Physics Universal Gravitation Study Guides

Chapter J Universal Gravitation J.1 - The Force of Gravity Introduction If Isaac Newton had merely written down his three laws of motion he would probably still be known as the most important physicist of all time. Instead, he went far beyond that; he solved all elementary problems involving mechanics and he found the force law describing gravity. His theory of gravity is the topic of this chapter. The history of physics is largely a history of the unification of the fundamental forces. Before Newton there were separate notions of gravity. What was known as gravity before Newton was the force holding things to the surface of the Earth; we will call this terrestrial gravity. It was not clear that this force was the same as what was responsible for the motion of planets, moons and comets; this we will refer to as astronomical gravity. Newton, with his theory of universal gravitation showed that these two forces were actually the same; he unified terrestrial and astronomical gravity. Origin of the Inverse Square Law r m1 F 21 F 12 m2 Newton was trying to understand the force law between point masses m1 and m2 separated by a distance r. The weight of a body is the gravitational force on it. Since the weight is proportional to the mass, the force on m2 must be proportional to m2 . F ∝ m2 Because of Newton's third law the magnitude of the force on m1 is the same. Thus the force on m2 must also be proportional to m1 . F ∝ m1 How does the force vary with distance? It should also decrease with the distance r and should go to zero as r approaches infinity. Some possibilities are F∝ 1 r , F∝ 1 r2 , F∝ 1 or F ∝ r3 e-α r r2 . Newton settled on an inverse square law F∝ 1 r2 . He deduced the inverse square law by comparing the ratio of the acceleration of the moon toward the earth to the acceleration due to gravity at the surface of the earth to the ratio of the radius of the moon's orbit to the radius of the earth. Suppose the force has the form F ∝ r p for some power p, where an inverse square law gives p = -2. Since the acceleration of a body is proportional to the force F ∝ a we can conclude a ∝ r p . From this proportionality we can write the ratio a2 a1 = r2 r1 p . We can find the power p using logs. p= ln(a2 / a1 ) ln(r2 / r1 ) Let a2 and r2 refer to the moon's orbit. At Newton's time the acceleration of the moon toward the earth could be calculated. The period of the moon's orbit is T2 = 27.32 days. The earth-moon distance, which was measured by parallax, is r2 = REM = 3.84 × 108 m. This gives 2 | Chapter J - Universal Gravitation a2 = 2π T2 2 r2 = 2 2π 3.84 × 108 m = 0.002721 27.32 × 24 × 3600 s m s2 . The acceleration at the surface of the earth is a1 = g = 9.80 m  s2 at a distance given by the earth's radius r1 = RE = 6.37 × 106 m. Combining these expressions gives p= ln(a2 / a1 ) ln(r2 / r1 ) ln 0.002721  9.80 ln 3.84×108  6.37×106 = = -2.00, which verifies the inverse square law. Newton realized that there was a crucial flaw in his logic here. He wanted a force law between particles; by a particle it is meant that the distance to an object is large compared to the size of the object. Clearly this is not correct for an object on the surface of the earth. Newton felt that the gravitational force for a spherical body, as long as one is outside of the body, is the same as if all of the mass is at the center. To prove this Newton had to first invent integral calculus and then do a difficult integral. We will refer to this result as the shell theorem and will discuss it shortly. Newton's Law of Universal Gravitation Newton's law of gravity is the attractive inverse square law between point masses. If m1 and m2 are point masses separated by distance r the magnitude of the force between them is F=G m1 m2 r2 . (J.1) This can be written as a vector expression. Let r12 be the vector from mass 1 to mass 2 and let F 12 be the force on mass 2 due to mass 1. F 12 = -G m1 m2  r12 2 r12  r12 is the unit vector in the direction of the vector r12 . r12 r12  r12 =  =  r12  r12 r12 m1 m2  r12 F 12 The constant G is known as Newton's universal gravitational constant. Gravitational forces between everyday objects are incredibly small. This is reflected in the small value of G in our SI system of units. Newton never knew the value of this constant. It was eventually measured to be G = 6.673 × 10-11 N · m2 kg2 . The Force on a Mass Due to a Distribution of Masses m m F1   r1 r1 r3 m1 F3 F2  r1 r3 m1 r2 r2  r3 F m3 m3  r2 m2 m2 Chapter J - Universal Gravitation | 3 Suppose we want to find the force on a mass m due to a discrete