Lecture11 - Gravita.on Before Newton: Ancient Greeks believed: Things fall because they are part of the element ‘earth’

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Unformatted text preview: Gravita.on Before Newton: Ancient Greeks believed: Things fall because they are part of the element ‘earth’ and are seeking their natural res.ng place. Air and Fire rise because their res.ng place is in the heavens. Stars and Planets are the realm of God… Later: Angels are pushing the planets in their orbits. Newton (1687): All these phenomena (things falling, hot air rising, the moon circling the earth, the earth circling the moon, projec.le mo.on etc.) can all be explained by one force: Gravita4on. A cartoonist thinks we would have laws of gravita5on a li6le earlier! Chapter 13 Newton’s law of universal gravita.on In 1687 in his ‘Principia’, Newton stated the law of universal gravita.on. This is called Newton’s law of universal gravita.on and can be stated as: Every par.cle in the universe aSracts every other par.cle with a force that is directly propor.onal to the product of their masses and inversely propor.onal to the square of the distance between them. If the par.cles have masses m1 and m2 and are separated by a distance r, the magnitude of the graçl force is: mm F = G 12 2 r where G is a constant, called the universal gravita5onal constant. In SI units, its value is G = 6.67 × 10-11 N m2/kg2. Direc.on of the force is toward the center of the other par.cle. m1 r m2 Newton’s law of universal gravita.on (contd…) We can write Newton’s law of gravita.on in vector form. Then it gives both direc.on and magnitude. mm ˆ F12 = G 1 2 2 r12 ...........(1) r , i.e., ˆ r where r is the unit vector along r ˆ r= r m1 F12 r12 r F21 r21 m2 Again note that my order of index used to label force is opposite to that of the textbook. F12 here is the force on the body 1 by body 2. If we use textbook’s conven.on there should be a minus sign on the right of equa.on (1). Accelera.on of free fall, g Consider a par.cle of mass m at a height h from the surface of the earth. Earth’s gravita.onal pull (force) on that object is: F=G MEm MEm =G r2 ( RE + h )2 m Where ME and RE are the mass and radius of the earth. From the observed value g of the accelera.on free fall, we find that the force on the body is F = mg Thus we get mg = G ME RE ( RE + h ) MEm 2 ⇒g=G ( RE + h ) GM E 2 RE ME 2 Al.tude (h in km) 0 g (m/s2) 9.83 9.80 9.71 8.70 0.225 Example Sea level Mount Everest Highest Balloon ride Space ShuSle Geo ­ sta.onary satellite On the surface of the Earth (h=0) g= 8.8 36.6 Note that above the surface of the earth g falls as 1/r2 ( where r = RE + h). 400 35700 How does g vary inside the earth? Let’s try to get an answer by modeling the earth as a sphere of uniform density. This highly idealized model since earth’s core is much more dense than its crust and the earth is not a perfect sphere. Imaging that we dig a deep hole ver.cally down from the surface and put a body of mass m there. The mass of the earth contribu.ng to the gravita.onal force is the mass inside the outer shell (shaded darker color in the figure). Once again equa.ng the gravita.onal force and and mg at that loca.on (g is a variable here), we get mg = G M in m GM ⇒ g = 2 in ..........(1) r2 r 3 mg r Min E ME M RE RE Assuming constant density inside the earth, we have (1) & (2) give g = ⎛ G ⎞ ⎛ M E r 3 ⎞ ⇒ g = ⎛ GM E ⎞ r , that is g ∝ r ⎜ 2⎟⎜ ⎜ R3 ⎟ ⎝r ⎠⎝ RE ⎟ ⎠ ⎝ E⎠ Thus assump.on of constant constant density, gives g propor.onal to r inside the earth – g increases linearly with r. Outside the earth, we know g decreases 1/r2. ⎛ ⎞ ⎛ ME ⎞ r3 ⎜ ME ⎟ ⎛ 4 3⎞ M in = ρVin = ⎜ Vin = ⎜ ⎟ ⎜ 3 π r ⎟ = M E R 3 .............