10. Universal Gravitation

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Robyn Beasley
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1 10. Univesal Gavitation Hee it is folks, the end of the echanics section of the couse! This is an appopiate place to complete the study of mechanics, because with his Law of Univesal Gavitation, Newton took dynamics out of the ealm of the local and extended it acoss the univese. Pio to Newton, it was thought that the laws on eath wee sepaate fom the laws in the heavens. We ll appoach Univesal Gavitation in the same way we did SH: the basics and then some applications. The applications will involve (a) single body obits, (b) binay systems, and (c) special gavitational effects. I) BASIC GAVITATIONAL QUANTITIES (1) Foce: F g = Gm/d 2 G = 6.67 x Nm 2 /kg 2 This value was detemined by Lod Cavendish 100 yeas afte Newton. Cavendish s expeiment enabled the mass of the eath to be detemined (see p.103)! d = the cente to cente distance between and m. It is not necessaily a adius! Newton deived his fomula using a) the equation fo centipetal foce b) Newton s 3 d Law c) Keple s 3 d Law b) the values fo the adius of the eath, adius of the moon, and the adius of the moon s obit, as detemined by Eastothenes & Aistachus ~ B.C. ajo esult: F g is an invese-squaed elationship F g α (1/d 2 ) So, if d doubles, F g is educed by a facto of 4; if d is educed by a facto of 3, F g will incease by a facto of 9. You dig? d m (2) Acceleation : Since F g is the net foce in gavitational situations, we also get that fo m a g = G/d 2 (Typically, ag is found fo a mass that epesents a moon, a planet, o a sta not something small, like a peson o a satellite.) a g is often expessed as simply g it s also efeed to as gavitational field stength this is also an invese-squaed elationship: a g α (1/d 2 ) this expession is always on the ultiple Choice section of the echanics Exam!

2 (3) Gavitational Potential Enegy (U g ) Last yea, and ealie this yea, we used mgh to epesent U g. Not anymoe! Why? Because in tems of the big pictue, gavitational enegy needs a fixed efeence point in the univese that eveyone can agee on. That point, physicists agee, should be at infinity, whee m is an infinite distance away fo. Hee, Ug is abitaily defined to be 0. As m is moved close and close to, the potential enegy is deceasing. Theefoe, U g must get moe and moe (-) as m gets close to. To find the expession fo U g at some sepaation d between the masses, we must integate fom to d: d W = F(x) dx U g = - F(x) dx = - (-Gm/x 2 ) dx d In ode to be mathematically consistent, foces of attaction (like gavity and unlike chages) ae given a (-) sign. In this way, when one integates fom a futhe sepaation to a close sepaation, the wok done (i.e., the ΔKE) will be positive, as it should be unde those cicumstances fo a consevative foce (which, of couse, gavity is). d U g = - (Gm/x) U g = -Gm/d Note: The deivation shown above is what I efe to as a stand alone, that is, only done one way. It was on the 1994 Exam! (4) Total echanical Enegy (TE) Yup! We e using (K + U) again. Only this time, we ll use ou new expession fo U g. TE is used, as you ll see, in vitually evey gavitational situation. TE = K - Gm/d

3 II) SINGLE BODY OBITS Waning: Single body obits ae eally an appoximation. Thee s eally no such thing! We e assuming that the mass that is obited aound (the lage mass) is at est. Based on Newton s 3 d Law, that s impossible! But, fo these situations, the cente of mass of the system is so fa inside the lage mass, that we can appoximate it as being at the cente of the lage mass in othe wods, we e assuming the lage mass is at est (elative to the obiting mass). Cicula Obits (Pinciples: Newton s 2 nd Law, Keple s 3 d Law & Consevation of echanical Enegy) (1) Newton s 2 nd Law: If a mass m is in a cicula obit of adius aound, then: Fg = Fc (F g is the net foce!) Gm/ 2 = ma c = mv 2 / i. Obital Speed: Using the 2 nd Law, we can find an expession fo obital speed v: v = (G/) ii. Obital Peiod: Since v = 2p/T, then Solving fo T yields is the mass at the cente of the obit!!! Since d is the obital adius, then d = fo this situation (and this situation only!). 2p/T = (G/) T = 2π (/G) (2) Keple s 3 d Law: If we e-aange the peiod expession to isolate 3 /T 2, we get 3 /T 2 = G/4π 2 Aha! This means that peiod (T) and obital adius () can be used to detemine the mass of a planet o sta! (3) Total echanical Enegy: TE = K + U = (mv 2 /2) + (-Gm/) = (Gm/2) + (-Gm/) TE = -Gm/(2) Aha-ha! When its TE is (-), a mass m is said to be gavitationally bound.

