CHAPTER Halley s comet has a period of about 76 y. What is its mean distance from the sun? R mean = (1 AU)(76) 2/3 (see Problem 3)

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1 CHAPTER 11 1* True or false: (a) Kepler s law of equal areas implies that gravity varies inversely with the square of the distance. (b) The planet closest to the sun, on the average, has the shortest period of revolution about the sun. (a) False (b) True If the mass of a satellite is doubled, the radius of its orbit can remain constant if the speed of the satellite (a) increases by a factor of 8. (b) increases by a factor of. (c) does not change. (d) is reduced by a factor of 8. (e) is reduced by a factor of. (c) 3 One night, Lucy picked up a strange message on her ham radio. Help! We ran away from Earth to live in peace and serenity, and we got disoriented. All we know is that we are orbiting the sun with a period of 5 years. Where are we? Lucy did some calculations and told the travelers their mean distance from the sun. What is it? Use Equ. 11-; R = R ES (5/1) /3 ; R ES = m R = ( )(5) /3 m = m 4 Halley s comet has a period of about 76 y. What is its mean distance from the sun? R mean = (1 AU)(76) /3 (see Problem 3) R mean = /3 m = m 5* A comet has a period estimated to be about 41 y. What is its mean distance from the sun? (41 y was the estimated period of the comet Hale Bopp, which was seen in the Northern Hemisphere in early Gravitational interactions with the major planets that occurred during this apparition of the comet greatly changed its period, which is now expected to be about 38 y.) R mean = (1 AU)(41) /3 (see Problem 4) R mean = /3 m = m 6 The radius of the earth s orbit is m and that of Uranus is m. What is the period of Uranus? Use Equ. 11-; T U = (1 y)(r U /1 AU) 3/ T U = ( / ) 3/ y = 83.7 y 7 The asteroid Hektor, discovered in 197, is in a nearly circular orbit of radius 5.16 AU about the sun. Determine the period of this asteroid.

2 T H = (1 y)(5.16/1) 3/ (see Problem 6) T H = 11.7 y 8 The asteroid Icarus, discovered in 1949, was so named because its highly eccentric elliptical orbit brings it close to the sun at perihelion. The eccentricity e of an ellipse is defined by the relation d p = a(1 - e), where d p is the perihelion distance and a is the semimajor axis. Icarus has an eccentricity of.83. The period of Icarus is 1.1 years. (a) Determine the semimajor axis of the orbit of Icarus. (b) Find the perihelion and aphelion distances of the orbit of Icarus. (a) Use Kepler s third law; a = (1 AU)(T I ) /3 (b) d p = a(1 - e); d a + d p = a; d a = a - d p a = /3 m = m d p = ( ) m = m; d a = m 9* Why don t you feel the gravitational attraction of a large building when you walk near it? The mass of the building is insignificant compared to the mass of the earth. 1 Astronauts orbiting in a satellite 3 km above the surface of the earth feel weightless. Why? Is the force of gravity exerted by the earth on them negligible at this height? They feel weightless because the force of gravity provides just their centripetal acceleration. The force of gravity is not negligible. (see Example 11-) 11 The distance from the center of the earth to a point where the acceleration due to gravity is g/4 is (a) R E. (b) 4R E. (c) R E /. (d) R E. (e) none of the above. (d) g 1/R. 1 At the surface of the moon, the acceleration due to the gravity of the moon is a. At a distance from the center of the moon equal to four times the radius of the moon, the acceleration due to the gravity of the moon is (a) 16a. (b) a/4. (c) a/3. (d) a/16. (e) none of the above. (d) a 1/R. 13* One of Jupiter s moons, Io, has a mean orbital radius of m and a period of s. (a) Find the mean orbital radius of another of Jupiter s moons, Callisto, whose period is s. (b) Use the known value of G to compute the mass of Jupiter. (a) Use Equ. 11-; R C = R I (T C /T I ) /3 (b) Use Equ ; M J = 4π R I 3 /GT I R C = ( )(14.4/1.53) /3 m = m M J = 4π ( ) 3 /[ ( ) ] = kg 14 The mass of Saturn is kg. (a) Find the period of its moon Mimas, whose mean orbital radius is m. (b) Find the mean orbital radius of its moon Titan, whose period is s. (a) Use Equ ; T = (4π R 3 /GM S ) 1/ T M = [4π ( ) 3 /( )] 1/ s = s (b) Use Equ. 11- R T = (138/8.18) /3 m = m 15 Calculate the mass of the earth from the period of the moon T = 7.3 d, its mean orbital radius

3 r m = m, and the known value of G. Convert T to s: T = s; Use Equ M E = 4π ( ) 3 /G( ) kg = kg Note: This neglects the mass of the moon; consequently, the mass calculated here is slightly too great. 16 Use the period of the earth (1 y), its mean orbital radius ( m), and the value of G to calculate the mass of the sun. Proceed as in Problem y = s; substituting in the numerical values yields M S = kg. 17* An object is dropped from a height of m above the surface of the earth. What is its initial accelera-tion? Since R = R E, a = g E /4 = (9.81/4) m/s =.45 m/s. 18 Suppose you leave the solar system and arrive at a planet that has the same mass per unit volume as the earth but has 1 times the earth s radius. What would you weigh on this planet compared with what you weigh on earth? 1. Find M P, mass of the planet, in units of M E. Find g P in units of g E 3. Weight on planet = 1 weight on earth For same ρ, M R 3 ; M = 1 3 M E Since g M/R, g P = g E 1 3 /1 = 1g E 19 Suppose that the earth retained its present mass but was somehow compressed to half its present radius. What would be the value of g, the acceleration due to gravity, at the surface of this new, compact planet? Since g is proportional to 1/R, g on planet would be m/s = 39.4 m/s. A planet moves around a massive sun with constant angular momentum. When the planet is at perihelion, it has a speed of m/s and is m from the sun. The orbital radius increases to m at aphelion. What is the planet s speed at aphelion? L = constant = mv p r p = mv a r a ; v a = v p r p /r a v a = ( / ) m/s = m/s 1* A comet orbits the sun with constant angular momentum. It has a maximum radius of 15 AU, and at aphelion its speed is m/s. The comet s closest approach to the sun is.4 AU. What is its speed at perihelion? v p = v a r a /r p (see Problem ) v p = /.4 m/s = 65 km/s The speed of an asteroid is km/s at perihelion and 14 km/s at aphelion. Determine the ratio of the aphelion to perihelion distance. r a /r p = v p /v a (see Problem ) r a /r p = /14 = A satellite with a mass of 3 kg moves in a circular orbit m above the earth s surface. (a) What is the gravitational force on the satellite? (b) What is the speed of the satellite? (c) What is the period of the satellite? (a) F g = Mg(R) = Mg [R E /(R E + h)] ; R E = m (b) Mv /R = F g ; v = (F g R/M) 1/ (c) T = πr/v F g = (6.37/56.37) N = N v = ( /3) 1/ m/s =.657 km/s T = π /657 s = s = 37 h

