used data taken by someone else, Tycho Bahe, of the positions of the planets in the sky as functions of time Fom this he ealized that the cicula obits of Copenicus and Galileo could not fit the data He settled on elliptical obits and summaized his esults with thee laws Afte Keple, Newton found his thee laws of motion and his law of univesal gavitation He was able to deive Keple's thee law fom his wok This was a emakable feat He was also able to pove that Keple's laws implied that the gavitational foce law had to be an invese squae law Keple's Fist Law - Planets move in elliptical obits about the sun, which is at a focal point of the ellipse Given two points, called focal points, an ellipse is the set of points in a plane such that the sum of the two distances fom each focus is a constant + a focus focus b a c Inteactive Figue a is called the semimajo axis; it is half the lagest distance between two points on the ellipse The othe dimension of the ellipse is the mino axis b Each focus is a distance of c fom the cente, whee c e a, with e known as the eccenticity The eccenticity is the deviation fom being a cicle, 0 e < e 0 is the cicula case, whee the two foci join and a b The vaiables a, b, c and e ae elated by c e a and b a - c a - e Poving that obits ae elliptical is not difficult but is a bit beyond the scope of this couse Keple's Second Law - The adial vecto fom the sun to the planet sweeps out equal aeas in equal time Sun da v dt This is a simple consequence of the consevation of angula momentum The aea of a paallelogam fomed by two vectos is A B sin θ which is just the magnitude of the coss poduct of the vectos In a time d t a planet moves by v d t This taces out a tiangle; this is half a paallelogam and thus has the infinitesimal aea

9 Chapte J - Univesal Gavitation 9 Since L p is constant, it follows that d A / d t is constant This poves the esult d A v d t d A L d t v m p m constant Sun Inteactive Figue Keple's Thid Law - Peiod Semimajo axis 3 o T a 3 Define the majo axis as the lagest distance between two points on an elliptical obit The semimajo axis, a, is half this Fo a cicula obit, this is the same as the adius a We saw that T 3 fo a cicula obit The thid law is then a genealization of this esult, which we will not pove 4 π T G M a3 e 0 e 0 3e 0 6 e 0 9 Sun a a a a Diffeent obits with the same peiod The eccenticities vay but the semimajo axis stays the same