(2 ) 4 ⎝ ⎠ 3 ⎝ VE ⎟ ⎠ E ⎜ π RE ⎟ ⎝3 ⎠ g∝r g∝ 1 r2 Example Problem 4, Ch 13. (Chalkboard stuff) Example In some Science Fic5on stories, people drill a hole through the en5re Earth and drop something in it. If you could really do that, what would happen ? Force on the body is ⎛ GM ⎞ F = mg = m ⎜ 3 E ⎟ r ⎝ RE ⎠ This force is directed to the center of the earth. To include this direc.on, we may just put minus sign the right hand side. ⎛ GM ⎞ F = − m ⎜ 3 E ⎟ r ⇒ F = − kr ⎝ RE ⎠ where k= GM E m R3 E This is clearly the restoring force of simple harmonic mo.on (like the spring force F =  ­kx. So the object moves back and forth in the tunnel. Gravita.onal poten.al energy Recall: W = ∫ F ( r )i dr = − ΔU ri rf m rf or, ΔU = − ∫ F ( r )i dr .....(1) ri Consider a mass m moving in the gravita.onal field of the earth. The force on m is GM m F=− r E 2 r R ˆ r M Suppose we are moving from along r away from the center. Then GM E m ⎛ GM E m ˆ ⎞ ˆ F i dr = − ⎜ r ⎟ i( rdr ) = − dr 2 ⎝ r2 ⎠ r With this equa.on (1) gives rf ΔU = U f − U i = ∫ ri GM E m r2 rf dr ⇒ U f − U i = GM E m ∫ ri ⎛ 1 1⎞ 1 dr ⇒ U f − U i = −GM E m ⎜ − ⎟ r2 ⎝ r f ri ⎠ This is the change in gravita.onal poten.al energy of the the par.cle ­earth system. The reference point from zero poten.al energy is arbitrary. Let’s take that point to be ri = ∞. This gives the poten.al energy func.on at point r: U (r ) = − GM E m r Gravita.onal poten.al energy (contd…) Gravita.onal poten.al energy of a par.cle of mass m at a distance r ≥ RE from the center of the earth: GM E m U (r ) = − .....(1) r In deriving this expression we have assumed that gravita.onal poten.al energy at infinite distance (where there is no influence of the earth) is zero. The nega.ve sign means that the par.cle is bound to the system by gravita.on and is not free. Recall that in Chapter 7, we had this rela.on for poten.al energy: U = mgh This rela.on is applicable near the surface of the earth (where g can be taken as constant throughout the height h). Also, zero poten.al energy point is h = 0. Again recall that force is nega,ve gradient of poten,al energy: F = − dU For U(r) given by (1): F = − d ⎛ GM E m ⎞ GM E m ⎜− ⎟ =− dr ⎝ r⎠ r2 dr we get back the gravita.onal force, as expected. (Nega.ve sign: force is directed toward mass ME.) Escape Speed When you throw something it usually falls back to Earth. How fast does an object have to go to be able to leave the influence of the gravita5onal force of a planet ? Assume an object reaches a distance of no gravita.onal influence (infinity) of the planet. At this point we’ll assume it has lost all its kine.c energy and just gets free from the gravita.onal pull. Therefore its total minimum energy is: Since this is a conserva.ve system, this total energy is conserved. So at the planet’s surface, where the object will have its required speed (escape speed): Examples: Body Mass (kg) 1.17x1021 7.36x1022 5.98x1024 1.90x1027 Radius (km) 380 1,740 6,370 71,500 Escape speed (km/s) 0.64 2.38 11.2 59.5 Solve for v: v= 2GM R Ceres (Asteroid) Moon Earth Jupiter Note that escape speed does not depend on object, only on the planet’s mass & radius. To get the space shuSle into orbit you do not need to go as fast as the escape speed. Just fast enough to enter orbit around earth (about 7.2 km/s = 16,000 mph). ...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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