Gravitation KEPLER'S LAWS. Tycho Brahe, into three simple laws that describe the motion of planets.

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1 Newton's Law oi Gravitation KEPLER'S LAWS Johannes Kepler spent years of exhaustive study distilling volumes of data collected by his mentor, Tycho Brahe, into three simple laws that describe the motion of planets. 123

2 Kepler's First Law Every planet moves in an elliptical orbit, with the Sun at one focus. planet semimajor axis Kepler's Second Law As a planet moves in its orbit, a line drawn from the Sun to the planet sweeps out equal areas in equal time intervals., These areas are equal. planet Kepler's Third Law If T is the period and a is the length of the semimajor axis of a planet's orbit, then the ratio T2/a3 is the same for all the planets. NEWTON'S LAW OF GRAVITATION Newton eventually proved that Kepler's first two laws imply a law of gravitation: Any two objects in the universe exert an attractive force on each other called the gravitational force whose strength is proportional to the product of the objects' masses and inversely proportional to the square of the distance between them. If we let G be the universal gravitational constant, then the strength of the gravitational force is given by the equation: 124 CRACKING THE AP PHYSICS B & C EXAMS

3 The forces F1-on.2 and F2Hm., act along the line that joins the bodies and form an action/reaction pair. The first reasonablyaccurate numerical value for G was determined by Cavendish more than one hundred years after Newton's Law was published. To three dedmal places, the currently accepted value of G is G = 6.67 x 10-" N-mVkg2 Kepler's Third Law then follows from Newton's Law of Gravitation. We'll show how this works, for the case of a circular orbit of radius R (which can be considered an elliptical orbit with eccentricity zero). If the orbit is circular, then Kepler's Second Law says thatthe planet's orbit speed, v, must be constant. Therefore, the planet executes uniform circular motion/and centripetal force is provided by the gravitational attraction of the Sun. If we let M be the mass of the Sun and m the mass of the planet, then this last statement can be expressed mathematically as: mv2 _rmm "^"-G~Rr (1) The period of a planet's orbit is the time it requires to make one revolution around the Sun, so dividing the distance covered, 2 nr, by the planet's orbit speed, v, we have Equation (1) implies that o2 = GM/R. Squaring both sides of Equation (2) and then substituting = GM/R, we find that v v2 ~GM/R~GM' Therefore, T2 4tc2 t =, a constant R3 GM which is Kepler's Third Law for a circular orbit of radius R. NEWTON'S LAW OF GRAVITATION 125

4 THE GRAVITATIONAL ATTRACTION DUE TO AN EXTENDED BODY Newton's Law of Gravitation is really a statement about the force between two point particles; objects that are very small in comparison to the distance between them. Newton also proved that a uniform sphere attracts another body as if all of the sphere's mass were concentrated at its center. m For this reason, we can apply Newton's Law of Gravitation to extended bodies, that is, to objects that are not small relative to the distance between them. Additionally, a uniform shell of mass does not exert a gravitational force on a particle inside it. This means that if a spherical planet is uniform, then as we descend into it, only the mass of the sphere underneath us exerts, a gravitational force; the shell above exerts no force because we're inside it. ; Example 7.1 What is the gravitational force on a particle of inass hi at a! - distance * from the center of a spherically-symmetric planet of uniform density p> total-mass-% and radius R tot (b) x<r 126 CRACKING THE AP PHYSICS B & C EXAMS

5 Solution. (a) If x t R, then the planet can be treated as a point particle with all its mass concen trated at its center, and (b) x However, if x < R, then only the mass within the sphere of radius x exerts a gravita tional force on the particle. Since the volume of such a sphere is (4/3) nx3, its mass is (4/3) nx3 p; well denote this by Mwlttlin x. Since the mass of the entire planet is (4/3) nr3 p, we see that M jcc3p x3., x3.. = - =7 = ^3 =* Mwithinx=^3^ Therefore, the force that this much mass exerts on the particle of mass m is (x<r) In summary then, ' Mm, _ G j-x for x < R *grav Mm,. D G 5- forx^r x grav die mass of the earth. lo*m> determine Solution. Consider a small object of mass m near the surface of the earth (mass M). Its weight is mg, but its weight is just the gravitational force it feels due to the earth, which is GMm/R2. Therefore, -.Mm.. gr2 m*=glf "* M g NEWTON'S LAW OF GRAVITATION 127