4 4 At the airport, a physics student weighs 8 N. The student boards a jet plane that rises to an altitude of 95 m. What is the student s loss in weight? 1. Find w at 9.5 km; w(h) = w E [R E /(R E + h)]. w = w E - w(h) w(h) = 8(637/6379.5) N = N w =.4 N 5* Suppose that Kepler had found that the period of a planet s circular orbit is proportional to the square of the orbit radius. What conclusion would Newton have drawn concerning the dependence of the gravitational attraction on distance between two masses? Take F = CR n, where C is a constant. Then, for a stable circular orbit, v /R = F = CR n. The period of the orbit is given by T = πr/v, and so T = πr/c 1/ R (n+1)/. Therefore, if T R, 1 - (n + 1)/ =, n = -3, and F 1/R 3. 6 A superconducting gravity meter can measure changes in gravity of the order g/g = (a) Estimate the maximum range at which an 8-kg person can be detected by this gravity meter. Assume that the gravity meter is stationary, and that the person s mass can be considered to be concentrated at his or her center of gravity. (b) What vertical change in the position of the gravity meter in the earth s gravitational field is detectable? (a) Find the gravitational field, g(r) due to mass m Solve for r; r = R E (1 11 m/m E ) 1/ (b) Write g(r) = C/r and differentiate Find dr = r for r = R E g(r) = Gm/r ; given: g(r) = 1-11 g E = 1-11 (GM E /R E ) r = (8 1 1 /6 4 ) 1/ m = 7.36 m dg/dr = -C/r 3 ; dg/g = - dr/r = 1-11 r = 1/ m» m =.3 mm 7 During a solar eclipse, when the moon is between the earth and the sun, the gravitational pull of the moon and the sun on a student are in the same direction. (a) If the pull of the earth on the student is 8 N, what is the force of the moon on the student? (b) What is the force of the sun on the student? (c) What percentage correction due to the sun and moon when they are directly overhead should be applied to the reading of a very accurate scale to obtain the student s weight? (a) F due to moon is F M = GmM M /R EM. Write F M in terms of F E = GmM E /R E = 8 N and evaluate F M. (b) Likewise, F S = 8(M S /M E )(R E /R ES ) (c) Percentage correction = (-.485/8) 1 F M = 8(M M /M E )(R E /R EM ) N F M = 8(7.35/598)(6.37/384.4) N =.7 N F S = 8( /5.98)(6.37/ ) =.483 N correction = -.61% 8 Suppose that the attractive interaction between a star of mass M and a planet of mass m << M were of the form F = KMm/r, where K is the gravitational constant. What would be the relation between the radius of the planet s circular orbit and its period? KMm/r = mv /r; so v = KM, independent of r T = πr/v = πr/(km) 1/, i.e., proportional to r 9* The mass of the earth is kg and its radius is 637 km. The radius of the moon is 1738 km. The acceleration of gravity at the surface of the moon is 1.6 m/s. What is the ratio of the average density of the moon to that of the earth?

5 1. Write g E and g M in terms of ρe and ρm g E = G(4πρER E 3 /3)/R E ; g M = G(4πρMR M 3 /3)/R M. Find g M /g E and solve for and evaluate ρm/ρe g M /g E = ρmr M /ρer E ; ρm/ρe = (1.6/9.81)(6.37/1.738) =.65 3 A plumb bob near a large mountain is slightly deflected from the vertical by the gravitational attraction of the mountain. Estimate the order of magnitude of the angle of deflection using any assumptions you like. First, we note that the two force fields are approximately at right angles. Their magnitudes are GM E /R E and Gm m /D m, where M m is the mass of the mountain and D m the distance to its center. Thus the angle of deflection is given by θ = tan -1 (M m R E /M E D ). Assume that the density of the mountain is the average density of the earth. Let the mountain be a cylinder 1 km in diameter and 4 km high. Its mass is ρeπr h. We consider the plumb bob to be adjacent to the mountain cylinder so that D = r = 5 km. Now M m /M E = πr h/(4πr E 3 /3)» and R E /D The angle of deflection is then approximately θ = tan -1 (5 1-4 ) = rad.3 o. 31 Why is G so difficult to measure? Measurement of G is difficult because masses accessible in the laboratory are very small compared to the mass of the earth. 3 The masses in a Cavendish apparatus are m 1 = 1 kg and m = 1 g, the separation of their centers is 6 cm, and the rod separating the two small masses is cm long. (a) What is the force of attraction between the large and small masses? (b) What torque must be exerted by the suspension to balance these forces? (a) Use Equs and 11-4 (b) τ = Fr, r =.1 m F = ( / ) N = N τ = N.m 33* The masses in a Cavendish apparatus are m 1 = 1 kg and m = 15 g, and the separation of their centers is 7 cm. (a) What is the force of attraction between these two masses? (b) If the rod separating the two small masses is 18 cm long, what torque must be exerted by the suspension to balance the torque exerted by gravity? (a) See Problem 3 F = ( / ) N = N (b) τ = Fr, r =.9 m τ = N.m 34 How would everyday life change if gravitational and inertial mass were not identical? The force required to accelerate an object (car, person, etc.) would not be proportional to its weight. Also, the period of a pendulum clock would depend on the mass of the bob. 35 If gravitational and inertial mass were not identical, what would change for (a) an offensive lineman on a football team? (b) a car? (c) a paperweight? (a) His effectiveness would depend on his mass rather than his weight. (b) The power requirement would not be determined by the car s weight, but by its mass. (c) There would be no significant effect. 36 A standard object defined as having a mass of exactly 1 kg is given an acceleration of.6587 m/s when a certain force is applied to it. A second object of unknown mass acquires an acceleration of m/s when the same force

