9
GRAVITATION
CHAPTER radius. Thus, R/RE = 1/ 3 = 57.7% gives the new, shrunken radius.
ActivPhysics can help with these problems: Activity 4.8 Section 9-2: The Law of Universal Gravitation Problem
1. Space explorers land on a planet with the same mass as Earth, but they nd they weigh twice as much as they would on Earth. What is the radius of the planet?
Problem
4. Calculate the gravitational acceleration at the surface of (a) Mercury and (b) Saturns moon Titan.
Solution
With reference to the rst two columns in Appendix E: (a) gMerc = G(0.3301024 kg)/(2.44106 m)2 = 3.70 m/s2 , and (b) gTitan = G(0.1351024 kg) 2 (2.58106 m)2 = 1.35 m/s .
Solution
At rest on a uniform spherical planet, a bodys weight is proportional to the surface gravity, g = GM/R2 . Therefore, (gp /gE ) = (Mp /ME )(RE /Rp )2 = 2. Since Mp /ME = 1, Rp = RE / 2.
Problem
5. Two identical lead spheres are 14 cm apart and attract each other with a force of 0.25 N. What is their mass?
Problem
2. Use data for the moons orbit from Appendix E to compute the moons acceleration in its circular orbit, and verify that the result is consistent with Newtons law of universal gravitation.
Solution
Newtons law of the universal gravitation (Equation 9-1), with m1 = m2 = m, gives m= F r2 = G = 8.57 kg. 0.25106 N (0.14 m)2 6.671011 Nm2 /kg2
1/2
Solution
The centripetal acceleration of the moon in its orbit around the Earth is approximately ac = (2/T )2 r = 2 (2/27.386,400 s)2 (3.85108 m) = 2.73103 m/s 2 (see Problem 4-43). (One could also use ac = v /r, but the orbital speed given in Appendix E is less accurate.) The acceleration of gravity at the moons approximate distance from the Earth is g(r) = GME /r2 = (6.671011 Nm2 /kg2 )(5.971024 kg) (3.85108 m)2 = 2.69103 m/s2 . Since the moons orbit is actually elliptical (with 5.5% eccentricity), these values based on circular orbits are not inconsistent.
(Newtons law holds as stated, for two uniform spherical bodies, if the distance is taken between their centers.)
Problem
6. The gravitational acceleration at the surface of 2 a planet is 22.5 m/s . Find the acceleration at a height above the surface equal to half the planets radius.
Problem
3. To what fraction of its current radius would Earth have to be shrunk (with no change in mass) for the gravitational acceleration at its surface to triple?
Solution
The surface gravity of a spherical planet is g = 1 2 GMp /Rp . At a height of 2 Rp above the surface, the 3 distance to the center of the planet is r = 2 Rp , so the acceleration of gravity at that height is g(r) = 2 2 4 4 GMp /r2 = 9 g = 9 (22.5 m/s ) = 10.0 m/s .
Solution
If the surface gravity of the Earth were three times its present value, with no change in mass, then 2 GME /R2 = 3GME /RE , where RE is the present
122
CHAPTER 9 can measure dierences in g as small as a few tenths of a milligal, where 1 milligal = 105 m/s2 is the unit used to measure gravity anomalies by geologists.)
Problem
7. What is the approximate value of the gravitational force between a 67-kg astronaut and a 73,000-kg space shuttle when theyre 84 m apart?
Problem
10. A roughly spherical volume of rock 1.3 km in radius is 30% denser than the surrounding rock, 3 whose density is 2700 kg/m . The denser rock is centered 2.1 km below Earths surface. By what percentage is the surface value of g directly above the denser rock increased due to its excess density? Hint: Treat the extra mass of the denser rock as a gravitating sphere, and calculate the gravitational acceleration it produces at Earths surface.
Solution
If we treat the astronaut and space shuttle as spherical bodies, Equation 9-1 can be used to calculate the approximate force: F = (6.671011 Nm2 /kg2 )(67 kg)(73,000 kg) (84 m)2 = 46.2 nN.
Problem
8. Astronauts put their spaceship into orbit about a planet. They nd that the acceleration of gravity at their orbital altitude is half that at the planets surface. How far above the planets surface are they orbiting? Answer in terms of the radius of the planet.
Solution
The extra mass of the denser rock (above the surrounding rock) is m = (4/3)R3 , where R = 1.3 km is the radius of the spherical volume and 3 = 30%2700 kg/m is the extra density. The extra gravitational acceleration produced at a distance (from the center) of 2.1 km is g = Gm/(2.1 km)2 = (6.671011 Nm2 /kg2 )(4/3)(1.3103 m)3 3 2 (810 kg/m )/(2.1103 m)2 = 1.13104 m/s = 5 1.1510 g = 11.3 milligal (see previous solution).
Solution
The acceleration of gravity at any altitude h, above the surface of a spherical planet of radius Rp , is 2 g(h) = GMp /(Rp + h)2 = g(0) Rp /(Rp + h)2 , where 1 g(0) is the value at the surface. If g(h) = 2 g(0), then Rp /(Rp + h) = 1/ 2, or h = ( 2 1)Rp = 0.414 Rp .
Problem
11. If youre standing on the ground 15 m directly below the center of a spherical water tank containing 4106 kg of water, by what fraction is your weight reduced due to the gravitational attraction of the water?
Problem
9. A sensitive gravimeter is carried to the top of Chicagos Sears Tower, where its reading for the 2 acceleration of gravity is 0.00136 m/s lower than at street level. Find the height of the building.