6 Since we know thatg = 9.8 m/s2 and G = 6.67 x KH1 N-mVkg2, we can substitute to find M _ gr2 _ (9.8 m/ s2)(6.37 x 106 m)2 24 G 6.67xlO-nN-m2/kg2-60x10 k8 Example 7.3 We can derive the expression GM/R2 by equating mg and GMm/R2 (as we did in the previous example), and this gives tile magni tude of tile absolute gravitational acceleration, a quantity that's sometimes denoted ^0. The notation ^ is acceleration, but with the spinning of the earth taken into account. Show that if an object is at the equator, its measured weight, mg, is less than its true weight, mgv and compute the weight difference for a person of mass m = 60 kg. Solution. Imagine looking down at the earth from above the North Pole. The net force toward the center of the earth is Fo - FN, which provides the centripetal force on the object. Therefore, u n R Since v = 2 JtR/T, where T is the earth's rotation period, we have or, since Fo = mg0 and FN = mg, 0 "~r{ T ) ~ T2 _ 4n2mR t Since the quantity 4 rpmr/t1 is positive, mg must be less than mg0. The difference between mg mg, for a person of mass m = 60 kg, is only: and 128 CRACKING THE AP PHYSICS B & C EXAMS

7 47i2mR = 4?t2(60 kg)(6.37xlo6 m) _ M T2 (24hx^x^)2 and the difference between gq and g is mg = 4!c2R= 4n2(6,37xl06 m) a Example 7.4 Communications satellites are often parked in geosynchro nous orbits above Earth's surface. Satellites have orbit periods that are equal to Earth's rotation period, so they remain above the sameposition on Earth's surface. Determine the altitude and the speed that a satellite must have to be in a geosynchronous orbit above a fixed point on Earth's equator. (The mass of the earth is 5.98 x 10" kg.) Solution. Let m be the mass of the satellite, M the mass of Earth, and R the distance from the center of Earth to the position of the satellite. The gravitational pull of Earth provides the centripetal force on the satellite, so The orbit speed of the satellite is 2nR/T, so G = which implies that G = R T2 Now the key feature of a geosynchronous orbit is that its period matches Earth's rotation period, T = 24 hr. Substituting the numerical values of G, M, and T into this expression, we find that _ 3K6^67xlQ-")(5.98xl024)( )2 4tc2 ~11 An2 = 4.23xl07m Therefore, if re is the radius of Earth, then the satellite's altitude above Earth's surface must be h = R-rE = (4.23 x 107 m) - (6.37 x 106 m) = 3.59 x 107 m which is equal to 5.6rE. The speed of the satellite in this orbit is, from the first equation in our calculations, regardless of the mass of the satellite (well, as long as m «M). NEWTON'S LAW OF GRAVITATION 129

8 Example 7.5 IC] A uniform, slender bar of mass M has length L Determine die gravitational force it exerts on the point particle of mass m shown below: m O I ".. - I M Solution. Since the bar is an extended body (and not spherically symmetric), we must calculate F using an integral. Select an arbitrary segment of length dx and mass dm in the bar, at a distance x from its left-hand end. 0 m o <---z---> -*- * dx Then, since the bar is uniform, dm = [M/L)dx, so the gravitational force between m and dm is M T Now, by adding (that is, by integrating) all of the contributions df, we get the total gravitational force, F: F=\dF J 1=0 L(z+x)2 1 dx x=q {Z + Xf j±t =L rmm\ -1 1 L 2(2+ L) mm = G Z(2 + L) 130 CRACKING THE AP PHYSICS B & C EXAMS

9 GRAVITATIONAL POTENTIAL ENERGY When we developed the equation U = mgh for the gravitational potential energy of an object of mass m at height h above the surface of the earth, we took the surface of the earth to be our U=0 reference level and assumed that the height, h, was small compared to the earth's radius. In.that case, the variation in g was negligible, so g was treated as constant. The work done by gravity as an object was raised to height h was then simply -F^ x As = -mgh, so U^ which by definition equals -W^^ was mgh. But now we'll take variations in g into account and develop a general equation for gravitational potential energy, one that isn't restricted to small altitude changes. Consider an object of mass m at a distance r, from the center of the earth (or any spherical body) moving by some means to a position r2: How much work did the gravitational force perform during this displacement? The answer is given by the equation: Therefore, since A U^ = -Wby^ we get GMm --- Let's choose our U = 0 reference at infinity. That is, we decide to allow U2 this equation becomes 0 as r2. Then GMm NEWTON'S LAW OF GRAVITATION 131