6 is applied to it. (a) What is the mass of the second object? (b) Is the mass that you determined in part (a) gravitational or inertial mass? (a) From F = ma, m = (.6587/1.175) kg =.714 kg. (b) It is the inertial mass of m. 37* The weight of a standard object defined as having a mass of exactly 1 kg is measured to be 9.81 N. In the same laboratory, a second object weighs 56.6 N. (a) What is the mass of the second object? (b) Is the mass you determined in part (a) gravitational or inertial mass? (a) As in Problem 36, m = (56.6/9.81) kg = 5.77 kg. (b) This is the gravitational mass of m determined by the effect on m of the earth s gravitational field. 38 (a) Taking the potential energy to be zero at infinite separation, find the potential energy of a 1-kg object at the surface of the earth. (Use m for the earth s radius.) (b) Find the potential energy of the same object at a height above the earth s surface equal to the earth s radius. (c) Find the escape speed for a body projected from this height. (a) Use Equ ; U(R E ) = -GM E m/r E = -gmr E (b) G, M E, and m are unchanged, U(R E ) = U(R E )/ (c) From (b), K esc (R E ) = 1/K esc (R E ); K v U(R E ) = -( ) J = J U(R E ) = J v esc (R E ) = v esc (R E )/ = (11./ ) km/s = 7.9 km/s 39 A point mass m is initially at the surface of a large sphere of mass M and radius R. How much work is needed to remove it to a very large distance away from the large sphere? W = U = U f - U i. U f =, U i = -GMm /R. So W = GMm /R. 4 Suppose that in space there is a duplicate earth, except that it has no atmosphere, is not rotating, and is not in motion around any sun. What initial velocity must a spacecraft on its surface have to travel vertically upward a distance above the surface of the planet equal to one earth radius? From the results of Problem 38, we know that the change in gravitational potential is exactly half the gravitational potential difference between R = R E and R =. Consequently, v = v esc / = 7.9 km/s. 41* An object is dropped from rest from a height of m above the surface of the earth. If there is no air resistance, what is its speed when it strikes the earth? 1. Use 1/mv = - U = U( R E ) - U(R E ). Solve for v 1/v = ( )[(1/ ) - (1/ )] J/kg v = 6.95 km/s 4 An object is projected upward from the surface of the earth with an initial speed of 4 km/s. Find the maximum height it reaches. 1. U f = K i + U i ; divide both sides by m. Solve for and evaluate h -GM E /(R E + h) = 1/v i - GM E /R E h = (Rv i )/[(GM E /R E ) - v i ] = m 43 A spherical shell has a radius R and a mass M. (a) Write expressions for the force exerted by the shell on a point mass m when m is outside the shell and when it is inside the shell. (b) What is the potential-energy function U(r) for this system when the mass m is at a distance r (r R) if U = at r =? Evaluate this function at r = R. (c) Using the general relation for du = -F dr = -F r dr, show that U is constant everywhere inside the shell. (d) Using the fact

7 that U is continuous everywhere, including at r = R, find the value of the constant U inside the shell. (e) Sketch U(r) versus r for all possible values of r. (a) F = m g; use Equ Outside: F = GMm /r, radially in; inside: F = (b) U(r) = - r F r dr for r > R U(r) = -GMm /r; U(R) = -Gmm /R (c), (d) F = for r < R, du/dr =, U = constant. Since U is continuous, then for r < R, U(r) = U(R) = -Gmm /R. (e) A sketch of U(r) is shown 44 Our galaxy can be considered to be a large disk of radius R and mass M of approximately uniform mass density. (a) Consider a ring element of radius r and thickness dr of such a disk. Find the gravitational potential energy of a 1-kg mass on the axis of this element a distance x from its center. (b) Integrate your result for part (a) to find the total gravitational potential energy of a 1-kg mass at a distance x due to the disk. (c) From F x = - du/dx and your result for part (b), find the gravitational field g x on the axis of the disk. Let σ = mass/unit area = M/πR. Let d = x + r be the distance from a point at a radius r from the center of the disk to the point a distance x along its axis. πrσ G dr du = - (a) Write an expression for du x + r R πrσ G dr (b) U = - = - πσ G ( x + R - x) x + r G M U(x) = - ( x + R - x) R (c) F x = -du(x)/dx F x = GM - 1 R - x x + R = g(x) 45* The assumption of uniform mass density in Problem 44 is rather unrealistic. For most galaxies, the mass density increases greatly toward the center of the galaxy. Repeat Problem 44 using a surface mass density of the form σ(r) = C/r, where σ(r) is the mass per unit area of the disk at a distance r from the center. First determine the constant C in terms of R and M; then proceed as in Problem 44.

8 (a) M = R πrσ dr = π C R dr = π C R C = M/πR GM dr (b) du = - R x + r U(x) = dr GM R + x + R = ln R x + r R x GM R (c) F x = -du/dx = g(x) GM F x = x x + R = g(x) 46 What is the effect of air resistance on the escape speed near the earth s surface? It causes an increase in the escape speed. 47 Would it be possible in principle for the earth to escape from the solar system? Yes, if it interacted with a huge comet. 48 If the mass of a planet is doubled with no increase in its size, the escape speed for that planet will be (a) increased by a factor of 1.4. (b) increased by a factor of. (c) unchanged. (d) reduced by a factor of 1.4. (e) reduced by a factor of. (a) See Equ * The planet Saturn has a mass 95. times that of the earth and a radius 9.47 times that of the earth. Find the escape speed for objects near the surface of Saturn. v e (Sat) = (GM S /R S ) 1/ = v e (earth)[(m S /M E )(R E /R S )] 1/ v e (Sat) = 11.(95./9.47) 1/ km/s = 35.5 km/s 5 Find the escape speed for a rocket leaving the moon. The acceleration of gravity on the moon is.166 times that on earth, and the moon s radius is.73r E. v e (moon) = (g m R m ) 1/ = v e (earth)[g m R m /g E R E ] 1/ v e (moon) = 11.( ) 1/ km/s =.38 km/s 51 A particle is projected from the surface of the earth with a speed equal to twice the escape speed. When it is very far from the earth, what is its speed? Since its speed is twice the escape speed, its initial kinetic energy is four times the energy needed to escape. Therefore, when it has escaped, it has an energy that is three times the escape energy and its speed is v e 3 = 19.4 km/s. 5 What initial speed should a particle be given if it is to have a final speed when it is very far from the earth equal to its escape speed? Using the same reasoning as in Problem 51, one finds that v i = v e = 15.8 km/s. 53* A space probe launched from the earth with an initial speed v i is to have a speed of 6 km/s when it is very far from the earth. What is v i?