Solution
The dierence in the acceleration of gravity between the bottom of the Sears Tower (distance R from the center of the Earth) and the top (distance R + h from the Earths center, where h is the height of the building) is g = GME 2h GME GME h(2R + h) = gbot . R2 (R + h)2 R2 (R + h)2 R
Solution
The fraction by which your weight is reduced is equal to the gravitational attraction of the water tank on you divided by your weight, or (Gmm /r2 )/mg = Gm /gr2 . Numerically, this is 2 (6.671011 Nm2 /kg2 )(4106 kg)/(9.8 m/s ) (15 m)2 = 1.21107 .
In the last step, gbot is the value of the acceleration of gravity at the bottom of the building, and since h R, we approximated the second fraction by neglecting h compared to R. If we use average values for gbot (about 9.80 m/s2 in Chicago) and R (approximately 6370 km) and the given g, then h (g/gbot )(R/2) = (0.00136/9.80)(6370 km/2) = 442 m. (The actual value of gbot , for example, depends on the shape of the earth, the altitude, the distribution and type of underlying rocks, etc. Present gravimeters
Section 9-3: Orbital Motion Problem
12. At what altitude will a satellite complete a circular orbit of the Earth in 2.0 hours?
Solution
Equation 9-4 gives r = (GME T 2 /4 2 )1/3 = [(6.671011 )(5.971024 )(23600 2)2 ]1/3 m = 8.06106 m. The altitude is h = r RE = (8.06 6.37)106 m 1690 km.
CHAPTER 9
123
Problem
13. Find the speed of a satellite in geosynchronous orbit.
Solution
Equation 9-3 and the radius of a geosynchronous orbit from Example 9-4 give v = GME /r = [(6.67 1011 Nm2 /kg2 )(5.971024 kg)/(42.2106 m)]1/2 = 3.07 km/s.
calculated in Problem 2. If the gravitational force varies as the inverse square of the distance, the ratio of the accelerations, 2.73103 /9.81 = 2.78104 , should equal (6.37106 /3.85108 )2 = 2.74104 . This is true, to within the accuracy of the data used. (See the solution to Problem 24, Chapter 6.)
Problem
17. During the Apollo moon landings, one astronaut remained with the command module in lunar orbit, about 130 km above the surface. For half of each orbit, this astronaut was completely cut o from the rest of humanity, as the spacecraft rounded the far side of the moon (see Fig. 9-34). How long did this period last?
Problem
14. Marss orbit has a diameter 1.52 times that of Earths orbit. How long does it take Mars to orbit the Sun?
Solution
Since the constant in Equation 9-4 relating T 2 and r3 is (approximately) the same for each planet, Keplers third law may be written as a ratio: (TMars /TE )2 = (rMars /rE )3 . Thus, TMars = (1.52)3/2 TE = 1.87 y.
130 km
Problem
15. Calculate the orbital period for Jupiters moon Io, which orbits 4.22105 km from the center of the 1.91027 -kg planet. figure 9-34 Problem 17.
Solution
From Equation 9-4 and the data given: T = = 4 2 r3 GM
1/2
Solution
The period of a circular orbit of altitude h = 130 km above the moons surface is (see Equation 9-4) T = 2 (Rm + h)3 /GMm = 2[(1.74 + 0.13)3 1018 m3 (6.671011 Nm2 /kg2 )(7.351022 kg)]1/2 = 7.26103 s = 121 min. The command module was cut o from Earth communications for roughly half of this period, or about 1 h.
4 2 (4.22108 m)3 (6.671011 Nm2 /kg2 )(1.91027 kg)
1/2
= 1.53105 s = 1.77 d.
Problem Problem
16. Given the orbital radius of 384,400 km and period of 27.3 days, calculate the acceleration of the moon in its circular orbit, and compare with the acceleration of gravity at Earths surface. Show that the acceleration of the moon is smaller by the ratio of the square of Earths radius to the square of the moons orbital radius, thus conrming the inverse-square law for the gravitational force. 18. A white dwarf is a collapsed star with roughly the mass of the Sun compressed into the size of the Earth. What would be (a) the orbital speed and (b) the orbital period for a spaceship in orbit just above the surface of such a white dwarf?
Solution
(a) The radius of a low orbit is approximately the radius of the white dwarf, or RE , so Equation 9-3 gives v = GM /RE = [(6.671011 Nm2 /kg2 ) (1.991030 kg)/(6.37106 m)]1/2 = 4.56106 m/s, or about 1.5% of the speed of light. (b) The orbital period, T = 2RE /v = 8.77 s, is correspondingly small.
Solution
The centripetal acceleration of the moon, calculated with this orbital radius, is the same, within the accuracy of three signicant gures, as that previously
124
CHAPTER 9 Sun, Equation 9-4, as a ratio, can be used to compare the mean orbital radii and periods, (T1 /T2 )2 = (r1 /r2 )3 (this is, in fact, an approximate statement of Keplers third law). The mean orbital radius of the Earth is 1 AU (an astronomical unit) 1.50108 km, and its period is 1 y (a sidereal year) 365 d, so (T /1 y)2 = (r/1 AU)3 for any satellite of the Sun. If T = 100 d, then (r/1 AU) = (100/365)2/3 = 0.422, or R = 0.422 AU = 6.3107 km. Of course, Equation 9-4 could be used directly: r = [GM (T /2)2 ]1/3 Nm2 = 6.671011 kg2 8.64106 s 2
2 1/3
Problem
19. Given that our Sun orbits the galaxy with a period of 200 My at a distance of 2.61020 m from the galactic center, estimate the mass of the galaxy. Assume (incorrectly) that the galaxy is essentially spherical and that most of its mass lies interior to the Suns orbit. To how many Sun-mass stars is your estimate equivalent?
Solution
For a circular orbit enclosing a spherically distributed mass M, Equation 9-4 gives M = 4 2 r3 /GT 2 . If we apply this to the Suns orbit around the galactic center (and use the rough conversion 1 y = 107 s), we can estimate the mass of the Milky Way galaxy as M 2 8 107 s 210
2
(1.991030 kg) = 6.31107 km.
