Universal Gravitation Unit 8

Lesson 1 : Newton’s Law of Universal Gravitation In 1687 Newton published Mathematical Principles of Natural Philosophy In this work he states : Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

G m1 m2 Fg = r2 G = 6.673 x 10-11 N.m2 / kg2 product of masses Universal Gravitational Constant r2 distance squared G has been measured experimentally. Its value is G = 6.673 x 10-11 N.m2 / kg2

Measuring the Gravitational Constant (G) Henry Cavendish (1798)

Gravitational attraction forms an “action-reaction” pair. F21 = -F12

Force exerted by the Earth on a mass m near the Earth’s surface. The gravitational force exerted by a finite-size, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center. Force exerted by the Earth on a mass m near the Earth’s surface. Fg = G ME m RE2

Newton’s Test of the Inverse Square Law The acceleration of the apple has the same cause as the centripetal acceleration of the moon. aM g = 1/rM2 1/RE2 = RE2 rM2 = (6.37 x 106 m)2 (3.84 x 108 m)2 = 2.75 x 10-4

ac = v2 r = (2prM / T)2 rM = 4p2rM T2 = 4p2(3.84 x 108 m) Centripetal acceleration of the Moon : (2.75 x 10-4)(9.80 m/s2) = 2.70 x 10-3 m/s2 Newton calculated ac from its mean distance and orbital period : ac = v2 r = (2prM / T)2 rM = 4p2rM T2 = 4p2(3.84 x 108 m) (2.36 x 106 s)2 = 2.72 x 10-3 m/s2

Example 1 Three 0.300 kg billiard balls are placed on a table at the corners of a right triangle. Calculate the gravitational force on the cue ball (m1) resulting from the other two balls.

Free-Fall Acceleration and Gravitational Force Equating mg and Fg : mg = G ME m RE2 g = G ME RE2 g = G ME (RE + h)2 Free-fall acceleration at a distance h above Earth’s surface :

Find the mass of the Earth and the average density of the Earth. Example 2 Find the mass of the Earth and the average density of the Earth.

Lesson 2 : Kepler’s Laws of Planetary Motion Kepler’s First Law All planets move in elliptical orbits with the Sun at one focus. F1 , F2 are foci located c from center r1 + r2 = constant 2a = major axis a = semimajor axis 2b = minor axis b = semiminor axis

Eccentricity (e) of an Ellipse describes the shape of the ellipse For planetary orbits, the Sun is at one focus. There is nothing at the other focus. Eccentricity (e) of an Ellipse describes the shape of the ellipse e = c a e = 0.25 0 < e < 1 For circles, c = 0, e = 0. Earth’s orbit e = 0.017 e = 0.97

Since t = r x F = 0, the angular momentum L is constant. Kepler’s Second Law The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. Since t = r x F = 0, the angular momentum L is constant.

dA = ½ (r x dr) = ½ (r x v dt) = In the time interval dt radius vector r sweeps out the area dA, which equals half the area (r x dr) of the parallelogram formed by vectors r and dr. dA = ½ (r x dr) = ½ (r x v dt) = L 2Mp dt constant The radius vector (r) from the Sun to any planet sweeps out equal areas in equal times.

Kepler’s Third Law The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit. Since gravitational force provides centripetal force, G MS MP r2 = MP v2 r Since orbital speed = 2pr / T, G MS r2 = (2pr / T)2 r

T2 = KS a3 Solving for T2 , 4p2 r3 T2 = GMS 4p2 is a constant KS = 2.97 x 10-19 s2/m3 Replacing r with a, T2 = KS a3

Example 1 Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the Sun is 3.156 x 107 s and its distance from the Sun is 1.496 x 1011 m.

Example 2 Consider a satellite of mass m moving in a circular orbit around the Earth at a constant speed v and at an altitude h above the Earth’s surface, as shown above.

a) Determine the speed of the satellite in terms of a) Determine the speed of the satellite in terms of G, h, RE (radius of Earth), and ME (mass of Earth). b) If the satellite is to be geosynchronous (that is, appearing to remain over a fixed position on the Earth), how fast is it moving through space ?

Lesson 3 : Gravitational Potential Energy Ug = mgh only valid when the mass is near the Earth’s surface, where the gravitational force is constant What is the general form of the gravitational potential energy function ?

Gravitational Force is Conservative Work done by gravitational force is independent of path taken by an object. As particle moves from point A to B, it is acted upon by a central force F, which is a radial force. Work done depends only on rf and ri.

