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Transcript
PH212 Chapter 13 Solutions 13.1.I DENTIFY and S ET U P : Use the law of gravitation, Eq. (13.1), to determine E XECUTE : the radius of the moon’s orbit around the earth is given in Appendix F as moon is much closer to the earth than it is to the sun, so take the distance sun to be The of the moon from the the radius of the earth’s orbit around the sun. E VALUATE : The force exerted by the sun is larger than the force exerted by the earth. The moon’s motion is a combination of orbiting the sun and orbiting the earth. 13.10.I DENTIFY : The force exerted by m on M and the force exerted by 2m on M are each given by Eq. (13.2) and the net force is the vector sum of these two forces. S ET U P : Each force is attractive. The forces on M in each region are sketched in Figure 13.10a. Let M be at coordinate x on the x-axis. E XECUTE : (a) For the net force to be zero, and must be in opposite directions and this is the case only for then requires and must be less than L, so (b) For as and as goes from 0 to 0.414L. For 0.414L to L. The graph of and as For For and and as increases from increases from 0 to to 0 as x as x goes from versus x is sketched in Figure 13.10b. E VALUATE : Any real object is not exactly a point so it is not possible to have both m and M exactly at or 2m and M both exactly at But the magnitude of the gravitational force between two objects approaches infinity as the objects get very close together. Figure 13.10 13.16.I DENTIFY : The gravity of Io limits the height to which volcanic material will rise. The acceleration due to gravity at the surface of Io depends on its mass and radius. S ET U P : The radius of Io is Use coordinates where maximum height, is upward. At the which is assumed to be constant. Therefore the constant- acceleration kinematics formulas apply. The acceleration due to gravity at Io’s surface is given by S OLVE : At the surface of Io, For constant acceleration (assumed), the equation applies, so Now solve for and The equation when gives E VALUATE : Even though the mass of Io is around 100 times smaller than that of the earth, the acceleration due to gravity at its surface is only about 1/6 of that of the earth because Io’s radius is much smaller than earth’s radius. 13.22.I DENTIFY : We can calculate the orbital period T from the number of revolutions per day. Then the period and the orbit radius are related by Eq. (13.12). S ET U P : and The height h of the orbit above the surface of the earth is related to the orbit radius r by E XECUTE : The satellite moves 15.65 revolutions in so the time for 1.00 revolution is gives and E VALUATE : The period of this satellite is slightly larger than the period for the satellite in Example 13.6 and the altitude of this satellite is therefore somewhat greater. 13.25.I DENTIFY : The orbital speed is given by where m is the mass of the star. The orbital period is given by S ET U P : The sun has mass E XECUTE : The orbit radius of the earth is (a) (b) (about two weeks). E VALUATE : The orbital period is less than the 88-day orbital period of Mercury; this planet is orbiting very close to its star, compared to the orbital radius of Mercury. 13.32.I DENTIFY : The gravitational potential energy of a pair of point masses is into infinitesimal pieces and integrate to find U. S ET U P : Divide the rod into differential masses rod. E XECUTE : Divide the rod at position l, measured from the right end of the (a) Integrating, For the natural logarithm is and (b) The x-component of the gravitational force on the sphere is with the minus sign indicating an attractive force. As the denominator in the above expression approaches and as expected. E VALUATE : When x is much larger than L the rod can be treated as a point mass, and our results for U and do reduce to the correct expression when 13.34.I DENTIFY : At the north pole, Sneezy has no circular motion and therefore no acceleration. But at the equator he has acceleration toward the center of the earth due to the earth’s rotation. S ET U P : The earth has mass radius Use coordinates for which the and rotational period direction is toward the center of the earth. The free-body diagram for Sneezy at the equator is given in Figure 13.34. The radial acceleration due to Sneezy’s circular motion at the equator is Sneezy. Figure 13.34 and Newton’s second law applies to E XECUTE : At the north pole Sneezy has exerted by the earth). Sneezy has mass and (the gravitational force At the equator Sneezy is traveling in a circular path and has radial acceleration law gives Newton’s second Solving for T gives E VALUATE : At the equator Sneezy has an inward acceleration and the outward tension is less than the true weight, since there is a net inward force. 13.53.I DENTIFY : From Example 13.5, the escape speed is Use to write this expression in terms of S ET U P : For a sphere E XECUTE : In terms of the density E VALUATE : 13.58.I DENTIFY : the ratio and so the escape speed is This is much less than the escape speed for the earth, 11,200 m/s. Use the measurements of the motion of the rock to calculate Then use this to calculate the mass of Mongo. For the ship, S ET U P : Take downward. the value of g on Mongo. and upward. When the stone returns to the ground its velocity is 12.0 m/s, The radius of Mongo is The ship moves in an orbit of radius E XECUTE : (a) and gives and (b) gives E VALUATE : and and so which agrees with the value calculated in part (a). 13.64. I DENTIFY : In part (a) use the expression for the escape speed that is derived in Example 13.5. In part (b) apply conservation of energy. S ET U P : In part (b) let point 1 be at the surface of the comet. E XECUTE : (a) The escape speed is so (b) (i) gives Solving for r gives and debris never loses all of its initial kinetic energy, but as (ii) The The farther the debris are from the comet’s center, the smaller is their kinetic energy. E VALUATE : The debris will have lost 90.0% of their initial kinetic energy when they are at a distance from the comet’s center of about ten times the radius of the comet. 13.71.I DENTIFY : Use Eq. (13.2) to calculate Apply Newton’s second law to circular motion of each star to find the orbital speed and period. Apply the conservation of energy expression, Eq. (7.13), to calculate the energy input (work) required to separate the two stars to infinity. (a) S ET U P : The cm is midway between the two stars since they have equal masses. Let R be the orbit radius for each star, as sketched in Figure 13.71. The two stars are separated by a distance 2R, so Figure 13.71 (b) E XECUTE : so And (c) S ET U P : Apply implies and to the system of the two stars. Separate to infinity E XECUTE : Thus the energy required is E VALUATE : The closer the stars are and the greater their mass, the larger their orbital speed, the shorter their orbital period and the greater the energy required to separate them.