(2.61020 m)3 (6.671011 Nm2 /kg2 )
2.61041 kg
Problem
22. Determine the orbital period of the Hubble Space Telescope, which orbits Earth at an altitude of 610 km.
This is about 1.31011 solar masses (m 21030 kg), which is not an unreasonable estimate of the number of stars in our galaxy. Astronomers plot the orbital velocity of objects (such as stars, clusters of stars, or clouds of hydrogen atoms) versus their distance from the galactic center to obtain the rotation curve for the galaxy. If most of the matter in a galaxy were concentrated within a certain distance of the center, then the rotation curve should fall o, as indicated in Equation 9-3, for greater distances. It doesnt. The identication of this so called dark matter, which is not seen in telescopes by light it gives o, but which is present due to its gravitational eects, is currently a topic of great interest in astronomy.
Solution
Using Equation 9-4 with r = RE + h = (6370 + 610) km we nd: T = 4 2 (6.98106 m)3 (6.671011 Nm2 /kg2 )(5.971024 kg)
1/2
= 5.81103 s = 96.8 min.
Problem
23. How far from the Suns center would a satellite in circular orbit be heliosynchronous? Hint: Consult Appendix E, and remember that a synchronous orbit must parallel the equator. So which rotation period is appropriate?
Problem
20. Satellites A and B are in circular orbits, with A twice as far from Earths center as B. How do their orbital periods compare?
Solution
The equitorial rotational period of the Sun is 27 d (from Appendix E). The radius of a circular orbit (for a body with mass much smaller than the Sun) with the same orbital period can be found from Keplers third law (Equation 9-4 in ratio form; see solution to Problem 21): r = (T /1 y)2/3 AU = (27/365)2/3 AU = 0.176 AU = 2.64107 km.
Solution
Taking the ratio of Equation 9-4 for the satellites, we nd (TA /TB ) = (rA /rB )3/2 = 23/2 = 2.83.
Problem
21. Where should a satellite be placed to orbit the Sun in a circular orbit with a period of 100 days?
Problem
24. Comets are thought to originate in a cloud of ice chunks in roughly circular orbits almost 1 light-year (the distance light travels in a year) from the Sun. Collisions among the ice chunks and gravitational eects of other stars occasionally
Solution
As long as the masses of the orbiting bodies (satellite or planet) are negligible compared to the mass of the
CHAPTER 9 send a chunk into an elliptical orbit to become an observed comet. What is the orbital period of the cometary ice chunks in their remote orbits?
125
Solution
It is convenient to write Keplers third law (for small objects orbiting the Sun) as a ratio, using the period (TE = 1 y) and mean radius rE = 1 AU) of the Earths orbit: (T /1 y)2 = (r/1 AU)3 . From Appendix C, 1 ly = (9.461015 /1.501011 ) AU = 6.31104 AU, so the period of distant cometary material is T = (6.31104 )3/2 y = 15.8 My.
figure 9-35 Problem 26 Solution.
Problem
25. The asteriod Pasacho orbits the Sun with a period of 1417 days. What is the semimajor axis of its orbit? Determine using Keplers third law in comparison with Earths orbital radius and period. the cluster since the universe began, some 15 billion years ago?
Solution
Our galaxys distance from the center of the Virgo Cluster is about r = (50106 ly)(9.461015 m/ly) = 4.731023 m. If most of the clusters mass, M 103 1011 (21030 kg) = 21044 kg, were spherically distributed inside this distance, and our galaxy were in a circular orbit around it, our period would be T = 2(r3 /GM )1/2 = 2[(4.731023 m)3 (6.671011 Nm2 /kg2 )(21044 kg)]1/2 = 1.771019 s 5.61011 y. This is a few tens times the age of the universe, so obviously a full orbit could not have been completed (at least under the present conditions). Even if the Virgo Cluster contained 100 times as much mass, perhaps as dark matter, our period would still be a few times greater than the age of the universe.
Solution
As in the solution to Problem 21, r = (T /1 y)2/3 (1 AU) = (1417/365)2/3 AU = 2.47 AU = 3.71108 km.
Problem
26. We derived Equation 9-4 on the assumption that the massive gravitating center remains xed. Now consider the case of two objects of equal mass M orbiting each other as shown in Fig. 9-35. Show that the orbital period is given by, 16 2 r3 , GM where r is the orbital radius (half the distance between the objects). T2 =
Solution
Under the conditions stated, Newtons second law for one of the objects is M v 2 /r = GM 2 /(2r)2 . Substituting v = 2r/T, we nd: v2 = GM = 4r 2r T
2
Section 9-4: Gravitational Energy Problem
28. Earths distance from the Sun varies from 1.471011 m at perihelion to 1.521011 m at aphelion because its orbit is not quite circular. Find the change in potential energy as Earth goes from perihelion to aphelion.
,
or T 2 =
16 2 r3 . GM
Problem
27. Our galaxy belongs to a large group of galaxies known as the Virgo Cluster. The cluster contains about 1000 times the mass of our galaxy, which in turn contains about 1011 times the Suns mass. Were roughly 50 million light-years from the center of the approximately spherical cluster. Could our galaxy have completed a full orbit of
Solution
1 1 From Equation 9-5, U = GM ME (rp ra ) = 2 11 2 30 (6.6710 Nm /kg )(1.9910 kg)(5.97 1024 kg)[(1.47)1 (1.52)1 ]1011 m = 1.771032 J.
Problem
29. How much energy does it take to launch a 230-kg instrument package on a vertical trajectory that peaks at an altitude of 1800 km?
