HW10 – gravitation: gravitation: in which we solve problems pertaining to gravitation, namely, Ch. 13: P10, P13, P16, P21, P26, P39, P45, P48, P54, P66. •• chapter 13, problem 10 (||12): Two dimensions. In Fig. 13-34, three point particles A B and C are fixed in place in an xy-plane. Particle A has mass m A , particle B has mass 2.00m A , and particle C has mass 3.00m A. A fourth particle D, with mass 4.00m A , is to be placed near the other three particles. In terms of distance d, at what (a) xcoordinate and (b) y-coordinate should particle D be placed so that the net gravitational force on
particle A from particles B,C, and D is zero?
The net gravitational force upon A due to B, C, and D is,
F
A
F AD
FAB FAC
FAB FAC FAD 0 G
m m GmA C 2 x B2 rAB r AC ˆ
This is the direction of
F AD .
m A mB r AB
y G ˆ
2
m A mC r AC 2
3.00mA 2.00mA y GmA x 2 d2 (1.5d ) ˆ
ˆ
( x) FAD
y
ˆ
tan
1 2.00 3.00/1.52
quad-3
ˆ
GmA2 d2
x cos56 2.4 y sin56
(1.1)
ˆ
ˆ
HOWEVER, it is NOT the LOCATION of particle D! Very important con cept for
work with Coulomb’s Law in Physics II. Basically, (1.1) tells us the angle at which this particle D must be placed, but not the distance from the origin! We must have
F AD
G m r m , which will tell us the distance from A
D
2
the origin of particle D. Then, we must have, F AD
Gm A mD r2
GmA 2 d2
This means r = (1.29d
2.4 x cos 56 y sin 56 ˆ
ˆ
GmA 2 d2
2.4
GmA 4.00mA r2
4.00
2.4
r
d 1.29 d ; (1.2)
– 56.3º); 56.3º); this is the location-vector of particle-D. Resolving this into compone nts,
(a) In ( x, y) y) notation, the x the x coordinate coordinate is x is x = = 0.716d 0.716d . (b) Similarly, the y the y coordinate coordinate is y is y = = 1.07d 1.07d . •• chapter 13, problem 13 (||16): Figure 13-37 shows a spherical hollow of diameter R inside a lead sphere of radius R = 4.00 cm; the surface of the hollow passes through the center of the sphere and “touches” the right side right side of the sphere. The mass of the sphere before hollowing was M = 2.95 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 0.431 kg that lies at a distance d = 9.00 cm from the center of the lead sphere, on the
straight line connecting the centers of the spheres and of the hollow?
2
If the lead sphere were not hollowed the magnitude of the force it exerts on m would be F 1 = GMm/d . Part of this force is due to material that is removed. Su perposition: We calculate the force exerted on m by a sphere that just fills the cavity, at the position of the cavity, and subtract it from the force of the solid sphere. Compute the mass of the hollowed-out sphere: The cavity has a radius r = R/2. The material that fills it has the 3 3 same density (mass to volume ratio) as the solid sphere, that is, M c/r = M / R , where M c is the mass that fills the cavity. The common factor 4/3 has been canceled. Thus, the mass is,
1 2 M 2
V2 V1
M1
4 3
R23
4 3
R13
M1
R2
3
R13
M1
( 12 R)3 R3
M
M 8
;
The center of the cavity is d r = d R/2 from m, so the force it exerts on m is G 18 Mm GM1m GMm GM 2 m 2 ; F 2 F1 ; 2 1 r1m2 d r 2m 2 d R 2
(1.3)
(1.4)
The force of the hollowed sphere on m is,
1 GMm 1 1 = 1 F = F1 F2 = GMm 2 2 d 8 d 1 R 2 8 1 1 R 2 d 2 2 d m (6.67 1011 N kg )(2.95 kg)(0.431 kg) 1 1 8.31 109 N ; 2 2 2 8 1 410 m (9.00 10 m) 2910 m 2
(1.5)
2
2
2
••• chapter 13, problem 16: In Fig. 13-39, a particle of mass m1 = 0.67 kg is a distance d = 23 cm from one end of a uniform rod with length L = 3.0 m and mass M = 5.0 kg. What is the magnitude of the gravitational force F on the particle from the rod?
