Unformatted text preview: Chapter 9 Solutions Problem
1. Space explorers land on a planet with the same mass as Earth, but they find they weigh twice as much as they would on Earth. What is the radius of the planet? Solution
At rest on a uniform spherical planet, a body's weight is proportional to the surface gravity, g = GM= . Therefore, R
2 M (g p= E ) = (M p= E )(RE= p) 2 = 2. Since M p= E = 1, Rp = RE = 2. g M R 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
If the surface gravity of the Earth were three times its present value, with no change in mass, then GM E= = 3GM E= E , R R where RE is the present radius. Thus, R RE = 1= 3 = 57.7% gives the new, shrunken radius. =
2 2 Problem
4. Calculate the gravitational acceleration at the surface of (a) Mercury and (b) Saturn's moon Titan. Solution
With reference to the first two columns in Appendix E: (a) gMerc = G(0.330 ! 10 (b) gTitan = G(0.135 ! 10
24 24 kg)= (2.44 ! 106 m)2 = 3.70 m/s2 , and kg)= (2.58 ! 106 m) 2 = 1.35 m/s2 . Problem
10. A roughly spherical volume of rock 1.3 km in radius is 30% denser than the surrounding rock, whose density is 2700 kg/m3 . The denser rock is centered 2.1 km below Earth's 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 Earth's surface. Solution
The extra mass of the denser rock (above the surrounding rock) is !m = (4= " R !# , where R = 1.3 km is the radius of 3) 3 the spherical volume and !" = 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 = G!m= (2.1 km)2 = (6.67 " 10 #1 1 N $ m2 /kg2 )(4= % (1.3 " 103 m) 3 (810 kg/m 3 )= 3)
3 (2.1 ! 103 m)2 = 1.13 ! 10"4 m/s2 = 1.15 ! 10"5 g = 11.3 milligal (see previous solution). 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 off from the rest of humanity, as the spacecraft rounded the far side of the moon (see Fig. 934). How long did this period last? FIGURE 934 Problem 17. Solution
The period of a circular orbit of altitude h = 130 km above the moon's surface is (see Equation 94) T = 2! (Rm + h) = m = 2! [(1.74 + 0.13) " 10 m = GM (6.67 " 10 N $ m /kg )(7.35 " 10 kg)] = 7.26 " 10 s = 121 min. The command module was cut off from Earth communications for roughly half of this period, or about 1 h. 3 3 18 3 #1 1 2 2 22 12 = 3 Problem
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 93 gives v = GM =RE = [(6.67 ! 10 N # m /kg )(1.99 ! 10 kg)= (6.37 ! 10 m)] = 4.56 ! 10 m/s , or about 1.5% of the speed of light. (b) The orbital period, T = 2! RE = = 8.77 s, is correspondingly small. v "1 1 2 2 30 6 1= 2 6 Problem
21. Where should a satellite be placed to orbit the Sun in a circular orbit with a period of 100 days? Solution
As long as the masses of the orbiting bodies (satellite or planet) are negligible compared to the mass of the Sun, Equa3 tion 94, as a ratio, can be used to compare the mean orbital radii and periods. (T1=2 ) = (r =2 ) (this is, in fact, an T 2 1r approximate statement of Kepler's third law). The mean orbital radius of the Earth is 1 AU (an astronomical unit) ! 1.50 ! 108 km, and its period is 1 y (a sidereal year) ! 365 d, so (T= y) 2 = (r= AU) 3 for any satellite of the Sun. If 1 1
7 1 365) 2=3 = 0.422, or R = 0.422 AU = 6.3 ! 10 km. Of course, Equation 94 could be T = 100 d, then (r= AU) = (100= used directly: 3 r = [GM (T= ! ) 2 ]1= 2 2 +% 2( 6 ( . %  6.67 " 10#1 1 N $ m (1.99 " 103 0 kg) 8.64 " 10 s 0 = ' * ' * 0 2! kg2 ) & ) 0 & , / 13 = = 6.31 " 10 7 km. Problem
11 28. Earth's distance from the Sun varies from 1.47 ! 10 m at perihelion to 1.52 ! 10 m at aphelion because its orbit is not quite circular. Find the change in potential energy as Earth goes from perihelion to aphelion.
11 Solution
From Equation 95, !U = GM M E (rp " ra ) = (6.67 # 10
"1 "1 "1 1 N $ m 2 /kg2 )(1.99 # 103 0 kg) # (5.97 # 102 4 kg) # [(1.47) !1 ! (1.52) !1 ] " 10!1 1 m = 1.77 " 103 2 J. 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 Earth's orbit. At what speed would it hit the Sun? Solution
Apply the conservation of energy to the dropped canister, as in Examples 96 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 of the Sun at r = R . Then !GM m= = 1 mv 2 ! GM m= , or r R 2 v = 2GM (R!1 ! r !1 ) = [2(6.67 " 10 !1 1 N # m 2 /kg2 )(1.99 " 103 0 kg)(( 6.96 " 108 m) !1 ! (1.50 " 101 1 m)!1 )]1=2 . = 616 km/s. Problem
31. A rocket is launched vertically upward from Earth's 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 96. Conservation of energy gives 2 1 or (R 2 mv 0 ! GM E m=RE = !GM E m= E + h),
2 " 1 v0 % h= $ ! ' $ ' # R E 2GM E & !1 ! RE = 530 km, when the proper values are substituted. Problem
41. The escape speed from a planet of mass 2.9 ! 10
24 kg is (7.1) km/s. What is the planet's radius? Solution
Equation 97 implies R = 2GM= esc = 2(6.67 ! 10 v
2 "1 1 N # m 2 /kg2 )(2.9 ! 102 4 kg)= (7.1 ! 10 3 m/s) 2 = 7.67 ! 10 6 m. Problem
42. Determine the escape speed from (a) Jupiter's moon Callisto, with mass 1.07 ! 10 (b) a neutron star, with the Sun's mass crammed into a sphere 6.0 km in radius.