126
CHAPTER 9
Solution
If we neglect any kinetic energy dierences associated with the orbital or rotational motion of the Earth or package, the required energy is just the dierence in potential energy of the Earths gravity given by 1 Equation 9-5, U = GME m [RE (RE + h)1 ]. In terms of the more convenient combination of constants 2 GME = gRE , U = mgRE h/(RE + h) = (2309.81 N)(63701800 km)/(8170) = 3.17 GJ.
Solution
If only the vertical motion relative to the Earth at rest is considered, the rst equation in Example 9-6 can be solved for v0 .The calculation is simplied by using
Problem
30. One proposal for dealing with radioactive waste is to shoot it into the Sun. Suppose a waste canister were simply dropped, starting from rest in the vicinity of Earths orbit. At what speed would it hit the Sun?
Solution
Apply the conservation of energy to the dropped canister, as in Examples 9-6 and 7; that is, the gravitational potential energy of the canister at rest (relative to the Sun) at a distance r = 1 AU equals its mechanical energy as it enters the visible surface 1 of the Sun at r = R . Then GM m/r = 2 mv 2 GM m/R , or v=
1 2GM (R r1 ) = [2(6.671011 Nm2 /kg2 )
(1.501011 m)1 )]1/2 = 616 km/s.
(1.991030 kg)((6.96108 m)1
Problem
31. A rocket is launched vertically upward from Earths surface at a speed of 3.1 km/s. What is its maximum altitude?
Solution
If we consider the Earth at rest as approximately an inertial system, then a vertically launched rocket would have zero kinetic energy (instantaneously) at its maximum altitude, and the situation is the same as Example 9-6. Conservation of energy gives 1 2 2 mv0 GME m/RE = GME m/(RE + h), or
2 v0 1 RE = 530 km, RE 2GME when the proper values are substituted. 1
h=
Problem
32. What vertical launch speed is necessary to get a rocket to 1100 km altitude?
CHAPTER 9
2 r = RE + h and GME = gRE . Then 2 v0 = 2GME
127
1 1 RE RE + h
=
2ghRE RE + h
height that is in error by 1%? Would those methods over- or underestimate the height?
2(9.81 m/s2 )(1100 km)(6370 km) , = (7470 km) or v0 = 4.29 km/s.
Solution
If the rocket has an initial vertical speed v0 , we can nd the height, h, to which it can rise (where its kinetic energy is instantaneously zero) from the conservation of energy: K 0 + U0 = GME m GME m 1 =K+U = mv 2 2 0 RE RE + h
Problem
33. Find the energy necessary to put 1 kg, initially at rest on Earths surface, into geosynchronous orbit.
Solution
The energy of an object at rest on the Earths surface is U0 = GME m/RE (neglect diurnal rotational energy, etc.), while its total mechanical energy in a 1 circular orbit is E = 2 U = GME m/2r (Equation 9-9). The energy necessary to put a mass of m = 1 kg into a circular geosynchronous orbit with r = 4.22107 m (see Example 9-4) is the dierence of these energies, E = 1 U U0 , or 2 E = GME (1 kg) = 57.8 MJ. 1 1 6.37106 m 2(4.22107 m)
2 Before solving for h, we replace GME by gRE , and let 2 h = v0 /2g. (h is the maximum height for constant gravity, from Equation 2-11.) After some algebra, we nd h = h RE /(RE h ). Since the factor multiplying h is > 1, h > h , and the equations of constant gravity underestimate the height. (This could have been anticipated because the force of gravity decreases with increasing altitude.) h diers from h by 1% ((h h ) h = 0.01), if h = 0.99h. This occurs for h = 0.99hRE (RE 0.99h), or h = RE /99 = 64.3 km.
Problem
36. What is the total mechanical energy associated with Earths orbital motion?
Problem
34. At its perihelion in February 1986, Comet Halley was 8.79107 km from the Sun and was moving at 54.6 km/s. What will be its speed when it crosses Neptunes orbit, in 2006?
Solution
If we assume a circular orbit and apply Equation 9-9, we obtain
Etot = M ME 1 U = G 2 2r (6.671011 Nm2 /kg2 )(1.991030 kg)(5.971024 kg) = 2(1.501011 m) = 2.641033 J.
Solution
The total energy at any point in the comets orbit around the Sun is the same, as in Example 9-7, where we can take v0 and r0 at perihelion as given, and nd v at the distance r = 4.50109 km of Neptunes orbit (see Appendix E). Then
1 2 v = [v0 2GM (r0 r1 )]1/2 2 2
(Alternatively, from Equation 9-8, Etot = K = 1 2 ME (2r/T )2 = 2.661033 J, consistent with the accuracy of the data used in Appendix E.)
Problem
37. Drag due to small amounts of residual air causes satellites in low Earth orbit to lose energy and eventually spiral to Earth. What fraction of its orbital energy is lost as a satellite drops from 300 to 100 km, assuming its orbit remains essentially circular?
=
54.6
km s
2 6.671011
Nm kg2
(1.991030 kg) 1 4.50109 km
1 8.79107 km
1/2
= 4.48 km/s.
Solution
The fractional dierence in orbital energy is 1 E/E = (E E )/E, where E = 2 U = GME m 2r 1/(RE + h) is the orbital energy, and h is the altitude of the circular orbit (see Equation 9-9).
Problem
35. Neglecting air resistance, to what height would you have to re a rocket for the constant acceleration equations of Chapter 2 to give a
128
CHAPTER E/E 9
Thus = (r1 r1 )/r1 = (h h) (RE + h ) = (100 300) km/(6370 + 100) km = 2/64.7 = 3.09102 3%. (The energy dierence is negative because energy is lost going from a higher to a lower orbit.)