distribution of masses, m1 , m2 , ... . Define the vector ri to be from mi to m. The force of mi to m is F i = -G m mi  ri ri2  where ri = ri / ri . The total force is the sum of all these parts F = ∑i F i . This gives F = -G m  i mi  ri . ri2 This is just using the idea that force is a vector and that forces add as vectors. In our discussion of moments of inertia we dealt with continuous distributions. Break up a continuous distribution into an infinite number of infinitesimal pieces; take ⅆ m to be one of these infinitesimal pieces. The vector r is from ⅆ m to m. F = -G m   r r2 ⅆm . Example J.1 - Net Force on Moon What is the magnitude net force on the moon, due to the earth and sun, at a (i) full moon, (ii) new moon and (iii) half moon? (iii) half moon Moon (ii) new moon Sun RES Earth (i) full moon REM (not to scale) The masses of the earth, the moon and sun, and the earth-moon and the earth-sun distances are: ME = 5.97 × 1024 kg , MS = 1.99 × 1030 kg , MM = 7.35 × 1022 kg , REM = 3.85 × 108 m and RES = 1.50 × 1011 m Here assume a circular orbit. Looking at their numerical values we can also see that REM << RES , we will assume this, as well. Solution The magnitude of the force of the earth on the moon is FE = G ME M M R2EM = 1.975 × 1020 N . The sun-moon distance varies, but with the REM << RES assumption we can set the moon-sun distance to the earth-sun distance, RMS = RES . The magnitude of the sun’s force on the earth becomes FS = G MS M M R2ES = 4.338 × 1020 N . (i) For the case of the full moon both forces F S and F E are in the same direction, to the left in the diagram. The magnitude of the net force is the sum of the two magnitudes. Fnet = FS + FE = 6.31 × 1020 N (ii) With the new moon, the forces F S and F E are in opposite directions, in the diagram: to the left for the sun and to the right for the earth. The magnitude of the net force is now the difference. Fnet = FS - FE = 2.36 × 1020 N (iii) At the half moon phase, the forces F S and F E are perpendicular, since in the limit REM << RES the hypotenuse of the earthmoon-sun triangle becomes parallel to the long side. The magnitude of the net force found by the Pythagorean theorem. Fnet = FS2 + FE2 = 4.77 × 1020 N Note that since FS > FE the moon’s trajectory must always curve toward the sun. 4 | Chapter J - Universal Gravitation The Shell Theorem and Spherical Distributions Consider a uniform spherical shell of mass M and radius R. If we consider a point mass m a distance r from the center then the shell theorem states that: M M m r r R m R r < R case r > R case for r < R, F = 0 and Mm  for r > R, F = -G r r2 To state this in words: When m is outside the shell (r > R) the shell behaves as if all the mass is at the center. When m is inside the shell (r < R) the total gravitational force on the mass is zero. Clearly, by symmetry, at the exact center of the shell the force should be zero. It is a remarkable property of the inverse square law that it is zero everywhere inside. We will not prove the shell theorem this semester. It will be proven in the second semester course in the analogous case of electrostatics. As a consequence of the shell theorem we can find the force on a mass m a distance r from the center of any spherical distribution. Break up the sphere into thin concentric shells. M R M R If the shell is inside the radius r the shell behaves as if all its mass is at the origin. If it is outside it doesn't contribute to the force. This gives the result: F = -G m Menclosed  r. r2 where Menclosed is the total mass enclosed by a sphere of radius r. M M m r R r R M enclosed M enclosed = M r < R case r > R case m Chapter J - Universal Gravitation | 5 The Gravitational Field We will refer to the acceleration due to gravity at some position as the gravitational field g; typically this no longer will be uniform. Since the force on a mass m is F = m g we can define the field at some position by the following procedure. Add a test mass m0 at some position. Define the field at that position to be the force divided by the test mass. g= F m0 The result is then independent of the test mass. Now consider a point mass M at the origin. The gravitational field at position r is found by placing a test mass m0 there. The force on m0 is F = -G m0 M  r. r2 Dividing m0 into this gives g = -G M  M r or g =  g  = G . r2 r2 This also applies outside of a spherical body. At the surface of a spherical planet of mass M and radius R, the magnitude of the field is g=G M R2 . g g g g g g g g g g g g g g g g g g g g g g g g Gravitational field at the surface and around a spherical planet Example J.2 - Weight