9 1/mv i = 1/mv f + 1/mv e ; v i = (v f + v e ) 1/ v i = ( ) 1/ km/s = 61.4 km/s 54 (a) Calculate the energy in joules necessary to launch a 1-kg mass from the earth at escape speed. (b) Convert this energy to kilowatt-hours. (c) If energy can be obtained at 1 cents per kilowatt-hour, what is the minimum cost of giving an 8-kg astronaut enough energy to escape the earth s gravitational field? (a) E =1/mv e E = 1/( ) J = 6.7 MJ (b) 1 kw.h = 3.6 MJ E = (6.7/3.6) kw.h = 17.4 kw.h (c) Cost = $ (M.1 E) Cost = $ Show that the escape speed from a planet is related to the speed of a circular orbit just above the surface of the planet by v e = v c, where v c is the speed of the object in the circular orbit. From Example 11-6, the kinetic energy of a mass m in a circular orbit is K = 1/ U. Note that this result is true for any circular orbit of a mass m about a massive center. But U is the escape energy of 1/mv e. Consequently, K = 1/mv c = 1/(1/mv e ) and v c = v e /. 56 Find the speed of the earth v c as it orbits the sun, assuming a circular orbit. Use this and the result of Problem 55 to calculate the speed v es needed by the earth to escape from the sun. 1. v c = πr ES /T; R ES = m, v c = (π / ) m/s = 9.9 km/s T = s. v e = v c v e = 4. km/s 57* If an object has just enough energy to escape from the earth, it will not escape from the solar system because of the attraction of the sun. Use Equation with M S replacing M E and the distance to the sun r S replacing R E to calculate the speed v es needed to escape from the sun s gravitational field for an object at the surface of the earth. Neglect the attraction of the earth. Compare your answer with that in Problem 56. Show that if v e is the speed needed to escape from the earth, neglecting the sun, then the speed of an object at the earth s surface needed to escape from the solar system is given by v e,solar = v e + v es, and calculate v e,solar. From Equ , v es = (GM S /r S ) 1/ = [ ( ) ( 1 3 )/ ] 1/ = 4. km/s; this is the same speed calculated in Problem 56, as it should be. The energy needed to escape the solar system, starting from the surface of the earth, is the sum of the energy needed to escape the earth s gravity plus that required to escape from the sun, starting at the earth s orbit radius. These energies are proportional to the squares of the corresponding escape velocities. Therefore, v e,solar = v e + v es ; v e,solar = ( ) 1/ km/s = 43.7 km/s. 58 Why is it reasonable to neglect the other planets in calculating the speed needed to escape from the solar system? Would you expect the actual value of this speed to be greater or less than that calculated in Problem 57? The masses of the planets are only about.1% of the sun s mass. The actual value of v e,solar is slightly greater because the total mass of the solar system is slightly larger than that of the sun. 59 An object is projected vertically from the surface of the earth. Show that the maximum height reached by the object is H = R E H /(R E - H ), where H is the height that it would reach if the gravitational field were constant. H = v /g. To determine H use energy conservation: 1/mv = - U = GMm[1/R E - 1/(R E +H)] or v = gr E [1/R E - 1/(R E +H)]. So H = R E H/(R E + H) and, solving for H, H = R E H /(R E - H ).

10 6 An object (say, a newly discovered comet) enters the solar system and makes a pass around the sun. How can we tell if the object will return many years later, or if it will never return? If the path is an ellipse, it will return; if its path is hyperbolic or parabolic, it will not return. 61* A spacecraft of 1 kg mass is in a circular orbit about the earth at a height h = R E. (a) What is the period of the spacecraft s orbit about the earth? (b) What is the spacecraft s kinetic energy? (c) Express the angular momentum L of the spacecraft about the earth in terms of its kinetic energy K and find its numerical value. We will use the result of Problem 55, i.e., v c = v e / surface. (a) 1. K(3R E ) = K(R E )/3; v = v c / 3 = v e / 6. T = π(3r E )/v (b) K = 1/mv (c) K = L /I; L = (KI) 1/ ; I = m(3r E ), where v c is the speed of a circular orbit just above the v = 11./ 6 km/s = 4.57 km/s T = π 3 637/( ) h = 7.3 h K = 5 ( ) = J L = 3R E (Km) 1/ = 3R E mv = J.s 6 Many satellites orbit the earth about 1 km above the earth s surface. Geosynchronous satellites orbit at a distance of m from the center of the earth. How much more energy is required to launch a 5-kg satellite into a geosynchronous orbit than into an orbit 1 km above the surface of the earth? Let E O be the total energy of a satellite in orbit. Then, from Problem 55, E O = K O + U O = 1/U O. E = E geo - E 1 = 1/(U geo - U 1 ) E = -1/GM E m(1/ / ) = J 63 It is theoretically possible to place a satellite at a position between the earth and the sun on the line joining them, where the gravitational forces of the sun and the earth on the satellite combine in such a way that the satellite will execute a circular orbit around the sun that is synchronous with the earth s orbit around the sun. (In other words, the satellite and the earth have the same orbital period about the sun, even though they are at different distances from the sun. The satellite always remains on the line joining the earth and the sun.) Write an expression that relates the appropriate circular orbital speed v of a satellite in such a situation to its distance r from the sun. Your expression may also contain quantities shown in Figure 11- plus the gravitational constant G. Write F net = F centrip = mv /r = GmM S /r - GmM E /(D - r) ; v = G M S /r - G ME r/(d - r ). 64 A 3-kg mass experiences a gravitational force of 1 N i at some point P. What is the gravitational field at that point? Use Equ g = (1/3) i N/kg = 4 i N/kg 65* The gravitational field at some point is given by g = N/kg j. What is the gravitational force on a mass of 4 g at that point? Use Equ F = mg = ( ) j N = 1-8 j N

