Phys 2101 Gabriela González

Size: px

Start display at page:

Download "Phys 2101 Gabriela González"

Jessica Barker
5 years ago
Views:

1 Phys 2101 Gabriela González

2 Newton s law : F = Gm 1 m 2 /r 2 Explains why apples fall, why the planets move around the Sun,

3 sciencebulletins.amnh.org And in YouTube!

4 Explains just as well as Newtons why things fall and planetary motion Einstein s messengers, National Science Foundation video einsteinsmessengers.org.. but it also predicts traveling away from moving masses!

5 Newton s gravitation law (not F=ma!!): F = G m 1 m 2 r 2 G = 6.67 x N m 2 /kg 2 5

6 We can add gravitational forces. We have to add them as vectors!! All particles have the same mass. What s the direction of the force on the labeled particle in each arrangement? Which arrangement has the largest force on the labeled particle? 6

For solid bodies, we have to add up the forces from all parts in the body: integrals!! Luckily, for spherical shells, things get simplified: also solid spheres!

7 For solid bodies, we have to add up the forces from all parts in the body: integrals!! Luckily, for spherical shells, things get simplified: also solid spheres! A uniform spherical shell of matter attracts a particle that is outside the shell as if all of the shell s mass were concentrated in the center. A uniform spherical shell of matter exerts no net gravitational force on a particle located inside it. 7

8 F N =mg GMm/R 2 Normal force of earth on us is weight, and if we are just standing on a scale, it is equal to Earth s attractive gravitational force. g=gm/r 2 =6.67x10-11 Nm 2 /kg 2 x 5.98 x10 24 kg /(6.37x10 6 m) 2 = 9.8 m/s 2 At the Moon, g=gm/r 2 =6.67x10-11 Nm 2 /kg 2 x 7.35 x10 22 kg /(1.74x10 6 m) 2 = 1.6 m/s 2 If objects fall from a height h, the force is F = GMm/r 2 = GMm/(R+h) 2 = ma g, with free fall acceleration a g =GM/(R+h) 2. Notice that the Earth also falls towards the apple, but with a much smaller acceleration! The Hubble telescope is 559km above the Earth: a g =6.67x10-11 Nm 2 /kg 2 x 5.98 x10 24 kg /(6.37x10 6 m+5.59x10 5 m) 2 = 8.3 m/s 2 8

Would you weigh the same in the North pole and in Quito, Ecuador? F N =mg At the North pole, g=a g =GM/R 2 GMm/R 2 =ma g Everywhere else, people just standing are moving in circles!

9 Would you weigh the same in the North pole and in Quito, Ecuador? F N =mg At the North pole, g=a g =GM/R 2 GMm/R 2 =ma g Everywhere else, people just standing are moving in circles! Because the planet is rotating and we are rotating with it, we know the net force on us resulting from gravity and from Earths normal force is a centripetal force pointing towards the center of the circular trajectory. GMm/R 2 =ma g F N =mg = mrω 2 At the Equator, g = GM/R 2 Rω 2 With R=6.37x10 6 m and ω=2π/86400 rad/s, we get Rω 2 = m/s 2 (less than 0.5% of 9.8 m/s 2 ) But the difference would be much larger if the Earth rotated faster can the weight be zero?? 9

10 Neutron stars are extremely dense stars, that can have rotation with large angular velocities. A neutron star is rotating at 1200 rpm, has a radius of 15km, and a mass of 1.4 solar masses. What is the gravitational acceleration on the star s surface? How fast a rotation would make a particle weightless at the surface? Fastest spinning pulsar: The scientists discovered the pulsar, named PSR J ad, in a globular cluster of stars called Terzan 5, located some 28,000 light-years from Earth in the constellation Sagittarius. The newly-discovered pulsar is spinning 716 times per second, or at 716 Hertz (Hz), readily beating the previous record of 642 Hz from a pulsar discovered in For reference, the fastest speeds of common kitchen blenders are Hz. 10

11 Potential energy = work done by gravitational forces. Work required to move a mass m from R to : W = R F(r) idr = GMm dr = GMm r 2 r Potential energy of a system of masses: R ΔU = W U U(R) = W = GMm R If U = 0, then U(r) = GMm r R = GMm R 11

12 If no other forces are acting, we know that mechanical energy is conserved: ΔKE+ΔU=0. When an object is launched from Earth with some kinetic energy, the kinetic energy is transferred to potential energy as it goes up: E = K + U = 1 2 mv 2 0 GMm R = 1 2 mv2 GMm (R + h) If initial energy is positive, the projectile can reach infinity with some speed. The minimum velocity for the energy to be positive is the escape speed : v e 2 = 2GM R 12

13 What is the escape speed in Earth? How far from the surface will a particle thrown at 112m/s go? With what speed will an object it the Earth if it is dropped from the space shuttle orbiting at 400km above the Earth? 13

14 Three laws, but all consequences of just one: Newton s law of gravitation! LAW OF ORBITS: All planets move in elliptical orbits, with the Sun at one focus. x 2 a + y2 2 b = 1 2 a = semimajor axis b = semiminor axis Sun s distance from center =ea e= eccentricity If e=0, orbit is circular. Animations by by Bill Drennon, Physics Teacher Central Valley Christian High School Visalia, CA USA, 14

15 1 AU = meters 15

16 LAW OF AREAS: A line that connects the planet to the Sun sweeps out equal areas in the plane of the planet s orbit in equal times. Areal velocity da/dt is constant da dt = d(rθr / 2) dt = 1 2 r 2 ω Angular momentum: L = r p = rmv = rmωr = mr 2 ω da dt = L 2m Conservation of angular momentum! 16

17 LAW OF PERIODS: The square of the period of any planet is proportional to the cube of the semimajor axis of the orbit. T 2 = 4π 2 GM a3 17

In spring 2002 S2 was passing with the extraordinary velocity of more than 5000 km/s at a mere 17 light hours distance -- about three times the size of our solar system -- through the perinigricon,

18 In spring 2002 S2 was passing with the extraordinary velocity of more than 5000 km/s at a mere 17 light hours distance -- about three times the size of our solar system -- through the perinigricon, the point of closest approach to the black hole. By combining all measurements of the position of S2 made between spring 1992 and summer 2002, we have obtained enough data in order to determine a unique keplerian orbit for this star, presented in Figure 1. It is highly elliptical (eccentricity 0.87), has a semimajor axis of 5.5 light days, a period of 15.2 years and an inclination of 46 degrees with respect to the plane of the sky. From Kepler's 3rd law we can determine the enclosed mass in a straightforward manner to be 3.7±1.5 million solar masses. Therefore at least 2.2 million solar masses have to be enclosed in a region with a radius of 17 light hours. Schödel, R. et al. A star in a 15.2-year orbit around the supermassive black hole at the centre of the Milky Way. Nature, 419, , (2002). 18

19 Energy conservation: kinetic + potential = constant! Newton s Gravitational law (circular orbits): U = GMm/r F=ma GMm/r 2 =m(v 2 /r) Then, K = ½ m v 2 = GMm/2r = U/2 E = K+U = U/2+U= U/2 = GMm/2r Elliptical orbits: E= GMm/2a 19

Chapter 13. Gravitation

Chapter 13. Gravitation Chapter 13 Gravitation 13.2 Newton s Law of Gravitation Here m 1 and m 2 are the masses of the particles, r is the distance between them, and G is the gravitational constant. G =6.67 x10 11 Nm 2 /kg 2