434 CHAPTER 13 Gravitation

13.70 ... If a satellite is in a sufficiently low orbit, it will encounterair drag from the earth’s atmosphere. Since air drag does negativework (the force of air drag is directed opposite the motion), themechanical energy will decrease. According to Eq. (13.13), if Edecreases (becomes more negative), the radius r of the orbit willdecrease. If air drag is relatively small, the satellite can be consid-ered to be in a circular orbit of continually decreasing radius. (a) According to Eq. (13.10), if the radius of a satellite’s circular orbitdecreases, the satellite’s orbital speed increases. How can you rec-oncile this with the statement that the mechanical energy decreases?(Hint: Is air drag the only force that does work on the satellite as theorbital radius decreases?) (b) Due to air drag, the radius of a satellite’scircular orbit decreases from r to where the positive quantity

is much less than r. The mass of the satellite is m. Show that the increase in orbital speed is that thechange in kinetic energy is that thechange in gravitational potential energy is

and that the amount of work done by the force ofair drag is Interpret these results in light ofyour comments in part (a). (c) A satellite with mass 3000 kg is initiallyin a circular orbit 300 km above the earth’s surface. Due to air drag,the satellite’s altitude decreases to 250 km. Calculate the initial orbitalspeed; the increase in orbital speed; the initial mechanical energy; thechange in kinetic energy; the change in gravitational potential energy;the change in mechanical energy; and the work done by the force ofair drag. (d) Eventually a satellite will descend to a low enough alti-tude in the atmosphere that the satellite burns up and the debris falls tothe earth. What becomes of the initial mechanical energy?13.71 . Binary Star—Equal Masses. Two identical stars withmass M orbit around their center of mass. Each orbit is circular andhas radius R, so that the two stars are always on opposite sides ofthe circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit.(c) How much energy would be required to separate the two starsto infinity?13.72 .. CP Binary Star—Different Masses. Two stars, withmasses and are in circular orbits around their center ofmass. The star with mass has an orbit of radius the star withmass has an orbit of radius (a) Show that the ratio of theorbital radii of the two stars equals the reciprocal of the ratio oftheir masses—that is, (b) Explain why the twostars have the same orbital period, and show that the period Tis given by (c) The twostars in a certain binary star system move in circular orbits. Thefirst star, Alpha, has an orbital speed of The second star,Beta, has an orbital speed of The orbital period is 137 d.What are the masses of each of the two stars? (d) One of the bestcandidates for a black hole is found in the binary system calledA0620-0090. The two objects in the binary system are an orangestar, V616 Monocerotis, and a compact object believed to be a blackhole (see Fig. 13.27). The orbital period of A0620-0090 is 7.75hours, the mass of V616 Monocerotis is estimated to be 0.67 timesthe mass of the sun, and the mass of the black hole is estimated tobe 3.8 times the mass of the sun. Assuming that the orbits are cir-cular, find the radius of each object’s orbit and the orbital speed ofeach object. Compare these answers to the orbital radius andorbital speed of the earth in its orbit around the sun.13.73 ... Comets travel around the sun in elliptical orbits withlarge eccentricities. If a comet has speed when at adistance of from the center of the sun, what is itsspeed when at a distance of 5.0 * 1010 m?

2.5 * 1011 m2.0 * 104 m>s

12.0 km>s.36.0 km>s.

T = 2p1R1 + R223>2>2G1M1 + M22.R1>R2 = M2>M1.

R2.M2

R1;M1

M2,M1

W = -1GmE m>2r 22 ¢r.-1GmE m>r 22 ¢r;

¢U = -2 ¢K =¢K = +1GmE m>2r 22 ¢r;¢v = +1¢r>22 2GmE>r 3;

¢rr - ¢r,

v

13.74 .. CP An astronaut is standing at the north pole of a newlydiscovered, spherically symmetric planet of radius R. In his handshe holds a container full of a liquid with mass m and volume V. Atthe surface of the liquid, the pressure is at a depth d below thesurface, the pressure has a greater value p. From this information,determine the mass of the planet.13.75 .. CALC The earth does not have a uniform density; it is mostdense at its center and least dense at its surface. An approximation of its density is where and

Use for the radiusof the earth approximated as a sphere. (a) Geological evidence indi-cates that the densities are and at theearth’s center and surface, respectively. What values does the linearapproximation model give for the densities at these two locations?(b) Imagine dividing the earth into concentric, spherical shells.Each shell has radius r, thickness dr, volume andmass By integrating from to showthat the mass of the earth in this model is (c) Show that the given values of A and B give the correct mass ofthe earth to within 0.4%. (d) We saw in Section 13.6 that a uniformspherical shell gives no contribution to g inside it. Show that

inside the earth in this model. (e) Verifythat the expression of part (d) gives at the center of the earthand at the surface. (f) Show that in this model gdoes not decrease uniformly with depth but rather has a maximumof at 13.76 .. CP CALC In Example 13.10 (Section 13.6) we saw thatinside a planet of uniform density (not a realistic assumption for theearth) the acceleration due to gravity increases uniformly with dis-tance from the center of the planet. That is, where is the acceleration due to gravity at the surface, r is the distancefrom the center of the planet, and R is the radius of the planet. Theinterior of the planet can be treated approximately as an incom-pressible fluid of density (a) Replace the height y in Eq. (12.4)with the radial coordinate r and integrate to find the pressure insidea uniform planet as a function of r. Let the pressure at the surface bezero. (This means ignoring the pressure of the planet’s atmosphere.) (b) Using this model, calculate the pressure at the center of the earth.(Use a value of equal to the average density of the earth, calcu-lated from the mass and radius given in Appendix F.) (c) Geologistsestimate the pressure at the center of the earth to be approximately

Does this agree with your calculation for the pressureat What might account for any differences?13.77 ... CP Consider a spacecraft in an elliptical orbit aroundthe earth. At the low point, or perigee, of its orbit, it is 400 kmabove the earth’s surface; at the high point, or apogee, it is 4000 kmabove the earth’s surface. (a) What is the period of the spacecraft’sorbit? (b) Using conservation of angular momentum, find the ratioof the spacecraft’s speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and thespeed at apogee. (d) It is necessary to have the spacecraft escapefrom the earth completely. If the spacecraft’s rockets are fired atperigee, by how much would the speed have to be increased toachieve this? What if the rockets were fired at apogee? Whichpoint in the orbit is more efficient to use?13.78 . The planet Uranus has a radius of 25,560 km and a surfaceacceleration due to gravity of at its poles. Its moonMiranda (discovered by Kuiper in 1948) is in a circular orbit aboutUranus at an altitude of 104,000 km above the planet’s surface.Miranda has a mass of and a radius of 235 km. (a) Calculate the mass of Uranus from the given data. (b) Calculate

6.6 * 1019 kg

11.1 m>s2

r = 0?4 * 1011 Pa.

r

r.

gsg1r2 = gsr>R,

r = 2A>3B = 5640 km.4pGA2>9B = 10.01 m>s2

g = 9.85 m>s2g = 0

g1r2 = 43pGr1A - 3

4Br2

M = 43pR31A - 3

4BR2.r = R,r = 0dm = r1r2dV.

dV = 4pr 2 dr,

2400 kg>m313,100 kg>m3

R = 6.37 * 106 mB = 1.50 * 10-3 kg>m4.A = 12,700 kg>m3r1r2 = A - Br,

p0;