10 Notice that, according to this equation (and our choice of U = 0 when r = «>), the gravitational potential energy is always negative. This just means that energy has to be added to bring an object (mass m) bound to the gravitational field of M to a point very far from M, at which U = 0. [C] A PROOF OF EQUATION (*) Because the gravitational force is not constant over the displacement, the work done by this force must be calculated using a definite integral: W F dt Displacement can be broken into a series of infinitesimal steps of two types: Those that are at a constant distance from the earth's center and those that are along a radial line going away from the earth's center. The result is that the displacement is equivalent to the sum of two curves, C, + C2: The gravitational force along C, does no work, because ifs always perpendicular to the displace ment. Along C2, the inward gravitational force is in the opposite direction from the outward displace ment, so along this ray, Fdr = Fdrcosl80 = -Fdr GMm, = = dr 132 CRACKING THE AP PHYSICS B & C EXAMS

11 Therefore, = J Fdr _ n GMm = GMm f='2 = GMm\ --\ \r2 h) Example 7.6 With what minimum speed must an object of mass m be launched in order to escape Earth's gravitational field? (This is called escape speed, v^j)> Solution. When launched, the object is at the surface of the earth (r. = re) and has an upward, initial velocity of magnitude z>,. To get it far away from the earth, we want to bring its gravitational potential energy to zero, but to find the minimum launch speed, we want the object's final speed to be zero by the time it gets to this distant location. So, by Conservation of Energy, 1 2 -GMvm n n -mvf + = '. re which gives Substituting the known numerical values for G, ME, and re gives us: 2GME _ 2(6.67xl0-u)(5.98xl024) h = 1.12x10* m7s (Note: This is the minimum speed needed to escape Earth's gravitational"field but, in this situation, a projectile would also have to contend with the Sun's gravitational field, etc.) NEWTON'S LAW OF GRAVITATION B 133

12 Example 7.7 A satellite of mass m is in a circular orbit of radius R around the earth (radius re, mass M). (a) What is its total mechanical energy (where U m is considered zero as R approaches infinity)? (b) How much work would be required io move the satellite into a new orbit, with radius 2R? Solution. (a) The mechanical energy, E, is the sum of the kinetic energy, K, and potential energy, U. You can calculate the kinetic energy since you know that the centripetal force on the satellite is provided by the gravitational attraction of the earth: Therefore, mv2 GMm 2 GMm, 2 GMm = ^ => mv = => K = ~ mv = R R2 R 2R r _ v n-~ GMnt -GMm GMm 2R R 2R (b) From the equation Ki + Ui + W = Kf+ Uv we see that Therefore, the amount of work necessary to effect the change in the satellite's orbit radius from R to 2R is _ GMm (~GMm\ 2(2R) I 2R j GMm 4R [C] A NOTE ON ELLIPTICAL ORBITS The expression for the total energy of a satellite in a circular orbit of radius R [derived in Example 7.7(a)] is: GMm t- (circular orbit) and this also holds for a satellite traveling in an elliptical orbit, if the radius R is replaced by a, the length of the semimajor axis: GMm t = (elliptical orbit) 134 CRACKING THE AP PHYSICS B & C EXAMS

13 Example 7.8 An asteroid of mass m is in an elliptical orbit around the Son (mass M). Assume that m«m. aphelion orbit of asteroid perihelion (a) What is the total energy of the asteroid? (b) What is the ratio of p, (the asteroid's speed at aphelion) to p2 (the asteroid's speed at perihelion)? Write your answer in terms of r, ' andrr (c) What is the time necessary for the asteroid to make a complete orbit around the Sun? Solution. (a) The total energy of the asteroid is equal to -GMm/2a, where a is the semimajor axis. However, notice in the figure above that 2a = r, + rr Therefore, the total energy of the asteroid is E = - GMm (b) One way to answer this question is to invoke Conservation of Angular Momentum. When the asteroid is at aphelion, its angular momentum (with respect to the center of the Sun) is Lx = rxmvy When the asteroid is at perihelion, its angular momentum is Lj = r2mv2. Therefore, L = L, => rxmv, = r2mv2 => = n r This tells us that the asteroid's speed at aphelion is less than its speed at perihelion (because rjrx < 1), as implied by Kepler's Second Law (a line drawn from the Sun to the asteroid must sweep out equal areas in equal time intervals). The closer the asteroid is to the Sun, the faster it has to travel to make this true. NEWTON'S LAW OF GRAVITATION 135