Since the rod is an extended object, we cannot apply Equation 13-1 directly to find the force. Instead, we consider a small differential element of the rod, of mass dm of thickness dr at a distance r from m1 . The gravitational force between dm and m1 , using dm ( M / L)dr , is, Gm Gm M Gm1 m Gm1 2 dm 2 1 dr 2 1 dr; , 2 r r L r r
dF d
where we have substituted M / L since mass is uniformly distributed. The direction of d F is to the right (see figure). The total force can be found by integrating over the entire length of the rod:
F dF
Gm1M L
L d
dr
d
r2
Gm1 M 1 Gm1 M 1 . L L d d d ( L d )
Substituting the values given in the problem statement, we obtain
F
Gm1M
d ( L d )
(6.67 1011 m3 /kg s2 )(0.67 kg)(5.0 kg) (0.23 m)(3.0 m 0.23 m)
3.0 1010 N .
Chapter 13, problem 21: Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, what must be its minimum mass so that material on its sur-face remains in place during the rapid rotation? From Eq. 13-14, we see the extreme case is when “ g” becomes 2 zero, and plugging in Eq. 13-15 leads to 0 g ar . Putting in symbolic expressions for g = 9.81 m/s and the 2
radial acceleration ar = v /R yields what we want. 0=
GM R 2
R M = 2
R3 2 G
.
24
24
Thus, with R = 20000 m and = 2 rad/s, we find M = 4.7 10 kg 5 10 kg. Chapter 13, problem 26: Consider a pulsar, a collapsed star of extremely high density, with a mass M equal to
that of the Sun ( 1.98 1030 kg ), a radius R of only 12 km, and a rotational period T of 0.041 s. By what percentage does the free-fall acceleration g differ from the gravitational acceleration a g at the equator of this spherical star? We are basically seeing how the star’s orbital-rotation-speed, which elicits a radial acceleration a r opposite that of gravity, subtracts from the natural pull of gravity. The difference between free-fall acceleration g and the gravitational acceleration a g at the equator of the star is (see Equation 13.14) due to a radial force-balance, a g g 2 R where
2 T
2 0.041s
153rad/s
is the angular speed of the star. The gravitational acceleration at the equator is a g
GM R
2
(6.67 1011 m3 /kg s2 )(1.981030 kg) (1.2 10 m) 4
2
9.17 1011 m/s2 .
Therefore, the percentage difference is
a g g a g
2 R ag
(153rad/s)2 (1.2104 m) 9.17 10 m/s 11
2
3.06 104 0.031%.
•• chapter 13, problem 39 (||40): (a) What is the escape speed on a spherical asteroid whose radius is 500 km 2 and whose gravitational acceleration at the surface is 3.0 m/s ? (b) How far from the surface will a particle go if it leaves the asteroid’s surface with a radial speed of 1000 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?
(a) the expression for escape velocity is for there to be zero kinetic energy at infinite distance,
r 2
K E U
F(r) dr F (R) dR
r
2GM
v E
r
r1
r r
GMm
r
R
2
dR 0
GMm r
GMm r
1
mvE 2 2
(1.6)
2ag r 2(3.0 sm ) (500 103 m) 1.7 103 ms ; 2
(b) revisiting the above expression (1.6), but this time integrating to h, the unknown height, h r
K E U
r
h
GMm R 2
r 2 GM v2 r
r r 1
1 GMm GMm GMm hr 2 dR h r r r h 1 2 mv r
r 1 v2
2a g r 1
500 103 m 2(3.0 m2 )(500103 m ) s
(1000 sm )2
(1.7)
2.5 10 m ; 5
1
(c) this time, we write as, h r
K E U
r
GMm R 2
a g r 2 1 1 1 3 m dR mv GMm ; (1.8) m ag r v 2ag r 1 h 1.4 10 2 r h 1 s r r h r 1
2
Chapter 13, problem 45: The Martian satellite Phobos travels in an approximately circular orbit of radius
r 9.4 106 m with a period of 7 h 39 min. Calculate the mass of Mars from this information. 2
2
3
The period T and orbit radius r are related by the law of periods: T = (4 /GM )r , where M is the mass of Mars. 4 The period is 7 h 39 min, which is 2.754 10 s. We solve for M : 4 2 r 3 4 2 (9.4 106 m)3 6.5 1023 kg. M 2 2 GT (6.67 1011 m3 / s 2 kg) 2.754 104 s Chapter 13, problem 48: The mean distance of Mars from the Sun is 1.52 times that of Earth from the Sun. From Kepler’s law of periods, calculate the number of years required for Mars to make one revolution around the Sun; compare your answer with the value given in Appendix C.