11 66 A point mass m is on the x axis at x = L and a second equal point mass m is on the y axis at y = L. (a) Find the gravitational field at the origin. (b) What is the magnitude of this field? (a), (b) Use Equ and vector addition g = (Gm/L )(i + j); g = Gm/L 67 Five equal masses M are equally spaced on the arc of a semicircle of radius R as in Figure A mass m is located at the center of curvature of the arc. (a) If M is 3 kg, m is kg, and R is 1 cm, what is the force on m due to the five masses? (b) If m is removed, what is the gravitational field at the center of curvature of the arc? (a) By symmetry, g x = ; write expression for F y Substitute numerical values (b) g = F/m F y = (GMm/R )( sin 45 o + 1) F = N j g = N/kg j 68 A point mass m 1 = kg is at the origin and a second point mass m = 4 kg is on the x axis at x = 6 m. Find the gravitational field at (a) x = m, and (b) x = 1 m. (c) Find the point on the x axis for which g =. The configuration is shown below. (a) g = g1 + g; g1 = -Gm 1 /4 i; g = Gm /16 i (b) g = (-Gm 1 /144 - Gm /36) i (c) g = when /x = 4/(6-x) ; solve for x g = G(1/4-1/) i = i N/kg g = -G(1/7 + 1/9) i = -G/8 i = i N/kg x + 7x - 36 = ; x =.484 m, m. From the diagram it is clear that only at x =.484 m is g =. 69* (a) Show that the gravitational field of a ring of uniform mass is zero at the center of the ring. (b) Figure 11- shows a point P in the plane of the ring but not at its center. Consider two elements of the ring of length s 1 and s at distances of r 1 and r, respectively. 1. What is the ratio of the masses of these elements?. Which produces the greater gravitational field at point P? 3. What is the direction of the field at point P due to these elements? (c) What is the direction of the gravitational field at point P due to the entire ring? (d) Suppose that the gravitational field varied as 1/r rather than 1/r. What would be the net gravitational field at point P due to the two elements? (e) How would your answers to parts (b) and (c) differ if point P were inside a spherical shell of uniform mass rather than inside a plane circular ring? Let λ = mass per unit length of the ring. (a) g of opposite elements of mass Rλ dθ cancel (b) 1. m 1 = r 1λ dθ; m = r λ dθ. g = Gm/r = Grλ dθ/r = Gλ dθ/r 3. By symmetry, g points along OP (c) Take x along OP (d) For g 1/r, g 1 = g λ dθ By symmetry g = at center m 1 /m = r 1 /r r 1 < r, therefore g 1 > g g points toward m 1, i.e., in direction of OP By symmetry, g y = ; g points in the direction OP g1 = -g; g =

12 (e) Now m 1 and m r, so g 1 = g g1 = -g; g = ; note: g = everywhere inside the shell 7 Show that the maximum value of g x for the field of Example 11-7 occurs at the points x = ±a/. From Example 11-7, g x = -GMx/(x + a ) 3/. To find maximum, differentiate with respect to x and equate to. dg x /dx = -GM[(x + a ) -3/ - 3x (x + a ) -5/ ] =. Solve for x: x = ±a/. 71 A nonuniform stick of length L lies on the x axis with one end at the origin. Its mass density λ (mass per unit length) varies as λ = Cx, where C is a constant. (Thus, an element of the stick has mass dm = λ dx.) (a) What is the total mass of the stick? (b) Find the gravitational field due to the stick at a point x > L. (a) L L 1 M = λ dx= C x dx= C L. (b) dg = -G dm/(x - x) i; g = - G C L x dx G M = ( x - x ) L x L ln - i. x - L x - L 7 A uniform rod of mass M and length L lies along the x axis with its center at the origin. Consider an element of length dx at a distance x from the origin. (a) Show that this element produces a gravitational field at a point x on the x axis (x > 1/L) given by GM dgx = - dx L( x - x ) (b) Integrate this result over the length of the rod to find the total gravitational field at the point x due to the rod. (c) What is the force on an object of mass m at x? (d) Show that for x >> L, the field is approximately equal to that of a point mass M. (a), (b) See Example (c) F = m g = -GMm /(x - L /4) i N. (d) For x >> L, g x = -GM/x, which is the result for a point mass M at the origin. 73* Explain why the gravitational field increases with r rather than decreasing as 1/r as one moves out from the center inside a solid sphere of uniform mass. g is proportional to the mass within the sphere and inversely proportional to the radius, i.e., proportional to r 3 /r = r. 74 A spherical shell has a radius of m and a mass of 3 kg. What is the gravitational field at the following distances from the center of the shell: (a).5 m; (b) 1.9 m; (c).5 m? (a) g = [see Problem 69(e)]. (b) g =. (c) g = GM/r = N/kg. 75 A spherical shell has a radius of m and a mass of 3 kg, and its center is located at the origin of a coordinate system. Another spherical shell with a radius of 1 m and mass 15 kg is inside the larger shell with its center at.6 m on the x axis. What is the gravitational force of attraction between the two shells?

13 The gravitational attraction is zero. The gravitational field inside the m shell due to that shell is zero; therefore, it exerts no force on the 1 m shell, and, by Newton s third law, that shell exerts no force on the larger shell. 76 Two spheres, S 1 and S, have equal radii R and equal masses M. The density of sphere S 1 is constant, whereas that of sphere S depends on the radial distance according ρ( to r ) = C / r. If the acceleration of gravity at the surface of sphere S 1 is g 1, what is the acceleration of gravity at the surface of sphere S? The accelerations of gravity are the same for both spheres; they depend only on M and R. 77* Two homogeneous spheres, S 1 and S, have equal masses but different radii, R 1 and R. If the acceleration of gravity on the surface of sphere S 1 is g 1, what is the acceleration of gravity on the surface of sphere S? g M/r g = g 1 (R 1 /R ) 78 Two concentric uniform spherical shells have masses M 1 and M and radii a and a as in Figure What is the magnitude of the gravitational force on a point mass m located (a) a distance 3a from the center of the shells? (b) a distance 1.9a from the center of the shells? (c) a distance.9a from the center of the shells? (a) At r = 3a, both masses contribute to g F = Gm(M 1 + M )/9a (b) At r = 1.9a, g due to M = F = GmM 1 /3.61a (c) At r =.9a, g = F = 79 The inner spherical shell in Problem 78 is shifted such that its center is now at x =.8a. The points 3a, 1.9a, and.9a lie along the same radial line from the center of the larger spherical shell. (a) What is the force on m at x = 3a? (b) What is the force on m at x = 1.9a? (c) What is the force on m at x =.9a? The configuration is shown on the right. The centers of the spheres are indicated by the center-lines. The locations of the mass m for parts (a), (b), and (c) are shown by small dots along the x axis. (a) At x = 3a, g 1 = GM 1 /(.a) and g = GM /(3a) (b) At x = 1.9a, g 1 = GM 1 /(1.1a) and g = (c) At x =.9a, g 1 = g = F = (Gm/a )(M 1 / M /9) F = GmM 1 /1.1a F = 8 Suppose the earth were a sphere of uniform mass. If there were a deep elevator shaft going 15, m into the earth, what would be the loss in weight at the bottom of this deep shaft for a student who weighs 8 N at the surface of the earth? 1. Write an expression for M inside R = R E - 15 km M = M E [(R E - 15)/R E ] 3, where R E is in km. Write an expression for g at R = R E - 15 km g = GM/R = GM E (R E -15) 3 /[(R E -15) R 3 E ]