23 kg and radius 2.40 Mm, and Solution N # m 2=kg2 ) ! (1.07 ! 102 3 kg)= (2.40 ! 106 m)]1=2 = 2.44 km/s, and (b) v = esc 2 30 3 12 = 8 [2(6.67 ! 10 N # m =kg )(1.99 ! 10 kg)= ! 10 m)] = 2.10 ! 10 m/s, or about 70% of the speed of light. (6 (See Equation 97.)
(a) v esc = [2(6.67 ! 10
"1 1 2 "1 1 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 Earth's center (Fig. 936). (a) What is the speed of the first 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 Earth's vicinity? Solution
(a) Conservation of energy applied to the first meteoroid gives: or v =
1 2 2 mv0 ! G(M E m=0 ) = r 1 2 mv 2 ! G(M E m/RE ), 2 v 0 + 2GM E (1 E ! 1= 0 ). In the numerical evaluation, replace GM E by gR2 , to obtain: =R r E v =
(b) For the second meteoroid, km $ 2 km $ ! 6370 $ ! ! # 2.1 & + 2# 0.0098 2 &(6370 km)#1 ' & = 11.2 km/s. " " s % s % " 250, 000 % v = " 1 1 % 2 v 0 + 2 gR2 $ ! ' = 9.74 km/s. E # 8500 km 250, 000 km& (c) The escape velocity at a distance of r = 8500 km from the center of the earth is v esc = 9.67 km/s, so the second meteoroid will probably not return. Alternatively, v esc 2GM E= = 2gR 2 = = r E r at a distance of 250,000 km is 1.78 km/s. FIGURE 936 Problem 43 Solution. 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
If we consider just the gravitational field of the planet and neglect possible losses of energy (from atmospheric drag, etc.), then the conservation of energy, K + U = K 0 + U 0 , 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 v 0 and distance r0 at a particular point: v =
2 2
2 v 0 ! 2GM (r0!1 ! r !1 ) (see Example 97). If we set v 0 = (2vesc ) = 8GM = at r0 = R (from Equation 97) for the projectile in this problem, then R v = 2GM (4R !1 ! R !1 + r !1) = 2GM (3R!1 + r !1 ). 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? Solution
= GM E = For an orbital speed 10% higher, v ! 2 = (1.1v ) 2 = GM E = ! , or r ! = r= r. r (1.1)2 . The 2 (1.1) ]r = 0.174(6370 + 5500) km = 2060 km. (b) From Equation 94, decrease in altitude is !h = r " r# = [1 " 1= 2 2 3 2= 3 2 2 3 GM T = 4! r = E . For a period 10% shorter, (0.9T ) = 4! r" = E , or r " = (0.9) r. Again, the change is GM
(a) From Equation 93, v
3 !h = [1 " (0.9)2= ]r = (0.0678)(11,870 km) = 805 km. 2 Problem
57. A satellite is in an elliptical orbit at altitudes ranging from 230 to 890 km. At the high point it's moving at 7.23 km/s. How fast is it moving at the low point? Solution
The conservation of energy is applied to a satellite in an elliptical Earth orbit in Example 97 (where it is a good approximation to neglect the gravitational influence of other bodies, atmospheric drag, etc.) to relate the speed and distance at 2 2 !1 !1 perigee (the lowest point) to the same quantities at apogee (the highest point): v p = v a + 2GM E (rp ! ra ). The calculation can be simplified by expressing the distances in terms of altitude above the Earth's surface and using a little algebra. Then !1 !1 2 r R r = RE (1 + h= E ) and (rp ! ra ) = (ra ! rp )= a rp = (ha ! hp )=RE (1 + ha=RE )(1 + hp = E ). Since R
2 2 2 2 GM E = E = g, v p = (7.23 km/s) + 2(9.81 m/s )(890 km ! 230 km)= + 890= R [(1 6370)(1 + 230= 6370)], and v p = 7.95 km/s. (This result also follows from the conservation of angular momentum or Kepler's second law, which implies that v a ra = v p rp .) 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 97 for the mass and use the given data to obtain M = Rv 2 = = (14 km)(0.13 m/s)2 = 2(6.67 ! 10 "1 1 N # m 2 /kg2 ) = 1.77 ! 101 2 kg. esc 2G ...
View
Full Document
 Spring '07
 Hicks
 Mass, General Relativity, Astronomical unit

Click to edit the document details