Solution
2 Equation 9-7 implies R = 2GM/vesc = 2 11 2 24 2(6.6710 Nm /kg )(2.910 kg) (7.1103 m/s)2 = 7.67106 m.
Problem
38. Show that an object released from rest very far from Earth (r RE ) reaches Earths surface at essentially escape speed.
Problem
42. Determine the escape speed from (a) Jupiters moon Callisto, with mass 1.071023 kg and radius 2.40 Mm, and (b) a neutron star, with the Suns mass crammed into a sphere 6.0 km in radius.
Solution
Equating the energy of the object when at rest a distance r from the Earth, with its energy after falling to the Earths surface, we nd GME m/r = 1 2 2 mv GME m/RE . Solving for v, we get v = (2GME /RE )(1 RE /r) = vesc (1 RE /r). For r RE , the assertion in the problem follows.
Solution
(a) vesc = [2(6.671011 Nm2 /kg2 )(1.071023 kg) (2.40106 m)]1/2 = 2.44 km/s, and (b) vesc = [2(6.67 1011 Nm2 /kg2 )(1.991030 kg)/(6103 m)]1/2 = 2.10108 m/s, or about 70% of the speed of light. (See Equation 9-7.)
Problem
43. Two meteoroids are 250,000 km from Earth and moving at 2.1 km/s. One is headed straight for Earth, while the other is on a path that will come within 8500 km of Earths center (Fig. 9-36). (a) What is the speed of the rst meteoroid when it strikes Earth? (b) What is the speed of the second meteoroid at its closest approach to Earth? (c) Will the second meteoroid ever return to Earths vicinity?
Problem
39. By what factor must the speed of an object in circular orbit be increased to reach escape speed from its orbital altitude?
Solution
The escape speed at a distance r from the Earths center is just 2 times the speed in a circular orbit of the same radius. Compare Equations 9-3 and 9-7.
Problem
40. Astronomers discover a new comet. As it crosses Earths orbit the comet is moving at 45 km/s. Is the comet in an open or closed orbit?
Solution
The total energy of the comet, at a distance of the Earths orbit, is E = 1 mv 2 GM m/r 2 (6.671011 )(1.991030 ) m2 (45103 m/s)2 =m 2 1.501011 s2 = m[10.1 8.85]108 (m2 /s2 ) > 0. Since E is positive, the orbit is open (hyperbolic), and the comet is an aperiodic one. (See the paragraph preceding Example 9-6.)
figure 9-36 Problem 43 Solution.
Solution
(a) Conservation of energy applied to the rst 1 1 2 meteoroid gives: 2 mv0 G(ME m/r0 ) = 2 mv 2 2 G(ME m/RE ), or v = v0 + 2GME (1/RE 1/r0 ). In 2 the numerical evaluation, replace GME by gRE , to obtain:
v = 2.1 km s
2
+ 2 0.0098
km s2
(6370 km) 1
6370 250,000
= 11.2 km/s.
Problem
41. The escape speed from a planet of mass 2.91024 kg is 7.1 km/s. What is the planets radius?
(b) For the second meteoroid, v =
2 2 v0 + 2gRE
1 1 8500 km 250,000 km
= 9.74 km/s.
CHAPTER 9 (c) The escape velocity at a distance of r = 8500 km from the center of the earth is vesc = 2GME /r = 2 2gRE /r = 9.67 km/s, so the second meteoroid will probably not return. Alternatively, vesc at a distance of 250,000 km is 1.78 km/s.
129
and if our energy use remained constant, by how much would (a) the moons orbital radius and (b) its orbital period change in a century?
Solution
1 (a) The moons orbital energy is E = 2 U = GME Mm /2r (see Equation 9-9). If the orbital radius decreased, the moon would lose energy at a rate dE/dt = (GME Mm /2r2 )(dr/dt). (Dierentiate and use the chain rule; see Appendix A-2.) To supply the stated power needs of humanity, dE/dt = 1013 W, so
Problem
44. One component of the proposed star wars antimissile defense system calls for powerful laser beams aimed by mirrors in geosynchronous orbit. One simple countermeasure to this delicate technology is to put a truckload of rocks in the same orbit, going in the opposite direction. At what relative speed would the rocks hit the mirror?
2r 2 dr = dt GME Mm =
dE dt 2(3.85108 m)2 (1013 W) Nm2 /kg2 )(5.971024 kg)(7.351022 kg)
(6.671011
= 1.01107 m/s.
Solution
The speed of an object in a circular geosynchronous orbit is v = GMe /(4.22107 m) = 3.07 km/s (see Equation 9-3 and Example 9-4). Since the rocks and mirror have the same speed, but in opposite directions, their relative speed is 2v = 6.14 km/s.
Problem
45. Neglecting Earths rotation, show that the energy needed to launch a satellite of mass m into circular orbit at altitude h is RE + 2h GME m . RE 2(RE + h)
At this rate, in one century, the orbital radius would change by r = (1.01107 m/s)(3.156107 s/y) (100 y) = 320 m. (b) From Keplers third law, 3 T 2 r3 , so the fractional change in the period is 2 the fractional change in radius (dierentiate and divide): 3 T /T = 2 (r/r). For the century-long change in part 3 (a), T = 2 (320 m/3.85108 m)(27.3 d) = 5 3.4010 d = 2.94 s.
Problem
47. A projectile is launched vertically upward from a planet of mass M and radius R; its initial speed is twice the escape speed. Derive an expression for its speed as a function of the distance r from the center of the planet.
