Chap13. Uniesal Gaitation Leel : AP Physics Instucto : Kim 13.1 Newton s Law of Uniesal Gaitation - Fomula fo Newton s Law of Gaitation F g = G m 1m 2 2 F21 m1 F12 12 m2 - m 1, m 2 is the mass of the object, and is the distance between the two cente of each mass, not fom the suface - The gaitational foce F g between two objects is diectly elated to the mass and inesely elated to the distance squaed between the two objects - The geate the mass of the object the stonge the F g, the geate the distance between the two objects the weake the F g. - The foce the two masses exeting on each othe is always the same - G is called the uniesal gaitational constant, whee the alue is G=6.674 10-11 N m 2 /kg 2 Q1) Find the gaitational attactie foce between two objects of mass m 1=4kg and m 2=2kg, sepaated by a distance of 6m. (G=6.67 10-11 N m 2 /kg 2 ) a) 0.79G (N) b) 0.68G (N) c) 0.45G (N) d) 0.22G (N) Do Example 13.1 Billads, Anyone? see p.390
13.2 Fee-Fall Acceleation and the Gaitational Foce Weight is actually the gaitational foce F g acting on ou mass against the suface of the eath. So whethe we find ou weight F g using F g = mg o F g = G m 1m 2, the esult is the same. 2 Q2) A peson has a mass of 60kg, find the weight using F g = mg and F g = G m 1m 2. Mass of the eath Me=5.97 1024 kg, adius of the eath 2 R e=6.37 10 6 m, G=6.67 10-11 N m 2 /kg 2. Method I) using F g = mg Re= Method II) using F g = G m 1m 2 2 Ans) 588N fo both cases So fo objects close to the suface of the eath, R e, hence mg = G M em R e 2 g = G M e R e 2 =9.8m/s2 Howee, fo objects that is high altitudes aboe the suface of the eath, g deceases with inceasing altitude. Eentually, when h, g 0. If = R e + h, then So F g = G M em 2 = G M em (R e +h) 2 M e g= G (R e +h) 2
Q3) Find the alue of g when h=2000km using g= G (R e +h) 2, whee Re=6.37 106 m, M e=5.97 10 24 kg and G=6.67 10-11 N m 2 /kg 2. Check you answe on p.391 Table 13.1. M e Do Example 13.2 The Density of Eath see p.392 13.3 Analysis Model : Paticle in a Field (Gaitational) p.392 The pesence of a lage planetay object will poduce a gaitational field. So if anothe paticle of mass m o is placed at a point whee the gaitational foce exists, the paticle will expeience a gaitational foce. The gaitational field g is defined as g F m o = GM 2 Do Example 13.3 The Weight of the Space Station see p.394 Ex) If you wee on the space station aboe, how much would you weigh? 1pound = 0.4536kg and 1N=0.225lbs
**13.4 Keple s Laws and the Motion of Planets see p.394~ Impotant! Keple's thee laws ae ; 1. All planets moe in elliptical obits with the Sun at one focal point. *2. The adius ecto dawn fom the Sun to a planet sweeps out equal aeas in equal time inteals *3. The squae of the obital peiod of any planet is popotional to the cube of the semimajo axis of the elliptical obit See Fig.13.5 Keple s 2 nd Law and Conseation of Angula Momentum Conside a planet of mass M p moing aound the Sun. Thee is no toque acting on the planet, so τ = dl dt = 0 esulting in constant angula momentum. That is, L=constant L = p = M p = constant **Deiation of Keple s 3 d law *~ey impotant! ΣF=F g= M p 2 ( F g is the centipetal foce the esults in cicula motion) => GM SM p = M p 2 2 => GM S 2 = ( 2π T )² => T 2 = ( 4π2 GM S ) 3 => T 2 ~ 3 (emembe that = 2π T fo obital speed) MS Mp Keple s 3 d law states that the obital peiod of any planet is popotional to the cube of the semimajo axis of the elliptical obit Do Example 13.4 Mass of the Sun see p.398
**Do example 13.5 A Geosynchonous Satellite impotant~! see p.399 13.5 Gaitational Potential Enegy see p.400 The gaitational potential enegy U g=mgy is alid only when the object of mass m is nea the Eath s suface. Fo distance fa away fom Eath s suface, g deceases, accoding to g = M 2. If y, then U GMm g=.we can see that gaitational potential enegy goes to zeo as appoaches infinity. Now emembe that the change in potential enegy of a system is defined as the negatie wok done by the foce( see chapte 7, p.199) U = U f U i = f i F()d Since F() = G Mm 2, whee negatie sign means attactie foce, then U f U i= G Mm f d = GMm 1 f i 2 d = GMm[ 1 ] f i 2 i U f U i= GMm( 1 f 1 i ) If we choose U i at infinite distance away( ), the potential enegy will be zeo, so the gaitational potential enegy between two mass M and m sepaated by a distance can be U= GMm The aboe equation shows that the gaitational potential enegy fo any pai of paticles aies as 1/, wheeas the foce between them aies as 1/ 2.
