2. Circular Motion and Gravitation 7-5 Newton's Law of Gravitation The galaxy cluster Abell 2218 is so densely packed that its gravity bends light passing through it. The arc-shaped structures are stars that lie five times further away than the cluster itself. GRAVITY The most pervasive force in the universe.

3. Topic 2.4 Extended A – Newton ’ s law of gravitation  Any piece of matter in the universe will attract all other pieces of matter in the universe.  The gravitational force is the weakest of all the forces: GRAVITY STRONG ELECTROMAGNETIC WEAK + + nuclear force light, heat and charge radioactivity freefall Einstein spent years trying to find the “superforce” which has all four of the fundamental forces of nature as special manifestations. This process is called “unification of forces,” and Einstein was not successful. FYI: After his death, physicists showed that the weak and the electromagnetic forces were manifestations of a single force, called the “electro-weak force.” ELECTRO-WEAK WEAKEST STRONGEST

4. N EWTON'S L AW OR U NIVERSAL G RAVITATION Topic 2.4 Extended A – Newton ’ s law of gravitation  It turns out that the force of attraction F between two point masses m 1 and m 2 is given by F 12 = -G m1m2r2m1m2r2  where F 12 is the force on m 1 caused by m 2,  where G is the universal gravitational constant and has the value G = 6.67×10 -11 N·m 2 /kg 2, m1m1 m2m2  where r is the distance between the centres of the two masses and is measured in meters. r F 12 FYI: F 21, the force on m 2 caused by m 1 is equal and opposite (Newton’s 3 rd, action-reaction pair) F 21 FYI: The gravitational force obeys an inverse square law. We will find out that next year that the electric force also obeys an inverse square law: F = kq 1 q 2 / r 2. Newton’s Law of Universal Gravitation

5. Topic 2.4 Extended A – Newton ’ s law of gravitation  If a mass is attracted to more than one other mass, we simply sum all of the forces together (as vectors, of course).  Find the net force acting on m 1 (2 kg) caused by m 2 (4 kg) and m 3 (6 kg) F 12 F 13 m1m1 m2m2 m3m3 7 m 3 m FYI: We do not need to find the force between m 3 and m 2 because we are interested only on m 1.  We simply use Newton’s Law of Gravitation twice: |F 12 | = G m1m2r2m1m2r2 = G (2)(4) 7 2 = 0.163G F 12 = 0.163G x |F 13 | = G m1m3r2m1m3r2 = G (2)(6) 3 2 = 1.333G F 13 = 1.333G y F net = F 12 + F 13 = 0.163G x + 1.333G y N EWTON'S L AW OR U NIVERSAL G RAVITATION FYI: This example illustrates the principle of superposition which means that the total gravitational force on a mass is simply the vector sum of the gravitational forces of all the masses surrounding it.

6. Topic 2.4 Extended A – Newton ’ s law of gravitation  Consider a mass m located near the earth’s surface:  It will feel an attractive force caused by the earth’s mass M, given by F = G mM r 2  But from Newton’s 2 nd law we have F = ma g where a g is the acceleration due to gravity. where r is the distance from the center of the earth.  Equating the two forces we have ma g = G mM r 2 so that a g = GM r 2 Gravitational acceleration near surface of planet G RAVITATION AT A P LANET'S S URFACE

7. Topic 2.4 Extended A – Newton ’ s law of gravitation G RAVITATION AT A P LANET'S S URFACE  At the surface of the earth, this reduces to a g = GM r 2 (6.67×10 -11 )(5.98×10 24 ) (6.37×10 6 ) 2 = = 9.829878576 m/s 2 Question: Is this the expected result?  As an aside, you might be curious as to how the various constants were found.  The radius of the earth R E is an easy-to-find value, and it was known to a good approximation by the Greek astronomer and mathematician Eratosthenes (3 B.C.).  The value of the universal gravitational constant G was considerably more difficult to find. In 1798, an experimental physicist by the name of Henry Cavendish performed a very delicate experiment to determine G.  Even Newton did not know the value of G.  Finally, knowing the value of freefall acceleration, one can indirectly calculate the mass of the earth knowing G and R E.

8. Topic 2.4 Extended A – Newton ’ s law of gravitation G RAVITATION AT A P LANET'S S URFACE  Our formula for a g works for any spherical mass, such as the moon, or the sun, or any other planet, satellite, or star.  For an astronaut on the surface of the moon, for example, a g = GM r 2 (6.67×10 -11 )(7.36×10 22 ) (1.74×10 6 ) 2 = = 1.62 m/s 2. FYI: This is 9.8/1.6 = 1/6 th that of the freefall acceleration on the surface of earth.  For an astronaut in the space shuttle orbiting at an altitude of 200 km, a g = GM r 2 (6.67×10 -11 )(5.98×10 24 ) (6.37×10 6 + 200000) 2 = = 9.24 m/s 2. FYI: This is nearly that of the freefall acceleration on the surface of earth. Why are the astronauts considered to be “weightless?”