Solution
The energy of a satellite in a circular orbit is 1 E = 2 U = GME m/2r, where r = RE + h (see Equation 9-9). This is the energy in an Earth-centered, non-rotating reference frame, neglecting the gravitational inuence of any other body, e.g., the Sun. The energy of a satellite on the Earths surface depends on its location, because of the 1 2 Earths diurnal rotation, E0 = U0 + 2 mv0 , where U0 = GME m/RE , and v0 is the speed of the Earths surface at that location. If we neglect the extra kinetic energy associated with v0 (since v0 2RE /1d, 1 2 2 mv0 0.34% |U0 |), the energy required to launch the satellite into a circular orbit at altitude h is 1 1 E E0 = 1 U U0 = GME m[RE 2 (RE + h)1 ] = 2 GME m(RE + 2h)/2RE (RE + h) as claimed.
Solution
If we consider just the gravitational eld of the planet and neglect possible losses of energy (from atmospheric drag, etc.), then the conservation of energy, K + U = K0 + U0 , gives the speed of an object at a radial distance r (from the center of a spherical massive body) as a function of its speed v0 and distance r0 at
1 2 a particular point: v = v0 2GM (r0 r1 ) (see 2 Example 9-7). If we set v0 = (2vesc )2 = 8GM/R at r0 = R (from Equation 9-7) for the projectile in this problem, then v = 2GM (4R1 R1 + r1 ) = 2GM (3R1 + r1 ).
Problem
48. A spacecraft is in a circular Earth orbit at an altitude of 5500 km. By how much will its altitude decrease if it moves to a new circular orbit where (a) its orbital speed is 10% higher or (b) its orbital period is 10% shorter?
Problem
46. Humanity currently consumes energy at the rate of about 1013 W. Suppose a method were developed to extract usable energy from the moons orbital motion. If all our energy came from this source
130
CHAPTER 9 This is the tidal force, and your expression shows that it drops as the cube of the distance.
r m 2a m M
Solution
(a) From Equation 9-3, v = GME /r. For an orbital speed 10% higher, v 2 = (1.1v)2 = GME /r , or r = r/(1.1)2 . The decrease in altitude is h = r r = [1 1/(1.1)2 ]r = 0.174(6370 + 5500) km = 2060 km. (b) From Equation 9-4, T 2 = 4 2 r3 /GME . For a period 10% shorter, (0.9T )2 = 4 2 r3 /GME , or r = (0.9)2/3 r. Again, the change is h = [1 (0.9)2/3 ]r = (0.0678)(11, 870 km) = 805 km.
2
figure 9-37 Problem 50.
Problem
49. The Pioneer spacecraft left Earths vicinity moving at about 38 km/s relative to the Sun (this gure combines the eect of rocket boost and Earths orbital motion). How far out in the solar system could Pioneer get without additional rocket power or use of the gravitational slingshot eect?
Solution
All the bodies are in the same line in Fig. 9-37, so we need to consider only the magnitude of the gravitational forces of the large mass on each small mass. The dierence is GM m 4ra GM m = GM m 2 F = (r a)2 (r + a)2 (r a2 )2 = 4GM ma r3 1 a2 r2
2
Solution
If we consider just the gravitational eld of the Sun, which is the dominant source of potential energy for the Pioneer spacecraft in this problem, then the conservation of energy in the form K + U = K0 + U0 1 (as in Example 9-6) gives 2 mv 2 GM m/r = 1 1 2 2 2 1 . 2 mv0 GM m/r0 , or r = [r0 (v0 v )/2GM ] Here, v0 is given as 38 km/s, and r0 = 1 AU = 1.50108 km. We do not know the actual shape of the 2 satellites orbit, but v 2 v0 at the maximum distance 1 2 from the sun, so r < rmax = [r0 v0 /2GM ]1 = 11 1 3 2 [(1.5010 m) (3810 m/s) /2(6.67 1011 Nm2 /kg2 )(1.991030 kg)]1 = 8.151011 m = 5.43 AU. This is a little farther than the orbit of Jupiter.
In the approximation (a2 /r2 ) 1, the desired expression follows.
Problem
51. Show that the force of the Suns gravity on Earth is nearly 200 times that of the moons gravity, but that the tidal force of the moon on Earth (see preceding problem) is about twice that of the Sun.
Solution
The ratio of the magnitude of the direct gravitational forces is
2 GMS ME /rSE MS FSun = = 2 Fmoon GMM ME /rEM MM
Section 9-6: Tidal Forces Problem
50. To get a feel for the behavior of tidal forces, consider an object consisting of two small masses m separated by a distance 2a, a distance r from a body of mass M, as shown in Fig. 9-37. Use the law of universal gravitation to determine the gravitational force of the large mass on each of the small masses. Find an expression for the dierence between the two forces, and put your expression over a common denominator. Show that, in the approximation where a2 can be neglected compared with r2 , your expression for the force dierence can be written 4GM ma . F = r3
rEM rSE = 178
2
=
1.991030 7.351022
0.385 150
2
(see Appendix E), while the ratio of the dierential tidal forces, as given in the preceding problem, is (the factors relating to the Earth, like m and a, cancel out even in a more exact calculation) FSun /Fmoon = (MS /MM )(rEM /rSE )3 = 0.458.
Problem
52. Suppose the composite object in Fig. 9-37 is held together by the gravitational attraction of the two small masses. Show that the tidal force of Problem 50 is greater than the self-gravitational force of the object when r3 < 16M a3 /m. This denes the Roche limit for this simple example.
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131
Solution
The gravitational attraction of the two small masses in Fig. 9-37 has magnitude Gm2 /(2a)2 . The dierential tidal force found in Problem 50 is greater than this when 4GM ma/r3 > Gm2 /4a2 , or r3 < 16a3 M/m as stated.
Problem
56. Two rockets are launched from Earths surface, one at 12 km/s and the other at 18 km/s. How fast is each moving when it crosses the moons orbit?