14 3. w(r) = mg(r) = w(r E )[(R E - 15)/R E ] = g E (R E -15)/R E w = (8 N)(15/R E ) = (8 15/637) N = 1.88 N 81* A sphere of radius R has its center at the origin. It has a uniform mass density ρ, except that there is a spherical cavity in it of radius r = 1/R centered at x = 1/R as in Figure Find the gravitational field at points on the x axis for x > R. (Hint: The cavity may be thought of as a sphere of mass m = (4/3)πr 3 ρ plus a sphere of mass -m.) Write g(x) using the hint. That is, find the sum of g of the solid sphere plus the field of a sphere of radius 1/R of negative mass centered at x = 1/R. 4π ρ g(x)= G 3 R x 3 - ( x 3 R / 8 - ½ R ) 8 For the sphere with the cavity in Problem 81, show that the gravitational field inside the cavity is uniform, and find its magnitude and direction. 1. Find the x and y components of g1, where g1 is the field due to a solid sphere of radius R and density ρ.. Find the x and y components of g, where g is a sphere of radius 1/R and negative density ρ centered at 1/R. 3. Add the x and y components to obtain the components of the field g = g1 + g. 1. g 1 = 4πρGr 3 /3r = 4πρGr/3 g 1x = -g 1 cos θ = -g 1 (x/r) = -4πρGx/3; g 1y = -4πρGy/3 (the negative sign because the field points inward). g = 4πρGr 3 /3r = 4πρGr /3, where g x = g [(x - 1/R)/r ] = 4πρG(x - 1/R)/3; r = [(x - 1/R) + y ] 1/ g y = g (y/r ) = 4πρGy/3 3. g x = g 1x + g x ; g y = g 1y + g y g x = -πρgr/3; g y =. g = g x, a constant 83 A straight, smooth tunnel is dug through a spherical planet whose mass density ρ is constant. The tunnel passes through the center of the planet and is perpendicular to the planet s axis of rotation, which is fixed in space. The planet rotates with an angular velocity ω such that objects in the tunnel have no acceleration relative to the tunnel. Find ω. F g = 4πρGmr/3 (see Problem 8); set F r = mrω ω = 4πρG/3; ω = (4πρG/3) 1/ 84 The density of a sphere is given by p(r) = C/r. The sphere has a radius of 5 m and a mass of 111 kg. (a) Determine the constant C. (b) Obtain expressions for the gravitational field for (1) r > 5 m, and () r < 5 m. (a) 5 4π pr dr = 4πC 5 r dr = M πc 5 = 111 kg; C = kg/m (b) 1. For r > 5 m, g = GM/r g = /r = /r N/kg. For r < 5 m, g = G r 4 π Cr dr r g = πgc = N/kg (Note that g is continuous at r = 5 m)

15 85* A hole is drilled into the sphere of Problem 84 toward the center of the sphere to a depth of km below the sphere s surface. A small mass is dropped from the surface into the hole. Determine the speed of the small mass as it strikes the bottom of the hole. 1. Write the work per kg done by g between r = 5 m and r = 3 m; note that g points inward E =. Evaluate v 3 g dr = πgc (5-3) = J = 1/v 5 v =.14 mm/s 86 The solid surface of the earth has a density of about 3 kg/m 3. A spherical deposit of heavy metals with a density of 8 kg/m 3 and radius of 1 m is centered m below the surface. Find g/g at the surface directly above this deposit, where g is the increase in the gravitational field due to the deposit. 1. Determine g at r = m due to a sphere of M = 4π ρr 3 /3 = kg; g = GM/r radius 1 m with density ρ = 5 kg/m 3 g = g = N/kg. Evaluate g/g were g = 9.81 N/kg g/g = Two identical spherical hollows are made in a lead sphere of radius R. The hollows have a radius R/. They touch the outside surface of the sphere and its center as in Figure The mass of the lead sphere before hollowing was M. (a) Find the force of attraction of a small sphere of mass m to the lead sphere at the position shown in the figure. (b) What is the attractive force if m is located right at the surface of the lead sphere? 1. First find the force F S due to the solid lead sphere.. Find the force F C due to each sphere of negative mass. 3. Add the forces. (a) 1. Write the force due to the solid sphere. Write F C 3. Find the x and y component of the two forces F C 4. Find F = F x i = (-F S + F Cx ) i (b) Set d = R F S = Gmm/d ; F S acts in the negative x direction F C = G(M/8)m/(d + R /4) F Cx = F C cos θ = F C d/(d + R /4) 1/ in the positive x direction; y components add to by symmetry F = -(GMm/d )[1 - (d 3 /4)/(d + R /4) 3/ ] i F = -.81(GMm/R ) i 88 If K is the kinetic energy of the moon in its orbit around the earth, and U is the potential energy of the earth moon system, what is the relationship between K and U? K = -1/U. (see Example 11-6) 89* A woman whose weight on earth is 5 N is lifted to a height two earth radii above the surface of the earth. Her weight will (a) decrease to one-half of the original amount. (b) decrease to one-quarter of the original amount. (c) decrease to one-third of the original amount. (d) decrease to one-ninth of the original amount. (d) g depends on 1/r. 9 The mean distance of Pluto from the sun is 39.5 AU. Find the period of Pluto. Proceed as in Problem 7 T p = (39.5) 3/ y = 48 y

