Universal Gravitation Physics 221 Department of Physics The Citadel Lecture Notes S. Yost November 14, 2008 Universal Gravitation

Announcements Set 13 – Ch. 13 problems due next Wednesday. Next: Chapter 14 – fluids. Read sections 1-3 for Monday, 4-5 for Wednesday, 6-7 for Friday. I will be away until Thursday. Monday – Virtual Lecture: Log into web assign before 5PM Monday for lecture notes and a set of examples to be worked on-line. This will be treated as a homework set.

Universal Gravity Newton’s Law of Gravity describes a universal attractive force between all massive objects. F = G m1 m2 / R2 where R is the separation, and G is Newton’s gravitational constant: G = 6.67 x 10-11 Nm2/kg2 Gravity is very weak between objects on a human scale. In fact, it is the weakest of all known forces, by far. It is important only because it is always attractive, and is universal, so that it can be significant for large masses.

Mass of the Earth What is the Earth’s mass? Weight is the gravitational attraction of an object to the Earth, but the Earth is big – not at a single location. The weight is the total gravitational attraction of an object to all the earth’s parts: It looks like it should be complicated: The gravitational force is the sum of the forces of all the little pieces dM of the Earth. This gives an integral: mg = ∫ Gm dM r/r2 dM r m Fortunately, this integral is simple for spherically symmetric objects.

Gravitating Sphere Don’t worry, we won’t actually do the integral, but I’ll give the answer… Consider a hollow uniform spherical shell of mass M: a cosmic ping-pong ball. If m is outside a spherical shell of mass M, then the gravitational force is F = GmM/R2, where R is the distance from m to the center of the shell. This is as if the shell were collapsed down to a point at its center. M m R

Gravitating Sphere Interestingly, if you are inside the sphere, the gravitational force is zero. Intuitively, this is because, if you look at the mass enclosed by two back-to-back cones of equal angle, it is proportional to the square of the distance to the cone, so that m/r2 is the same for two cones in opposite directions, and the forces cancel. m1 If you are inside the earth, a distance r from the center, you only feel an attraction to the rock inside radius r, not outside. r1 m r2 m2 Gmm1/r 12= Gmm2/r22

Mass Of Earth Then the weight of an object on the Earth’s surface is mg = GMm/Re2 with Re = 6370 km. GM = gRe2. (Useful relation! It saves you from having to remember G and M in many problems) M = g Re2/G = (9.8 m/s2)(6.37x106 m) 6.67 x 10-11 Nm2/kg2 = 5.96 x 1024 kg. Actually, 5.98 x 1024 kg. (The earth isn’t exactly a sphere.)

Radius of the Earth Fun facts: You don’t have to look up or remember the radius of the Earth if you remember that the meter was defined so the North Pole is 10,000 km from the equator. This implies that (p/2)Re = 10,000 km, or Re = 20,000 km/p = 6370 km. Eratosthenes first measured the radius of the Earth in 225 BC by noting that the sun’s rays made an angle q = 1/8 radians at a location 800 km south of a point where it was directly overhead. This implies that 800 mi = Re/8, or Re = 6400 km. 10,000 km p/2 Re q d q Re d = Re q

Orbital Velocity Example We can use the inverse square law to find the orbital velocity of a spaceship at a height 2Re above the Earth’s surface. (Re = 6370 km) 2Re Re

Orbital Velocity Example The spaceship is 3 times as far from the center of the Earth as the ground is, so at this height, a = g/9 = 1.1 m/s2 Setting the gravitational acceleration equal to the centripetal acceleration gives a = g/9 = v2/R = v2 / (3Re). v = (3 x 6370x103 m x 1.1 m/s2) ½ = 4.59 km/s.