$Using Kepler's Third Law, with a = (r, + r:), we find T2 y CM T = \ GM "V GM i 2GM ORBITS OF THE PLANETS Kepler's First Law states that the planets' orbits are ellipses, but the ellipses that the$

14 (c) As you know, the time necessary for the asteroid to make a complete orbit around the Sun is the orbit period, T. Using Kepler's Third Law, with a = (r, + r:), we find T2 y CM T = \ GM "V GM i 2GM ORBITS OF THE PLANETS Kepler's First Law states that the planets' orbits are ellipses, but the ellipses that the planets in our solar system travel are nearly circular. The deviation of an ellipse from a perfect circle is measured by a parameter called its eccentricity. The eccentricity, e, is the ratio of c (the distance between the center and either focus) to a, the length of the semimajor axis, For every point on the ellipse, the sum of the distances to the foci (plural of focus) is a constant (and is equal to 2n in the figure below). Kepler's First Law also states that one of the foci of a planet's elliptical orbit is located at the position of the Sun. Actually, one of the foci is at the center of mass of the Sun-planet system, because when one body orbits another, both bodies orbit around their center of mass, a point called the barycenter. barycenter orbit of hi. orbit of nu 136 CRACKING THE AP PHYSICS B 8 C EXAMS

15 For most of the planets, which are much less massive than the Sun, this correction to Kepler's First Law has little significance, because the center of mass of the Sun and the planet system is close enough to the Sun's center. For example, let's figure out the center of mass of the Sun-Earth system. The mass of the Earth is m = 5.98 x 1024 kg, the mass of the Sun is M = 1.99 x 1030 kg, and the Sun- Earth distance averages R = x 10n m. Therefore, letting x = 0 be at the Sun's center, we have = M*Sun + m*earth = M(0)+(5.98xl024 kg)(1.496xlon m) = M+m (1.99X1030 kg)+(5.98xlo24 kg) So the center of mass of the Sun-Earth system is only 450 km from the center of the Sun, a distance of less than 0.1% of the Sun's radius. Example 7.9 Derive a corrected version of Kepler's Third Law for the following orbiting system. Both bodies have orbit period T. barycenter orbit of m. orbit of m-i Solution. The centripetal force on each body is provided by the gravitational pull of the other body, so = G_m1m2_ ana m2v\jn mjn2 -r and which imply m, and i\] ^i\jxi\2; R2 \Ri+R2) But, since both bodies have the same orbit period, T, we have 2tcR,.. 2nR2 NEWTON'S LAW OF GRAVITATION 137

16 Substituting these results into the preceding pair of equations gives us: (2^/T)2 _ m, = G "1., and v I" ' =Gwhich simplify to Adding this last pair of equations gives us the desired result 2 Notice that this final equation is a general version of Kepler's Third Law for a circular orbit derived earlier, T2/B? = 4 n2/gm, where it was assumed that the planet orbited at a distance R from the center of the Sun...^ :nrirt.'^f.:irtrv^v'"1 V':"^.;-/i;.:.'::ir" ^i:!^.=iw^1: :,;. ;?^,,^;rv^ni;.:;!-:!^.:;!!:^-;.:-^;;.-^..:^:-::^;^:?;^ ^t^wirr^tw)-^^ Solution. The centripetal force on the satellite is provided by Earth's gravitational pull. Therefore, mv2 _ r Mm Solving this equation for v yields Notice that the satellite's speed doesn't depend on its mass; even if it were a baseball, if its orbit radius were R, then its orbit speed would still be-^/gm/r. 138 CRACKING THE AP PHYSICS B & C EXAMS