So, work done along any path between points A and B = 0 Work done by force F is always perpendicular to displacement. So, work done along any path between points A and B = 0 The total work done by force F is the sum of the contributions along the radial segments. W = F(r) dr ò ri rf

ò DU = Uf – Ui = - F(r) dr GMEm F(r) = - r2 ri rf Negative of the work done by gravitational force F(r) = - GMEm r2 negative sign means force is attractive

Taking Ui = 0 at ri = infinity, ò ri rf dr 1 rf Uf – Ui = GMEm = GMEm [ - ] r2 r ri Uf – Ui = - GMEm ( - ) 1 rf ri Taking Ui = 0 at ri = infinity, U(r) = - GMEm r (for r > RE) Because of our choice of Ui, the function U is always negative.

U(r) = - GMEm r (for r > RE)

In General (for any two particles) U = - Gm1m2 r Gravitational potential energy varies as 1/r, whereas gravitational force varies as 1/r2. U becomes less negative as r increases. U becomes zero when r is infinite.

reduces to the familiar relationship DU = mgDy. Example 1 A particle of mass m is displaced through a small vertical distance Dy near the Earth’s surface. Show that in this situation the general expression for the change in gravitational potential energy given by reduces to the familiar relationship DU = mgDy. Uf – Ui = - GMEm ( - ) 1 rf ri

Lesson 4 : Energy in Planetary and Satellite Motion E = KE + U E = ½ mv2 - GMm r Newton’s Second Law applied to mass m : GMm r2 = ma = mv2 r

Multiplying both sides by r and dividing by 2, ½ mv2 = GMm 2r Substituting into E = ½ mv2 - GMm r , E = GMm r 2r - E = GMm 2r - (total energy for circular orbits)

total energy is negative GMm 2r - total energy is negative KE is positive and equal to half the absolute value of U. For elliptical orbits, we replace r with semimajor axis a, E = GMm 2a - (total energy for elliptical orbits)

In all isolated gravitationally bound two-object systems : Total energy is constant. Total angular momentum is constant.

Example 1 The space shuttle releases a 470 kg communications satellite while in an orbit 280 km above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit, which is an orbit in which the satellite stays directly over a single location on the Earth. How much energy does the engine have to provide ?

When object reaches rmax, vf = 0, and Escape Speed Minimum value of the initial speed needed to allow the object to move infinitely far away from the Earth. When object reaches rmax, vf = 0, and ½ mvi2 - GMEm RE = - Solving for vi2, vi2 = 2GME 1 RE rmax - ( )

If we let rmax  infinity, vi2 = 2GME 1 RE rmax - ( ) If we let rmax  infinity, vesc = 2GME RE

Example 2 Calculate the escape speed from the Earth for a 5000 kg spacecraft, and determine the kinetic energy it must have at the Earth’s surface in order to move infinitely far away from the Earth.

Example 3 (AP 1984 #2) Two satellites, of masses m and 3m, respectively, are in the same circular orbit about the Earth’s center, as shown in the diagram above. The Earth has mass Me and radius Re. In this orbit, which has radius 2Re, the satellites initially move with the same orbital speed vo, but in opposite directions.

a) Calculate the orbital speed vo of the satellites a) Calculate the orbital speed vo of the satellites in terms of G, Me, and Re. b) Assume that the satellites collide head-on and stick together. In terms of vo, find the speed v of the combination immediately after the collision.

c) Calculate the total mechanical energy of the c) Calculate the total mechanical energy of the system immediately after the collision in terms of G, m, Me, and Re. Assume that the gravitational potential energy of an object is defined to be zero at an infinite distance from the Earth.

Example 4 (AP 1992 #3) A spacecraft of mass 1,000 kg is in an elliptical orbit about the Earth, as shown above. At point A the spacecraft is at a distance rA = 1.2 x 107 m from the center of the Earth and its velocity, of magnitude vA = 7.1 x 103 m/s, is perpendicular to the line connecting the center of the Earth to the spacecraft. The mass and radius of the Earth are ME = 6.0 x 1024 kg and rE = 6.4 x 106 m, respectively.

Determine each of the following for the spacecraft when it is at point A. a) The total mechanical energy of the spacecraft, assuming that the gravitational potential energy is zero at an infinite distance from the Earth.

b) The magnitude of the angular momentum of the b) The magnitude of the angular momentum of the spacecraft about the center of the Earth. Later the spacecraft is at point B on the exact opposite side of the orbit at a distance rB = 3.6 x 107 m from the center of the Earth. c) Determine the speed vB of the spacecraft at point B.

Suppose that a different spacecraft is at point A, a distance rA = 1 Suppose that a different spacecraft is at point A, a distance rA = 1.2 x 107 m from the center of the Earth. Determine each of the following. d) The speed of the spacecraft if it is in a circular orbit around the Earth.

e) The minimum speed of the spacecraft at point A e) The minimum speed of the spacecraft at point A if it is to escape completely from the Earth.