Solution
The same reasoning as applied in the previous problem, with the same approximations and the 2 neglect of the Earths rotation, gives v 2 = v0 + 1 1 2GME (r r0 ). Here, v0 is the launch speed of the rocket, r0 = RE at the Earths surface, and r = 3.85105 km crossing the moons orbit, so 2 v 2 = v0 1.23108 (m/s)2 . For the values of v0 given, 12 and 18 km/s, v = 4.59 and 14.2 km/s, respectively.
Paired Problems Problem
53. An astronaut hits a golf ball horizontally from the top of a lunar mountain so fast that it goes into circular orbit. What is its orbital period?
Solution
The period of a grazing orbit, r R, around a spherical object of mass M and radius R (see Appendix E for lunar values) can be found from Equation 9-4, T = 2 R3 /GM = 2[(1.74106 m)3 (6.671011 Nm2 /kg2 )(7.351022 kg)]1/2 = 6.51103 s = 109 min.
Problem
57. A satellite is in an elliptical orbit at altitudes ranging from 230 to 890 km. At the high point its moving at 7.23 km/s. How fast is it moving at the low point?
Problem
54. Find the period of a spacecraft in circular orbit 150 km above Marss surface.
Solution
The conservation of energy is applied to a satellite in an elliptical Earth orbit in Example 9-7 (where it is a good approximation to neglect the gravitational inuence of other bodies, atmospheric drag, etc.) to relate the speed and distance at perigee (the lowest point) to the same quantities at apogee (the highest 2 2 1 1 point): vp = va + 2GME (rp ra ). The calculation can be simplied by expressing the distances in terms of altitude above the Earths surface and using a little 1 1 algebra. Then r = RE (1 + h/RE ) and (rp ra ) = 2 (ra rp )/ra rp = (ha hp )/RE (1 + ha /RE )(1 + hp /RE ). 2 2 Since GME /RE = g, vp = (7.23 km/s)2 + 2 2(9.81 m/s )(890 km 230 km)/[(1 + 890/6370)(1 + 230/6370)], and vp = 7.95 km/s. (This result also follows from the conservation of angular momentum or Keplers second law, which implies that va ra = vp rp .)
Solution
As in the previous problem, Equation 9-4 and Appendix E give T = 2[((3.38 + 0.15) 106 m)3 (6.671011 Nm2 /kg2 )(6.421023 kg)]1/2 = 6.37103 s = 106 min.
Problem
55. Two meteoroids are 160,000 km from Earths center and heading straight toward Earth. One is moving at 10 km/s, the other at 20 km/s. At what speed will they strike Earth?
Solution
If we consider only the dierence of potential energy of the Earths gravitational eld (since r0 = 160,000 km is small compared to the distance from the Sun, approximately 150106 km, the dierence of potential energy of the meteoroids in the Suns gravitational eld is less than 2% of the Earths), the conservation of energy can be applied as in Example 9-7 to determine the impact speed. (This estimate also neglects atmospheric eects or, alternatively, gives the speed of impact on the atmosphere, r RE , instead of the 2 2 Earths surface.) Then vimpact = v0 + 2GME (r1 1 2 r0 ) = v0 + 1.20108 (m/s)2 , where we used data from Appendix E. (a) For v0 = 10 km/s, vimpact = 14.8 km/s, and (b) for v0 = 20 km/s, vimpact = 22.8 km/s.
Problem
58. A missiles trajectory takes it to a maximum altitude of 1200 km. If its launch speed is 6.1 km/s, how fast is it moving at the peak of its trajectory?
Solution
The missiles trajectory is elliptical (if atmospheric drag and other gravitational perturbations than the Earths spherical eld are neglected) and intersects the Earths surface at launch and impact. The analysis of 2 Example 9-7 and the previous problem gives v 2 = v0 1 2GME (r0 r1 ). Here, v0 = 6.1 km/s, and r0 = RE
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CHAPTER 9
at launch, and v and r = RE + 1200 km refer to the point of maximum altitude. The same algebra as for 2 the previous solution shows that v 2 = v0 2gRE h 2 2 (RE + h) = (6.1 km/s) 2(0.00981 km/s )(1200 km) (6370/7570) = (4.17 km/s)2 .
Solution
For a small object (the spacecraft) in a circular orbit about a central massive body (the Sun), the total, kinetic, and potential energies are related by 1 Equations 9-8 and 9, E = K = 2 U. (b) Therefore 11 K = E = 5.310 J. (c) Since K = 1 mv 2 , v = 2 2K/m = [2(5.31011 J)/720 kg]1/2 = 38.4 km/s. (a) Since U = 2E = GM m/r, r = GM m/(2E) = (6.671011 Nm2 /kg2 )(1.991030 kg)(720 kg) (2 5.31011 J) = 9.021010 m = 0.601 AU.
Problem
59. To what radius would Earth have to be shrunk, with no loss of mass, for escape speed at its surface to be 30 km/s?
Solution
The escape speed from the surface of a spherical body of mass M (equal to ME in this case) is given by Equation 9-7, from which the radius can be found if 2 vesc is given: R = 2GME /vesc. In the numerical calculation, we could use the known escape speed of the earth, vesc = 11.2 km/s for R = RE , to eliminate the constants 2GME = (11.2 km/s)2 RE , obtaining R = (11.2/30)2RE = 0.139RE 888 km. (Other 2 combinations of constants, e.g., GME = gRE , give the same result to within the accuracy of the data.)
Problem
62. A 2500-kg boulder is among the myriad particles making up Saturns rings shown in Fig. 9-38. The boulder is in a circular orbit 95,000 km from the center of Saturn. Find (a) its total energy, (b) its orbital speed, and (c) its orbital period.