4 Elliptical Obits (Pinciples: Consevation of echanical Enegy & Consevation of Angula omentum) d a Hee, a is the length of the majo axis (1) Consevation of echanical Enegy: Because of the natue of elliptical obits, the K and the U ae constantly changing, but as always, the TE stays the same. Theefoe, the obital speed at any sepaation d fom the sta (which, as we know fom Keple s 1 st Law, is at one of the 2 foci) can be found by ou old eliable, the Consevation of echanical Enegy: K 1 - Gm/d 1 = K 2 - Gm/d 2 (1/2)mv 12 Gm/d 1 = (1/2)mv 22 Gm/d 2 (2) Consevation of Angula omentum This law povides the poof fo Keple's 2nd Law (the law of equal aeas). Since angula momentum must be conseved, the speed of a planet/satellite is geatest when it is closest to the sun/planet and it is least when it is futhest. mv a a = mv p p Aha! The Consevation of Angula omentum (above) allows us to detemine the obital speed at peihelion given aphelion and vice-vesa. Notice that F 2 is empty. Whee s F 1? That s whee the Sun is!!! v a peihelion p Sun F 2 a aphelion v p Waning: Fo obital speed, the expession G/ only woks fo cicula obits!!! The obital speed fo elliptical obits can only be found as shown above!

5 Question: What happens if the obital speed of a satellite is changed while the satellite is in a cicula obit about a sun/planet of mass? Why? v inceases at point A v deceases at point A B B Hee, the satellite will go into an elliptical obit with at the focus nea point A. Its geatest speed is at point A, while it is slowest at point B. Why? Since K A + U A = K B + U B and (1) U A < U B then (2) K A > K B ( v A > v B ) Hee, the satellite will go into an elliptical obit with at the focus nea point B. Its least speed is at point A, while it is fastest at point B. Why? Since K A + U A = K B + U B and (1) U A > U B then (2) K A < K B ( v A < v B ) Fo eithe of the ellipses, when the satellite is close, it goes faste; when its fathe away, it goes slowe. This esult is also explained by the Consevation of Angula omentum!!! The only point that is common to both the oiginal cicula obit and the elliptical obit is point A. If v inceases by at least 2, then the satellite will escape fom mass! This is known as the escape velocity, which we ll discuss a bit late. Anothe vague possibility: If v deceases by too much, the satellite will cash into! Exactly how much?, you ask. Enough that U B = -Gm/ planet!, I eply smugly.

6 III) BINAY SYSTES (Simulation: ast162 csep10 utk/visual Binaies) In binay systems, two stas ae obiting aound a common cente of mass. d = + Fo binay systems, d = + c.m. m 1 st step: detemine the expessions fo the obital adii! As it tuns out, obital adii ae easy to find, thanks to (1) Cente of mass: x c.m. = (0) + md ( + m) Hee, the oigin is chosen at the cente of. The x c.m. value gives you the adius of obit fo (see illustation above). Since d = +, this leads to: /m = / The 2 nd step: find the obital speeds, v and v m! (2) Since we ae assuming cicula obits, then we can use F g = F c to find the obital speed and/o centipetal acceleation of eithe mass. Gm/(d) 2 = v 2 / = mv m 2 / With eithe obital speed, you can detemine a LOT of stuff!!! i. Peiod (T) by using v = 2π/T o v m = 2π/T. Aha! Since the cente of mass doesn t move the peiod is the same fo both m and! ii. The othe obital speed; since the peiods ae the same, then: v /v m = / iii. Centipetal acceleation (a c ) by using a c = 4π 2 /T 2 fo o a c = 4π 2 /T 2 fo m iv. TE = K + K m + U g v. ω = 2π/T (same fo both & m) vi. L = v & L m = mv m (o, you can use L = I ω & L m = I m ω)

7 Example 1 A binay system consists of a mass and a mass 3 obiting aound a common cente of mass. The two masses ae sepaated by a distance, as shown below. In tems of G,, and find: (a) the obital speed of mass 3 (3/2) G/ (b) the peiod of evolution π /G (c) the centipetal acceleation of mass 3 (d) the angula speed of G/ 2 (e) the angula momentum of 3 2 G/ (3/8) G/ (f) the total enegy of the system -(3/2)G 2 /

8 Example 2 Two stas each of mass fom a binay sta system such that both stas move in the same cicula obit of adius. The univesal gavitational constant is G. (a) Find an expession fo the speed of eithe sta in tems of, G, and. (b) Expess the total enegy, E T, of the system in tems of, G, and. (1/2) G/ -(3/4)G 2 / Suppose, instead, one of the stas has a mass 2. (c) On the diagam below, sketch the cicula obits fo this sta system. 2 (d) Find the atio of the speeds, v 2 /v 1 : 2