16 91 The semimajor axis of Ganymede, a moon of Jupiter discovered by Galileo, is km, and its period is days. Determine the mass of Jupiter. Use the same procedure as for Problem 15 M J = 4π ( ) 3 /G( ) = kg 9 Calculate the mass of the earth using the known values of G, g, and R E. Use Equ. 11-7; M E = gr E /G M E = [9.81( ) / ] kg = kg 93* Uranus has a moon, Umbriel, whose mean orbital radius is m and whose period is s. (a) Find the period of another of Uranus s moons, Oberon, whose mean orbital radius is m. (b) Use the known value of G to find the mass of Uranus. (a) Use Equ. 11-; T O = T U (R O /R U ) 3/ (b) M = 4π R 3 /GT T O = ( s)(5.86/.67) 3/ = s M U = 4π ( ) 3 /G( ) kg = kg 94 Joe and Sally learn that there is a point between the earth and the moon where the gravitational effects of the two bodies balance each other. Being of a New Age bent, they decide to try to conceive a child free from the bondage of gravity, so they book an earth-to-moon trip. How far from the center of the earth should they try to conceive Zerog, the first zero-gravity baby? 1. If M E /r = M m /(R EM - r), g = ; solve for r r = βr EM /(1 + β), where β = (M E /M m ) 1/. Evaluate r for M E /M m = 81.36, r = m R EM = m 95 The force exerted by the earth on a particle of mass m a distance r from the center of the earth has the magnitude GM E m/r = mgr E /r. (a) Calculate the work you must do against gravity to move the particle from a distance r 1 to r. (b) Show that when r 1 = R E and r = R E + h, the result can be written W = mgr E [(1/R E ) - 1/(R E + h)]. (c) Show that when h << R E, the work is given approximately by W = mgh. (a) W r r = Fg dr = GMEm dr r = GM 1 m r 1 = GM r 1 1 m r1 E E 1 r r r1. (b) In the above expression, replace GM E m by mgr E, r 1 by R E, and r by R E + h to obtain the result given. (c) [(1/R E ) - 1/(R E + h)] = h/[r E (R E + h)]; if h << R E, the denominator» R E and W = mgh. 96 Suppose that the gravitational force of attraction depended not on 1/r but was proportional to the distance between the two masses, like the force of a spring. In a planetary system like the solar system, what would then be the relation between the period of a planet and its orbit radius, assuming all orbits were circular? If F = Cr, where C is a constant, then rω r. Thus ω = constant, and T is constant, independent of r. 97* A uniform sphere of radius 1 m and density kg/m 3 is in free space far from other massive objects. (a) Find the gravitational field outside of the sphere as a function of r. (b) Find the gravitational field inside the sphere as a

17 function of r. (a) g = -GM/r ; M = 4πρR 3 /3 (b) Use Equ M = kg; g = -.559/r g = r 98 Two spherical planets have identical mass densities. Planet P 1 has a radius R 1, and planet P has a radius R. If the acceleration of gravity at the surface of planet P 1 is g 1, what is the acceleration of gravity at the surface of planet P? g M/R R 3 /R = R g = g 1 R /R 1 99 Jupiter has a mass 3 times that of Earth and a volume 13 times that of Earth. A day on Jupiter is 9 h 5 min long. Find the height h above Jupiter at which a satellite must be revolving to have a period equal to 9 h 5 min. 1. Use Equ with M J replacing M S. Determine R J = R E (13) 1/3 3. Find R for T = 354 s; h = R - R J R = (T GM J /4π ) 1/3 ; M J = kg R J = (13) 1/3 m = m R = m; h = R - R J = m 1 The average density of the moon is ρ = 334 kg/m 3. Find the minimum possible period T of a spacecraft orbiting the moon. The minimum period is when the orbit radius equals the object s radius, i.e., orbit just above the surface of the moon. 1. Write the condition for a stable orbit, R o = R m = R mrω = mmg/r = 4πρR 3 mg/3r = 4πρmGR/3. Solve for T = 4π /ω and T T = 3π/ρG; T = (3π/ρG) 1/ 3. Evaluate T = T min T min = (3π/ ) 1/ s = 65 s = 1h 48 min 11* A satellite is circling around the moon (radius 17 km) close to the surface at a speed v. A projectile is launched from the moon vertically up at the same initial speed v. How high will it rise? 1. From Problem 55, v = v e /; 1/mv e = GmM m /R m. Solve for h 1/v = 1/GM m /R m = GM m [1/R m - 1/(R m + h)] h = R m = 17 m 1 Two space colonies of equal mass orbit a star (Figure 11-6). The Yangs in m 1 move in a circular orbit of radius 1 11 m with a period of y. The Yins in m move in an elliptical orbit with a closest distance r 1 = 1 11 m and a farthest distance r = m. (a) Using the fact that the mean radius of an elliptical orbit is the length of the semimajor axis, find the length of the Yin year. (b) What is the mass of the star? (c) Which colony moves faster at point P in Figure 11-6? (d) Which colony has the greater total energy? (e) How does the speed of the Yins at point P compare with their speed at point A? (a) 1. Find R, the semi-major axis of Yins. Use Equ. 11-; T Yin = T = T Yang (R /r 1 ) 3/ (b) From Equ , M = 4π r 3 /GT (c) Since R > r 1, E > E 1 ; U P = U 1 ; E = K + U (d) See above R = 1/ m = m T = (1.4) 3/ y = 3.31 y M = 4π 1 33 /[ ( ) ] kg = kg K > K 1 ; v > v 1 E > E 1 (e) L is conserved; r v A = r 1 v P v P = 1.8v A