Solution
(a) From Equation 9-9 for circular orbits and Appendix E, E = GM m/2r = (6.67 1011 Nm2 /kg2 )(5.691026 kg)(2.5103 kg) 2(9.5107 m) = 4.991011 J. (b) Equation 9-8 gives v = 2K/m = 2E/m = 2(4.991011 J)/(2.5103 kg) = 20.0 km/s. (c) The period is T = 2r/v = 2(9.5107 m) (2.00104 m/s) = 2.99104 s = 8.30 h.
Problem
60. What is the mass of a planetary moon whose radius is 14 km and whose surface escape speed is 0.13 m/s?
Solution
One can solve Equation 9-7 for the mass and use the given data to obtain
2 M = Rvesc /2G
Supplementary Problems Problem
63. Mercurys orbital speed varies from 38.8 km/s at aphelion to 59.0 km/s at perihelion. If the planet is 6.991010 m from the Suns center at aphelion, how far is it at perihelion?
= (14 km)(0.13 m/s)2 /2(6.671011 Nm2 /kg2 ) = 1.771012 kg.
Solution Problem
61. A 720-kg spacecraft has total energy 5.31011 J and is in circular orbit about the Sun. Find (a) its orbital radius, (b) its kinetic energy, and (c) its speed. Keplers second law implies rp = ra (va /vp ) = (6.991010 m)(38.8/59.0) = 4.601010 m (see the solution to Problem 57). Otherwise, an equation similar to that resulting from the conservation of 2 1 1 2 energy in Example 9-7, rp ra = (vp va ) 12 1 2GM = 7.4410 m , gives the same value of rp .
Problem
64. Show that the form U = mgr follows from Equation 9-5 when r1 r2 . Hint: Write r2 = r1 + r, and apply the binomial approximation (see Appendix A).
Solution
If we let r = r2 r1 r1 in Equation 9-5, and neglect r compared to r1 in the denominator of the
CHAPTER 9 last step, we obtain U = GM m 1 1 = GM m r1 r2 GM m r GM m = r. 2 r1 (r1 + r) r1 r2 r1 r1 r2 or, after canceling m, GME GM = 2 (r d) d2 2 TE
2
133
(r d).
2 Near the earths surface (r1 RE ), g GM/r1 is constant, so U mg r.
Problem
65. A black hole is an object so dense that its escape speed exceeds the speed of light. Although a full description of black holes requires general relativity, the radius of a black hole can be calculated using Newtonian theory. (a) Show that the radius of a black hole of mass M is 2GM/c2 , where c is the speed of light. What are the radii of black holes with (b) the mass of the Earth and (c) the mass of the Sun?
This equation could be solved for d numerically, using available PC-software and values for the other 1 quantities from Appendix E (except TE 365 4 d). The result for d is a little less than 1.64 million km, but in view of the initial crude assumptions about orbits, this is only an estimate at best. (An alternate approach would be to rewrite the equation for the centripetal acceleration as d r =
2
2 TE
ME M
2
1
d r d r
2
r3 GM
2
1
d r
3
.
Solution
(a) The so-called Schwarzchild radius of a black hole of mass M turns out to be the same as the radius given by Equation 9-7 with vesc = c = 2GM/rs . Thus rs = 2GM/c2 . (b) 2GME /c2 = 8.85 mm, and (c) 2GM /c2 = 2.95 km.
If we ignore the presence of Jupiter, etc., a more exact form of Keplers third law relates the constants in square brackets by 2 TE
2
r3 GM
=1+
ME . M
Then consideration of just the leading terms in the resulting equation, d r =
2
Problem
66. Find the point discussed in the Application: Perpetual Sunshine (page 221)that is, the point along the Earth-Sun line where an object will orbit the Sun with the same orbital period as does Earth. Express as a distance sunward of Earth, both in SI units and as a fraction of Earths orbital radius. Use data from Appendix E. You may need to use numerical or graphical techniques.
1+
ME M ME M
1 d r
2
d r
2
1
d r
3
,
gives (d/r) (ME /3M )1/3 1%; i.e., d is about 1.5 million km.)
Solution
Suppose that the Lagrangian point L1 is a distance d from the Earth, along the Earth-Sun line of length r = 1 AU, as shown. We assume that a small body at L1, with mass m, and the Earth are in approximately circular orbits about the much more massive and practically motionless Sun. The centripetal force on m, which has the same orbital period as the Earth and an orbital radius of r d, is m(2/TE )2 (r d) (see Equation 4-12), whereas the gravitational forces exerted by the Sun and the Earth are GM m/(r d)2 and GME m/d2 , respectively. As explained in the Application: Perpetual Sunshine, at L1, the dierence in the gravitational forces equals the centripetal force Problem 66 Solution.
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CHAPTER 9
Problem
67. Two satellites are in geosynchronous orbit, but in diametrically opposite positions (Fig. 9-39). Into how much lower a circular orbit should one spacecraft descend if it is to catch up with the other after 10 complete orbits? Neglect rocket ring times and time spent moving between the two circular orbits.
figure 9-39 Problem 67.
Solution
In a lower circular orbit (smaller r) the orbital speed is faster (see Equation 9-3). The time for 10 complete 1 orbits of the faster satellite must equal the time for 9 2 geosynchronous orbits, if the faster satellite is to catch up as described (it starts out one-half an orbit behind). Thus, 10T = 9.5(1 d), or T = 0.95 d, where T is the period of the lower, faster orbit. Keplers third law (Equation 9-4) then gives r = (GME T 2 /4 2 )1/3 = (0.95)2/3 rGS, since for the geosynchronous orbit, (GME /4 2 )1/3 = rGS /(1 d)2/3 , where rGS = 42,200 km (see Example 9-4). The dierence in the orbital radii is rGS, r = [1 (0.95)2/3 ](42,200 km) = 1420 km. (We neglected the time spent in changing orbits as suggested.)