17 CHAPTER 7 REVIEW QUESTIONS SECTION I: MULTIPLE CHOICE 1. If the distance between two point particles is doubled, then the gravitational force between them (A) decreases by a factor of 4. (B) decreases by a. factor of 2. (C) increases by a factor of 2.. (D) increases by a factor of 4. (E) Cannot be determined without knowing the masses 2. At the surface of the earth, an object of mass m has weight w. If this object is transported to an altitude that's twice the radius of the earth, then, at the r\ew location, 5. A satellite is currently orbiting Earth in a circular orbit of radius R; its kinetic energy is K,. If the satellite is moved and enters a new circular orbit of radius 2R, what will be its kinetic energy? (A) K./4 (B) K/2 (C) K, (D) 2K, (E) 4K, 6. A moon of Jupiter has a nearly circular orbit of radius R and an orbit period of T. Which of the following expressions gives the mass of Jupiter? (A) its mass is m/2 and its weight is w/2. (B) its mass is m and its weight is w/2. (C) its mass is m/2 and its weight is w/4. (D) its mass is m and its weight is a>/4. (E) its mass is m and its weight is w/9. (A) (B) (C) (D) (E) 2KR/T 21CR'/(GF) 4JC2R7(GT2) 4rt2R7(GT2) 3. A moon of mass m orbits a planet of mass 100m. Let the strength of the gravitational force exerted by the planet on the moon be denoted by F,, and let the strength of the gravitational force exerted by the moon on the planet be F2. Which of the following is true? (A) F, = 100F2 (B) F, = 10F2 (Q F, = F2 (D) F2 = 10F, (E) F2=100F, 7. Two large bodies, Body A of mass m and Body B of mass 4m, are separated by a distance R. At what distance from Body A, along the line joining the bodies, would the gravitational force on an object be equal to zero? (Ignore the presence of any other bodies.) (A) R/16 (B) R/8 (C) R/5 (D) R/4 (E) R/3 4. The planet Pluto has 1/500 the mass and 1/15 the radius of Earth. What is the value of g on the surface of Pluto? (A) (B) (C) (D) (E) 0.3 m/s* 1.6 m/s2 2.4 m/s2 4.5 m/s2 7.1 m/s2 8. The mean distance from Saturn to the Sun is 9 times greater than the mean distance from Earth to the Sun. How long is a Saturn year? (A) 18 Earth years (B) 27 Earth years (C) 81 Earth years (D) 243 Earth years (E) 729 Earth years NEWTON'S LAW OF GRAVITATION 139

18 9. The Moon has mass M and radius R. A 10. C] A planet.orbits the Sun in an elliptical small object is dropped from a distance of orbit of eccentricity e. What is the ratio of 3R from the Moon's center. The object's the planet's speed at perihelion to its' impact speed when it strikes the surface speed at aphelion? of the Moon is equal to JkGMjR for k = jb) j/o - <0 S(C) Vh 3/4 (E) (14 (D) 4/3 (E) 3/2 140 CRACKING THE AP PHYSICS B & C EXAMS

19 SECTION II: FREE RESPONSE 1. Consider two uniform spherical bodies in deep space. Sphere 1 has mass m, and Sphere 2 has mass mr Starting from rest from a distance R apart, they are gravitationally attracted to each other. (a) Compute the acceleration of Sphere 1 when the spheres are a distance R/2 apart. (b) Compute the acceleration of Sphere 2 when the spheres are a distance R/2 apart. (c) Compute the speed of Sphere 1 when the spheres are a distance R/2 apart. (d) Compute the speed of Sphere 2 when the spheres are a distance R/2 apart. Now assume that these spheres orbit their center of mass with the same orbit period, T. (e) Determine the radii of their orbits. Write your answer in terms of mv mv T, and fundamental constants. 2. [C] A satellite of mass m is in the elliptical orbit shown below around Earth (radius re, mass M). Assume that m«m. orbit of satellite apogee (a) Determine vv the speed of the satellite at perigee (the point of the orbit closest to Earth). Write your answer in terms of rv rv M, and G. (b) Determine vv the speed of the satellite at apogee (the point of the orbit farthest from Earth). Write your answer in terms of r,, rv M, and G. (c) Express the ratio vt/v2 in simplest terms. (d) What is the satellite's angular momentum (with respect to Earth's center) when it's at apogee? (e) Determine the speed of the satellite when it's at the point marked X in the figure. (f) Determine the period of the satellite's orbit. Write your answer in terms of r,, rv M, and fundamental constants. (g) What is the eccentricity of the satellite's orbit? Express your answer in terms of r, and ry NEWTON'S LAW OF GRAVITATION 141

Chapter 13. Gravitation

Chapter 13. Gravitation Chapter 13 Gravitation e = c/a A note about eccentricity For a circle c = 0 à e = 0 a Orbit Examples Mercury has the highest eccentricity of any planet (a) e Mercury = 0.21 Halley s comet has an orbit