on a Different Planet An astronaut has an earth weight of 120lb. What is her weight, in pounds, on a planet with twice the earth’s radius and five times its mass. Solution The gravitational field (or acceleration due to gravity) g on the surface of a spherical planet of mass M and radius R is: g=G M R2 . The weight W of an astronaut of mass m is related to g by W = mg. Both formulas apply to both cases, on the earth and the other planet. The astronaut’s mass m is the same on both planets. The two formulas apply to each case. For the weights we see that the ratio of the weights equals the ratio of the g values. W1 = m g1 and W2 = m g2 Similarly for g we see: W2 Dividing W1 = g2 g1 6 | Chapter J - Universal Gravitation g1 = G M1 R21 and g2 = G M2 R22 g2 Dividing g1 = M2 / M 1 (R2 / R1 )2 Take case 2 to be the new planet and case 1 the earth. W2 W1 = M 2 / M1 2 (R2 / R1 ) ⟹ W WE = M / ME 2 ( R / RE ) = 5 22 We can solve for the result. W= 5 2 2 WE = 5 22 × 120lb = 150lb The Cavendish Experiment Newton published his theory of universal gravitation in 1687. The value of G was not known at the time; it was not measured until more than a century later in 1798 by Henry Cavendish. Because spherical masses behave the same as point masses, Cavendish could use spherical masses to measure G for point masses. The experiment involved a torsion balance, which consisted of two spherical masses m at the end of a light rigid bar. The bar was suspended from a vertical wire at its center. Under a small torque, the balance would rotate and the amount of rotation was proportional to the torque; τ = κθ, where the constant of proportionality κ could be measured. Two larger spherical masses M were placed, as shown, and that small gravitational force between m an M caused a slight rotation. The small angle of rotation, gave the torque and that then gave the magnitude of the gravitational force. The Cavendish torsion balance viewed from above. At the left is the torsion balance with two smaller spherical masses and at the right is the same torsion balance rotated after two larger masses are placed. Although Cavendish is credited with measuring G for the first time, in reality what he did was find the density and mass of the Earth. Since g = GMearth  R2earth , that was equivalent to measuring G. The notation we use in (J.1) involving the constant G is relatively modern and was not introduced until around 1873. J.2 - Energy Considerations Potential Energy of Two Masses Recall that a conservative force is any force whose work is independent of path. We can define a potential energy for any conservative force by Δ U = -W = - F · ⅆ r We want to find the potential energy for two masses m and M separated by a distance r. To get this we will first consider M to be at a fixed position, which we will take as the origin, and move m in its presence along some path and find ⅆ U for that path. Chapter J - Universal Gravitation | 7  r ⅆr  ⅆr m r  r+ⅆr ri r f M Since the force on m due to M is F = -G Mm r2  r, we get  Δ U = - F · ⅆ r = -  rf -G ri Mm   r·ⅆr r2  Multiplying a vector by a unit vector gives the component of the vector in the direction of the unit vector. r · ⅆ r becomes the radial component of ⅆ r which is just ⅆ r, the infinitesimal change in the radial variable and note this this is not the same as the magnitude of ⅆ r.  r · ⅆ r = ⅆ r where ⅆ r = ⅆ  r  =  r + ⅆ r  -  r  ≠  ⅆ r   With the substitution of ⅆ r for r · ⅆ r, the limits of the integral depend only on the radial distances of the endpoints and it is now a simple singlevariable integral. rf ΔU = GMm  ri ⅆr r2 = -GMm 1 rf - 1 ri Clearly this result is independent of the path taken; this demonstrates the force is conservative and justifies the definition of the potential energy. We are looking for some function U (r), describing the potential energy as a function of position that satisfies Δ U = U r f  - U (ri ). This is unique up to an arbitrary constant. The simplest choice is U (r) = -G Mm r where we chose to make the potential energy zero at infinity. U (∞) = 0 or more precisely lim U (r) = 0 r→∞ U ri rf r Uf Ui Interactive Figure - Gravitational Potential Energy between Two Masses Potential Energy of a Configuration Now we consider the case of several point masses, m1 , m2 , ... . Take the distance between m1 and m2 to be r12 and generally ri j as the distance between the ith and jth masses. For two masses we have U = -G m1 m2 / r12 . For three masses we have this term plus energy terms between m1 and m3 , and between m2 and m3 . We get 8 | Chapter J - Universal Gravitation U = -G m1 m2 -G r12 m1 m3 r13 -G m2 