9 IV) SPECIAL GAVITATIONAL EFFECTS (1) Consevation of Enegy & Escape Velocity Q: If a ocket is launched fom the suface of a planet with a velocity v 0, how fa fom the planet (along a adial path) will it go befoe stopping? A: Use Consevation of Enegy, baby! All the indicatos ae in place: 2 points at which infomation about speed and height ae eithe given o asked fo. 1 2 h v 0 v = 0 d = + h K 1 + U 1 = K 2 + U 2 (1/2)mv (- Gm/) = 0 - Gm/(d) Special Case: Escape Velocity When a mass is gavitationally bound, it s TE is (-). In ode to escape gavity, the TE must be at least 0! Since KE 0 and U g 0, then the TE = 0 when the mass has enough initial velocity to stop eeeeaaaalllly fa away (i.e., when d = ). So, based on the Consevation of Enegy we can say K 1 + U 1 = K 2 + U 2 (1/2)mv e 2 + (-Gm/) = v e = (2G/) ( = adius of planet) v escape v = 0 d = (ho ho ho)

10 (2) Gavity inside the Eath (anothe echanics stand alone situation) e m P Fo a Stickman of mass m located at a point P which is a distance < e fom the cente of the eath, only the mass of the eath inside the sphee of adius contibutes to (a) the gavitational foce acting on Sticky, and (b) the gavitational potential enegy of Sticky Foce of Gavity (F g ) In geneal, the foce of gavity is expessed by : F g = Gm/d 2 Fo this situation, we (1) eplace d with, and (2) eplace with V Since we appoximate the eath s density as being unifom, then the eath s mass contained within the sphee of adius can be expessed as: = V = [(4/3)π 3 ] So, the foce of gavity can be expessed as F g = - Gm[(4/3)π 3 ]/ 2 Which simplifies to F g = - [Gm(4/3)π] Woo-hoo! This is a linea elationship!!! In othe wods, the foce of gavity is zeo at the cente of the eath, and inceases at a steady ate as one moves fom the cente to the suface. Gm/ 2 F g 2 - Gm/ 2

11 Gavitational Potential Enegy (U g ) The basic idea is to apply ou 2 stand alones! U g = - F g dx (stand alone #1) Hee, the limits of integation will be fom the suface () to some distance <: ΔU g = - F g dx = U f U 0 = U inside - U suface This is the same thing, mathematically, that we did on page 93! The only diffeence is that, instead of integating fom to d, we e integating fom to. Plugging in fo F g (fom stand alone #2) and eliminating the (-) signs yields: Since = /V = /[(4/3)π 3 ], then ΔU g = [Gm(4/3)π] x dx ΔU g = [Gm/ 3 ] x dx The mess in the backets is a constant, so this is an easy integation! ΔU g = [Gm] x 2 = [Gm] 2 - [Gm] = U inside - U suface = U inside - ( - Gm/) 2 3 Simplifying this to isolate U inside yields U inside = Gm 2 - (3/2) Gm/ Aha! This means that the U g at the cente of the eath (whee = 0) is: U cente = (- 3/2) Gm/ Aha-ha! Now we can find the speed of a mass, eleased fom est at the suface, when it eaches the cente of the eath: U suface + K suface = U cente + K cente Gm/ + 0 = (-3/2)Gm/ + (1/2)mv cente 2 Solving fo v cente yields: v cente = G/ Wow! This is the same as the obital speed fo a single body cicula obit!!

12 Sola System Data Object ass (kg) adius (m) otational Peiod (sec) Obital adius (m) evolution Peiod (sec) Sun 2 x x x 10 6 n/a n/a ecuy 3.28 x x x x x 10 6 Venus 4.83 x x x x x 10 7 Eath 5.98 x x x x x 10 7 as 6.37 x x x x x 10 7 Jupite 1.90 x x x x x 10 8 Satun 5.67 x x x x x 10 8 Uanus 8.80 x x x x x 10 9 Neptune 1.03 x x x x x 10 9 Pluto 6 x x x x x 10 9 oon 7.34 x x x x x 10 6

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

= 4 3 π( m) 3 (5480 kg m 3 ) = kg. CHAPTER 11 THE GRAVITATIONAL FIELD Newton s Law of Gavitation m 1 m A foce of attaction occus between two masses given by Newton s Law of Gavitation Inetial mass and gavitational mass Gavitational potential