18 13 In a binary star system, two stars orbit about their common center of mass. If the stars have masses m 1 and m and are separated by a distance r, show that the period of rotation is related to r by T = 4πr 3 /[G(m 1 + m )]. Take the coordinate origin at the center of mass. Then r 1 m 1 = r m and r = r 1 + r. The force holding m in orbit is Gm 1 m /(r 1 + r ) = m r ω. ω = Gm 1 /r (r 1 + r ). Now r = rm 1 /(m 1 + m ), so ω = 4π /T = G(m 1 + m )/r 3 and T = 4π r 3 /G(m 1 + m ). 14 Two particles of mass m 1 and m are released from rest with infinite separation. Find their speeds v 1 and v when their separation distance is r. 1. Use energy conservation. Use conservation of linear momentum 3. Use () in (1) 4. Solve for v 5. To find v 1 permute subscripts in (4) Gm 1 m /r = 1/m 1 v 1 + 1/m v (1) m 1 v 1 = -m v ; v 1 = -v (m /m 1 ) () v (m + m /m 1 ) = Gm 1 m /r (3) v = [Gm 1 /r(m 1 + m )] 1/ (4) v 1 = [Gm /r(m 1 + m )] 1/ 15* A hole is drilled from the surface of the earth to its center as in Figure Ignore the earth s rotation and air resistance. (a) How much work is required to lift a particle of mass m from the center of the earth to the earth s surface? (b) If the particle is dropped from rest at the surface of the earth, what is its speed when it reaches the center of the earth? (c) What is the escape speed for a particle projected from the center of the earth? Express your answers in terms of m, g, and R E. (a) From Equ. 11-7, F = GmM E r/r E 3 = gmr/r E W = R E F gm dr = RE R E gmr rdr = E (b) Use energy conservation; 1/mv = W v= gr E (c) E esc = W + 1/mv e = 1/mv esc v esc = gr E + gr E ; v esc =(3gR E ) 1/ = 13.7 km/s 16 A thick spherical shell of mass M and uniform density has an inner radius R 1 and an outer radius R. Find the gravitational field g r as a function of r for all possible values of r. Sketch a graph of g r versus r. 1. For r < R 1, g = (see Equ. 11-4b).. For r > R, g(r) is that of a mass M centered at the origin, i.e., g(r) = GM/r. 3. For R 1 < r < R, g(r) is determined by the mass within the shell of radius r. The mass density is ρ = 3M/[4π(R 3 - R 1 3 )], and the mass within a radius r is given by 4πρ(r 3 - R 1 3 )/3 = M(r 3 - R 1 3 )/(R 3 - R 1 3 ). So in this region g(r) =GM(r 3 - R 1 3 )/r (R 3 - R 1 3 ).

19 A graph of g(r) is shown alongside. Here we have set R 1 =, R = 3, and GM = (a) Sketch a plot of the gravitational field g x versus x due to a uniform ring of mass M and radius R whose axis is the x axis. (b) At what points is the magnitude of g x maximum? (a) The geometry of the problem is shown at the left above. Consider an element of the ring of length dl. The element of field at a point x is dg = Gλ dl/(x + R ). By symmetry, the y and z components of g vanish. The x component of dg = dg cos θ = Gλ dl[x/(x + R ) 3/ ]. Since all parts of the ring contribute equally and πλr = M, the field g(x) = Gmx/(x + R ) 3/. A plot of g(x) is shown to the right above. The curve is normalized for R = 1 and GM = 1. (b) Differentiate g(x) and set dg/dx = ; (x + R ) 3/ - 3x (x + R ) 1/ = ; x = ±R/ shown.. This agrees with the sketch 18 In this problem, you are to find the gravitational potential energy of the stick in Example 11-8 and a point mass m that is on the x axis at x. (a) Show that the potential energy of an element of the stick dm and m is given by G m dm G M m du = - = dx x - x L ( x - x) where U = at x =. (b) Integrate your result for part (a) over the length of the rod to find the total potential energy for the system. Write your result as a general function U(x) by setting x equal to a general point x. (c) Compute the force on m at a general point x from F x = -du/dx and compare your result with m g, where g is the field at x calculated in Example (a) Let U = at x =. Let λ = M/L. Then the potential energy of the masses m at x and dm = λ dx at x is

20 given by du = -Gm dm/r where r = x - x. Thus du = -Gmm dx/[l(x - x)]. L/ GMm dx GMm GMm (b) U = - = [ ( x - L/) - L ln x - x L x + L / ln( x + / )] = ln L. L x L / -L/ (c) Since x is a general point along the x axis, F(x ) = -du/dx = (Gmm /L)[1/(x + L/) - 1/(x - L/)]; this simplifies to F(x ) = -Gmm /(x - L /4) in agreement with the result of Example * A uniform sphere of mass M is located near a thin, uniform rod of mass m and length L as in Figure Find the gravitational force of attraction exerted by the sphere on the rod. (see Problem 7) We shall determine the force exerted by the rod on the sphere and then use Newton s third law. The sphere is equivalent to a point mass m located at the sphere s center. 1. Use the result of Problem 7; x = a + L/ F = GmM/[(a + L/) - L /4] = GmM/[a(a + L)]. By Newton s third law, this is the force on the rod 11 A uniform rod of mass M = kg and length L = 5 m is bent into a semicircle. What is the gravitational force exerted by the rod on a point mass m =.1 kg located at the center of the circular arc? The semicircular rod is shown in the figure. We shall use an element of length R dθ = (L/π) dθ of mass (M/π) dθ. By symmetry, F y =. So we first find df x and then integrate over θ from -π/ to π/ to determine F x. 1. Obtain an expression for df x. Integrate from θ = -π/ to π/ 3. Evaluate F x df x = [GMm/π(L/π) ]dθ cos θ F x = πgmm/l F x = N 111 Both the sun and the moon exert gravitational forces on the oceans of the earth, causing tides. (a) Show that the ratio of the force exerted by the sun to that exerted by the moon is M S r m /M m r S, where M S and M m are the masses of the sun and moon and r S and r m are the distances from the earth to the sun and to the moon. Evaluate this ratio. (b) Even though the sun exerts a much greater force on the oceans than the moon does, the moon has a greater effect on the tides because it is the difference in the force from one side of the earth to the other that is important. Differentiate the expression F = Gm 1 m /r to calculate the change in F due to a small change in r. Show that df/f = (- dr)/r. (c) During one full day, the rotation of the earth can cause the distance from the sun or moon to an ocean to change by at most the diameter of the earth. Show that for a small change in distance, the change in the force exerted by the sun is related to the change in the force exerted by the moon by F S / F m» (M S r 3 m )/(M m r 3 S ) and calculate this ratio. (a) Force on a mass m is F = GMm/r F S /F m = M S r m /M m r S = 179 (see Problem 7) (b) Find df/dr and df/f df/dr = -Gm 1 m /r 3 = -F/r; df/f = - dr/r

21 (c) F = -(F/r) r Insert numerical values F S / F m = (F S /F m )(r m /r S ) = (M S r m 3 )/(M m r S 3 ) F F S m ( = ( )( ) 11 )( ) =.46