m3 r23 . The general expression is U = -G  i< j mi m j ri j . The sum is over all indices i and j where we insist that i < j to avoid double counting. m3 r13 r23 m1 r12 m2 Interactive Figure - The potential energy for multiple point masses is found by summing over all pairs. Energy and Escape Speed The total energy is the sum of the kinetic energies of all masses and the total potential energy above. For a single mass m moving in the presence of a large mass M the total mechanical energy is E=K+U= 1 2 mv2 - G Mm r . The potential energy is negative, going to zero at infinity. To escape the gravitational pull of M, m must have enough energy to reach infinity, or zero energy. The escape speed is the critical speed needed to escape the gravity of a planet from the surface of the planet of mass M and radius R. This becomes 0=E= 1 2 mv2esc - G Mm R . Solving for the escape speed we get vesc = 2GM R . Example J.3 - Rocket Launched off a Spherical Planet A rocket is launched off the surface of a spherical planet of mass M and radius R with a speed v0 in a vertical (perpendicular to the surface) direction. (a) If the speed v0 is less than the escape speed, the rocket will reach some maximum distance from the center rmax and fall back. What is rmax ? Solution E= 1 2 mv2 - G Mm r m is the rocket’s mass. It must scale out of the problem. Take the initial position to be at the surface and the final to be at rmax . The final kinetic energy is zero. Chapter J - Universal Gravitation | 9 Ei = E f ⟹ 1 2 mv20 - G Mm Mm =0-G R rmax ⟹ rmax = 1 R - -1 v20 2GM (b) For rocket speeds greater than the escape speed, the rocket will continue to infinity and still have kinetic energy. What is v∞ , the speed at infinity? (To be precise, this should be discussed in terms of limits: as r → ∞, v → v∞ .) Solution The expression for the energy is the same as is the initial values. The final potential energy is now zero. Ei = E f ⟹ Note that this can be written as v∞ = 1 2 mv20 - G Mm R = 1 2 mv2∞ + 0 ⟹ v∞ = v20 - 2 G M R v20 - v2esc . J.3 - Orbits Circular Orbits m F=G mM r2 R M r h We will now consider circular orbits of a satellite of mass m orbiting a much larger object of mass M, m ≪ M. Take the larger object to be a sphere of radius R. If the satellite orbits at a height h above the surface, then the radius of the orbit is r = R +h. We now apply the second law to the satellite. The only force is gravity and the acceleration is centripetal. We get: Fnet,c = mac ⟹ G Mm r2 = mac . In this expression the satellite mass cancels; this is the essence of weightlessness. Suppose one studies the orbit of the international space station, then m is the station's mass. If an astronaut is floating in the station and not touching anything, then m is the astronaut's mass. The cancellation of m implies that the astronaut and station have the same orbit; he will float relative to the station. After canceling the masses we can then use the two expressions for the centripetal acceleration for uniform circular motion. This gives a pair of expressions, one that relates the speed and radius and the other relating the period and radius. G M r2 v2 r 2π 2   T = ac = r Solving these expressions for the speed and period gives. v= GM r 2 T2 = 4π GM r3 Example J.4 - Low-earth Orbit On a planet without an atmosphere, it is possible to orbit just above the surface. For the earth the orbit must be above the atmosphere. Although the atmosphere gets less dense with the height above the surface, there is a point where the drag is small enough for an orbit. Consider a low earth orbit to be 200 km (about 120 mi) above the surface. 10 | Chapter J - Universal Gravitation On a planet without an atmosphere, it is possible to orbit just above the surface. For the earth the orbit must be above the atmosphere. Although the atmosphere gets less dense with the height above the surface, there is a point where the drag is small enough for an orbit. Consider a low earth orbit to be 200 km (about 120 mi) above the surface. What are the period and speed of this low-earth orbit? Solution Rearth = 6.38 × 106 m , Mearth = 5.97 × 1024 kg h = 200 km = 2.0 × 105 m ⟹ r = Rearth + h = 6.58 × 106 m 4 π2 T2 = G Mearth r3 ⟹ T = 5310 s = 89 min Low-earth orbits can be higher but the height should not be much larger. These will have slightly longer periods, but low-earth orbits are close to 90 minutes. G Mearth v= r = 7780 m / s Example J.5 - Geostationary Orbit It is useful for communication satellites, weather satellites and spy satellites to maintain a fixed position in the sky relative to the rotating earth. This is called a