**13.6 Enegy Consideations in Planetay and Satellite Motion impotant~! Conside a planetay body of mass m obiting a mass body of M, like the eath and Sun. The total enegy of the system is the combination of kinetic enegy and gaitational potential enegy E= 1 2 m2 GMm If the satellite is bound to a planet, that is, in stable obit, then GMm 2 = m2 which becomes 1 2 m2 = GMm 2 Hence, the kinetic enegy K = 1 2 m2 can be expessed as GMm, so 2 E = GMm 2 GMm = GMm 2 see p.440 **p.403 Do example 13.7 Changing the Obit of a Satellite **~impotant!!! Fom example13.5, the necessay distance to place a satellite to be geosynchonous obit is =4.22 10 7 m. *~Re=6.37 10 6 m, ME=5.97 10 24 kg
Escape Speed see p.404 fo deiation esc= 2GM E R E Do example 13.8 Escape Speed of a Rocket Following questions ae peious AP multiple choice question. Answe the questions! 1. The mass of Planet X is one-tenth that of the Eath, and its diamete is one-half that of the Eath. The acceleation due to gaity at the suface of Planet X is most nealy (A) 2m/s 2 (B) 4m/s 2 (C) 5m/s 2 (D) 7 m/s 2 (E) 10 m/s 2 2. A satellite taels aound the Sun in an elliptical obit as shown aboe. As the satellite taels fom point X to point Y. which of the following is tue about its speed and angula momentum? Speed Angula Momentum (A) Remains constant Remains constant (B) Inceases Inceases (C) Deceases Deceases (D) Inceases Remains constant X SUN Y 3. A newly discoeed planet, "Cosmo," has a mass that is 4 times the mass of the Eath. The adius of the Eath is R e. The gaitational field stength at the suface of Cosmo is equal to that at the suface of the Eath if the adius of Cosmo is equal to (A) ½R e (B) R e (C) 2R e (D) R e (E) Re 2 4. Two atificial satellites, 1 and 2, obit the Eath in cicula obits haing adii R 1 and R 2, espectiely, as shown.if R 2 = 2R 1, the acceleations a 2 and a 1 of the two satellites ae elated by which of the following? (A) a 2 = 4a 1 (B) a 2 = 2a 1 (C) a 2 = a 1 (D) a 2 = a 1/2 (E) a 2 = a 1/4
5. A satellite moes in a stable cicula obit with speed o at a distance R fom the cente of a planet. Fo this satellite to moe in a stable cicula obit a distance 2R fom the cente of the planet, the speed of the satellite must be (A) o/2 (B) o / 2 (C) o (D) 2 o (E) 2 o 6. A satellite S is in an elliptical obit aound a planet P, as shown aboe, with 1 and 2 being its closest and fathest distances, espectiely, fom the cente of the planet. If the satellite has a speed 1 at its closest distance, what is its speed at its fathest distance? (A) 1 1 (B) 2 2 1 (C) ( 2 1 ) 1 1 (D) 1+ 2 2 1 (E) 2 1 1 + 1 2 7. A pendulum with a peiod of 1 s on Eath, whee the acceleation due to gaity is g, is taken to anothe planet, whee its peiod is 2 s. The acceleation due to gaity on the othe planet is most nealy (A) g/4 (B) g/2 (C) g (D) 2g (E) 4g 8. A satellite of mass M moes in a cicula obit of adius R with constant speed. Tue statements about this satellite include which of the following? I. Its angula speed is /R. II. Its tangential acceleation is zeo. III. The magnitude of its centipetal acceleation is constant. (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III 9. Two identical stas, a fixed distance D apat, eole in a cicle about thei mutual cente of mass, as shown. Each sta has mass M and speed. G is the uniesal gaitational constant. Which of the following is a coect elationship among these quantities? (A) 2 = GM/D (B) 2 = GM/2D (C) 2 = GM/D 2 (D) 2 = MGD (E) 2 = 2GM 2 /D