9. Topic 2.4 Extended A – Newton ’ s law of gravitation G RAVITATION AT A P LANET'S S URFACE  Now, it just so happens that g at the equator is less than g at the poles.  The reason for this is that the earth is rotating. RERE r  Each latitude has a different centripetal acceleration given by a c = rω 2  The earth has a rotational velocity given by ω = 2π rad 24 h 1 h 3600 s × ω = 7.3×10 -5 rad/s

10. Topic 2.4 Extended A – Newton ’ s law of gravitation G RAVITATION AT A P LANET'S S URFACE  A fbd for a mass on the equator like like this: RERE r W acac FBD mass on equator  We have: ΣF x = ma N - W = -ma c N N = W - ma c N = m(a g – a c )  The normal force is the apparent weight of the mass, so that W apparent = m(a g – a c ) mg apparent = m(a g – a c ) g apparent = a g – a c g apparent = – R E ω 2 GM R E 2 g apparent = – 6.37×10 6 · (7.3×10 -5 ) 2 (6.67×10 -11 )(5.98×10 24 ) (6.37×10 6 ) 2 g apparent = 9.795932846 m/s 2 = 9.829878576 m/s 2 Compare to g for a stationary earth… FYI: If a g = a c, the apparent weight of the mass becomes zero. Since it has no reason to stay in contact with the planet, it is free to “leave.” At this stage, scientists believe a planet (or star) would disintegrate or “explode.”

11. Topic 2.4 Extended A – Newton ’ s law of gravitation N EWTON’S S HELL T HEOREM  A uniform spherical shell of matter exerts no net gravitational force on a particle located inside it. m M F mM = 0 For a sphere, the net force from opposite conic sections exactly counter-balance one another… m M F mM = 0 More mass Farther away Less mass Closer No matter where inside the sphere the particle is located. Force on particle inside spherical shell F mM = 0

12. Topic 2.4 Extended A – Newton ’ s law of gravitation N EWTON’S S HELL T HEOREM  A uniform spherical shell of matter exerts a net force on a particle located outside it as if all the mass of the shell were located at its center. m M r FYI: Proof of Newton’s shell theorem required the use of calculus. This is one of the main reasons Newton invented integral calculus! Force on particle outside spherical shell F mM = GmM r 2

13. Topic 2.4 Extended A – Newton ’ s law of gravitation  Even though the earth is not homogeneous, we can use Newton’s shell theorem to prove that we were justified in treating the earth as a point mass located at its center in all of our calculations inner core M i outer core M o mantle M m crust M c  For a point mass m located a distance r from the center of the earth we have F mM = GmM c r 2 + GmM m r 2 + GmM o r 2 + GmM i r 2  so that F mM = Gm(M c +M m +M o +M i ) r 2 F mM = GmM E r 2 N EWTON’S S HELL T HEOREM

14. Topic 2.4 Extended A – Newton ’ s law of gravitation G RAVITATIONAL P OTENTIAL E NERGY  We’ve already discussed gravitational potential energy U = mgy.  The problem with this formula is that it is a local formula: It works only in the vicinity of the surface of the earth.  It is beyond the scope of this course to prove the following formula, but for point masses, or celestial- sized spherical masses, U = - GmM r Gravitational potential energy

15. Topic 2.4 Extended A – Newton ’ s law of gravitation G RAVITATIONAL P OTENTIAL E NERGY  Consider the three charges shown here, “assembled from infinity.” How much potential energy is stored in the configuration?  Find the potential energies in pairs, then sum them up. Since U is a scalar, this is easy. m3m3 m1m1 m2m2 r 12 r 23 r 13 U = - + + Gm 1 m 2 r 12 Gm 2 m 3 r 23 Gm 1 m 3 r 13

16. Topic 2.4 Extended A – Newton ’ s law of gravitation E SCAPE V ELOCITY  Escape velocity it the minimum velocity required to escape the gravitational force of a planet.  Consider a rocket of mass m on the surface of a planet of mass M, say the earth: M m R  We will use energy considerations to find the escape velocity. K + U = K 0 + U 0  If the rocket can reach to r = ∞, and come to a stop there, it has escaped the earth: + - = + - GmM r 0 GmM r 1212 mv 0 2 1212 mv 2 0 0 = GmM R 1212 mv 0 2 2GM R v esc = escape velocity

17. Circular Motion and Gravitation 7-5 Newton's Law of Gravitation E SCAPE V ELOCITY  For us, the escape velocity from the earth is 2(6.67×10 -11 )(5.98×10 24 ) 6.37×10 6 v esc = 2GM R v esc = v esc = 11191 m/s v esc = 25027 mph !

18. Topic 2.4 Extended A – Newton ’ s law of gravitation B LACK H OLES  Soon after publication of Newton’s law of gravity a mathematician by the name of Laplace postulated the existence of a black hole – a body so massive the even light cannot escape from it: 2GM R v esc = 2GM R c = 2GM c 2 R s = 2GM R c 2 = Schwarzchild radius of a black hole FYI: The Schwarzchild radius tells us what the radius of an object of mass M would have to be in order to become a black hole. FYI: Laplace used INCORRECT methods to derive his formula. It wasn't until shortly after 1915 and Einstein's publication of his general theory that Schwarzchild postulated the existence of a black hole, and used the general theory to derive the same formula (CORRECTLY). He therefore gets the honor. We have shown Laplace's method.

19. Topic 2.4 Extended A – Newton ’ s law of gravitation B LACK H OLES  For the earth, R S is given by 2GM c 2 Rs = Rs = 2(6.67×10 -11 )(5.98×10 24 ) (3×10 8 ) 2 Rs = Rs = R s = 0.00886 m = 8.86 mm