geostationary orbit. This requires an orbit directly over the equator with a period of 24 h. (a) Take the period of a low-earth orbit to be 1.5 hours and the orbital radius to be the earth’s radius, RE . Estimate the radius of a geostationary orbit as a multiple of RE . Solution T2 = 4 π2 GM r3 We will use ratios to solve this, as we did in Example-J.2. We will use the period equation above for two cases: r1 = RE with T1 = 1.5 h and find the unknown r2 with T2 = 24 h. M = Mearth for both cases. T12 = 4 π2 G Mearth r2 = T2 r13 and T22 = 2/3 r1 ≃ T1 24 h 1.5 h 4 π2 G Mearth r23 T2 Dividing T1 2 = r2 3 r1 2/3 RE = 162/3 RE ⟹ r2 ≃ 6.4 RE (b) Calculate this geostationary orbital radius precisely. While we are being precise, the orbital period is not one solar day but one sidereal day 23.93 h. (Note that a sidereal day is the time it takes for the earth to rotate once relative to the distant stars, as opposed to relative to the sun. There are 365.24 solar days in a year but, with one extra rotation relative to the distant stars, there are 366.24 sidereal days in a year.) Solution T2 = r = 6.673 × 10-11 4 π2 G Mearth N · m2 kg2 r3 ⟹ r = G Mearth 5.97 × 1024 kg T2 1/3 4 π2 (23.93 × 3600 s)2 4 π2 1/3 = 4.22 × 107 m Compare this with the radius of the earth. Rearth = 6.38 × 106 m ⟹ r = 6.61 Rearth Kepler's Laws Kepler came after both Copernicus and Galileo, and came before Newton. Kepler used data taken by someone else, Tycho Brahe, of the positions of the planets in the sky as functions of time. From this he realized that the circular orbits of Copernicus and Galileo could not fit the data. He settled on elliptical orbits and summarized his results with three laws. After Kepler, Newton found his three laws of motion and his law of universal gravitation. He was able to derive Kepler's three law from his work. This was a remarkable feat. He was also able to prove that Kepler's laws implied that the gravitational force law had to be an inverse square law. Chapter J - Universal Gravitation | 11 After Kepler, Newton found his three laws of motion and his law of universal gravitation. He was able to derive Kepler's three law from his work. This was a remarkable feat. He was also able to prove that Kepler's laws implied that the gravitational force law had to be an inverse square law. Kepler's First Law - Planets move in elliptical orbits about the sun, which is at a focal point of the ellipse. Given two points, called focal points, an ellipse is the set of points in a plane such that the sum of the two distances from each focus is a constant. r1 + r2 = 2 a r1 focus r2 focus 2b b = a 1-e2 c=ea 2a 2c Interactive Figure a is called the semimajor axis; it is half the largest distance between two points on the ellipse. The other dimension of the ellipse is the minor axis 2b. Each focus is a distance of c from the center, where c = e a, with e known as the eccentricity. The eccentricity is the deviation from being a circle, 0 ≤ e < 1. e = 0 is the circular case, where the two foci merge and a = b. As e approaches one, the ellipse approaches a parabola. Proving Kepler’s first law is not that difficult, but is a bit beyond the traditional scope of this course. Kepler's Second Law - The radial vector from the sun to the planet sweeps out equal areas in equal time. Sun Interactive Figure - The time to sweep out the gray area stays constant. Kepler’s second law describes how fast a planet moves in its orbit; when near the sun it moves rapidly and and far away it moves slowly. It sweeps out equal areas in equal time. This is a simple consequence of the conservation of angular momentum. 12 | Chapter J - Universal Gravitation v ⅆt ⅆA r Sun The area of a parallelogram formed by two vectors is A B sin θ which is just the magnitude of the cross product of the vectors. In a time ⅆ t a planet moves by v ⅆ t. This traces out a triangle; this is half a parallelogram and thus has the infinitesimal area ⅆA = 1    r⨯v ⅆt  2 Since L = r ⨯ p is constant, it follows that ⅆ A / ⅆ t is constant. ⅆA ⅆt = 1   1  L  r⨯v  =  r⨯ p  = = constant. 2 2m 2m This proves the result. Kepler's Third Law - Period2 ∝ (Semimajor axis)3 or T 2 ∝ a3 Define the major axis as the largest distance between two points on an elliptical orbit. The semimajor axis, a, is half this. For a circular orbit, this is the same as the radius a = r. We saw that T 2 ∝ r3 for a circular orbit. The third law is then a generalization of this result, which we will not prove. T2 = e=0 e =0 .3 e =0 .6 4 π2 GM a3 . e =0 .9 Sun 2a 2a 2a 2a Different orbits with varying eccentricities. The semimajor axis and thus the period stay the same.