Kepler’s law of periods, expressed as a ratio, is 3
2
a M TM T M 3 (1.52) 1y a E T E
2
where we have substituted the mean-distance (from Sun) ratio for the semi-major axis ratio. This yields T M = 1.87 y. The value in Appendix C (1.88 y) is quite close, and the small apparent discrepancy is not significant, since a more precise value for the semi-major axis ratio is a M /a E = 1.523, which does lead to T M = 1.88 y using Kepler’s law. A question can be raised regarding the use of a ratio of mean distances for the ratio of semi-major axes, but this requires a more lengthy discussion of what is mean t by a ”mean distance” than is appropriate here.
Chapter 13, problem 54 (Hunting a black hole): Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed v = 270 km/s, orbital period T = 1.70 days, and approximate mass m1 = 6M s , where M s is the Sun’s mass, 1.98 1030 kg . Assume that the visible star and its
companion star, which is dark and unseen, are both in circular orbits (Fig. 13-46). What multiple of M s gives the approximate mass m2 of the dark star? The two stars are in circular orbits, not about each other, but about the two-star system’s center of mass (denoted as O), which lies along the line connecting the centers of the two stars. The gravitational force between the stars provides the centripetal force necessary to keep their orbits circular. Thus, for the visible, Newton’s second law gives
F
Gm1m2
m1v 2
r2
r 1
where r is the distance between the centers of the stars. To find the relation between r and r 1 , we locate the center of mass relative to m1 . Using Equation 9-1, we obtain r1
m1 (0) m2 r m1 m2
m2 r
r
m1 m2
m1 m2 m2
r 1 .
On the other hand, since the orbital speed of m1 is v 2 r1 / T , then r1 vT / 2 and the expression for r can be rewritten as r
m1 m2 vT m2
2
.
Substituting r and r 1 into the force equation, we obtain
F
4 2Gm1m23 (m1 m2 )2 v2T 2
2 m1v T
or m23 (m1 m2 )
2
v3T 2 G
(2.7 105 m/s)3 (1.70 days)(86400 s/day) 11
2 (6.67 10
m /kg s ) 3
2
6.90 1030 kg
3.467 M s , 30 where M s 1.99 10 kg is the mass of the sun. With m1 6 M s , we write m2 M s and solve the following
cubic equation for
: 3
(6 ) 2
3.467 0 .
The equation has one real solution: 9.3 , which implies m2 / M s 9 .
Chapter 13, problem 66 (Attacking a satellite [pew, pew]): One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500 km above Earth’s surface collides with a pellet having mass 4.0 g. (a ) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ra tio of this kinetic energy to the kinetic energy of a 4.0 g bullet from a modern army rifle with a muzzle speed of 950 m/s?
(a) The pellets will have the same speed v but opposite direction of motion, so the relative speed between the pellets and satellite is 2v. Replacing v with 2v in Eq. 13-38 is equivalent to multiplying it by a factor of 4. Thus, 3 2 24 11 GM E m 2(6.67 10 m /kg s ) 5.98 10 kg 0.0040 kg K rel 4 (6370 500) 103 m 2r 4.6 105 J.
(b) We set up the ratio of kinetic energies: K rel K bullet
4.6 105 J
1 2
0.0040kg 950 m/s
2
2.6 102.