Binary Sun Example Suppose that the earth were replaced by another sun. What would be their orbital period? All the other planets are negligable in mass – this is a binary system. Each sun attracts the other. M R = 1AU M

Binary Sun Example The centripetal acceleration of either sun is a = v2/R. The force on either sun is GM2/(2R)2. So v2 = GM/(4R). Compare Earth’s velocity: ve2 = GM/R. v/ve = ½ . So the period is 2 earth years. M R = 1AU M

Gravitational Potential Energy The gravitational attraction is F(r) = -G Mm/r2. G = 6.67 x 10-11 Nm2/kg2. The gravitational potential energy is U(r) = - G Mm/r. The integration constant is chosen so that U(r) = 0 at r = ∞ F m r M

Gravitational Potential Energy If an object falls to Earth from rest at a very great distance, how fast does it enter the Earth’s atmosphere? “Very great distance” means infinite, or in practical terms, far enough away that it doesn’t affect the result. An object infinitely far away would, in reality, never get here, so don’t take this too literally.

Gravitational Potential Energy The initial energy is zero, since v = 0 implies K = 0 and r = ∞ implies U = 0. The final energy is then 0 = ½ mv2 + U(Re) = ½ mv2 – gmRe2/Re Since 0 = ½ mv2 – gmRe the entry velocity is v = √ 2gRe = √2(9.8 m/s2)(6.38 x 106 m) = 11 km/s.

Gravitational Potential Energy Conversely, v = √ 2gRe = 11 km/s is the velocity at which a projectile would have to be launched to get to r = ∞ with zero velocity. This is called the escape velocity.

Energy In Orbit What is the total energy of an orbiting satellite? K = ½ mv2. U = -GMm/R. But mv2/R = GMm/R2. (Centripetal force) Then the total energy of an orbiting satellite is U = - ½ GMm/R: half its potential energy.

Kepler’s Laws Before Newton, Kepler had analyzed a vast collection of observations by Tycho Brahe, and created a mathematical model of the motion of the planets. A mathematical model is a precise description of a phenomena, but one not necessarily intended to explain the phenomena. This model is known as Kepler’s Laws.

Kepler’s Laws Law 1: All planets move in elliptical orbits. P = perihelion A = aphelion

Elliptical Orbit Rp = perigee (perihelion) Ra = apogee (aphelion) Rp + Ra = 2R. R R Ra Rp

Kepler’s Laws Law 2: A line joining any planet to the sun sweeps out equal area in equal times.

Proof of 2nd Law Angular momentum conservation: Gravity acts along a line through the sun: no torque about the sun. L = r x mv, L/m = |r x v| Lt/m = r x vt = area of parallelogram with sides r and vt = 2 x area of triangle “swept out” = 2A Result: A = Lt/2m dA/dt = L/2m = constant vt A r r vt

Proof of 2nd Law The total area inside ellipse depends on L: A = LT/2m = pab. if T = period. More angular momentum gives a “fatter” orbit. a b A = pab area of an ellipse

Kepler’s Laws Kepler’s Third Law: The square of a planet’s year, divided by the cube of its semimajor axis, is the same for all planets: T2/R3 = TE2/RE3 If T is measured in Earth years, and R in Astronomical Units (the distance from the Earth to the sun), then TE = RE = 1, and for any other planet, T2 = R3.

Kepler’s Third Law 1 AU = 1.50 x 1011 m 1 year = 3.16 x 107 s Example: Saturn’s semimajor axis is R = 9.58 AU, giving an orbital period of T = (9.583) ½ = 29.7 years It is easy to prove Kepler’s Law for Circular orbits…

Kepler’s Third Law 4p2R3 = GMT2. Centripetal Acceleration: a = v2/R Newton’s Law: ma = GmM/R2 Then v2 = GM/R. v = 2pR/T gives Kepler’s Third Law: 4p2R3 = GMT2. But don’t bother remembering this – it’s how we got it that matters. For an ellipse, replace R by the semi-major axis to get the general form of Kepler’s Law. v R

Kepler’s Third Law 4p2R3 = GMT2. still holds for elliptical orbits if R is the semi-major axis. We won’t prove it, but it can be done algebraically by calculating the energy and angular momentum at the perigee and apogee, and using the fact that they are conserved, and the 2nd Law. You would also find that the total energy of the orbit is E = - ½ GM/R, generalizing the earlier result for the energy of an orbit. R