University Physics AI No. 4 The Gravitational Force and the Gravitational Field Class Number Name IChoose the Correct Answer 1. The magnitude of the force of gravity between two identical objects is given by F0. If the mass of each object is doubled but the distance between them is halved, then the new force of gravity between the objects will be （ A ） (A) 16 F0 (B) 4 F0 (C) F0 (D) F0/2 Solution: The magnitude of the gravitational force is 2rGMmFgrav=, according to the problem, we get 02222 1616)2/(4FrGm rGmFgrav= 2. A spherical symmetric nonrotating body has a density that varies appreciably with the radial distance from the center. At the center of the body the acceleration of free fall is ( C ) (A) Definitely larger than zero. (B) Possibly larger than zero. (C) Definitely equal to zero. Solution: According to the shell theorem 1, at the center of the body, the force acted on the body is zero, so the acceleration of the body is zero. 3. The acceleration due to gravity in a hole dug into a nonuniform spherically symmetric body ( C ) (A) will increase as you go deeper, reaching a maximum at the center. (B) will increase as you go deeper, but eventually reach a maximum, and then decrease until you reach the center. (C) can increase or decrease as you go deeper. (D) must decrease as you go deeper. Solution: For a nonuniform spherically symmetric body, the force of gravity depends on the distance from the center, which is related to how the density of the body changed with respect to the distance from the center, for instance: according to the Gausss law of gravity =VGMGsgisd44drrwhen2rA=, the acceleration grwill increase as you go deeper, when Ar= the acceleration grwill decrease as you go deeper. II. Filling the Blanks 1. Two masses m1 and m2 exert gravitational forces of equal magnitude on each other. Show that if the total mass M = m1+m2 is fixed, the magnitude of the mutual gravitational force on each of the two masses is a maximum when m1 = m2 ( Fill ). Solution: The magnitude of the gravitational force is 211 221)( rmMGm rmGmFgrav=, We get 20)2(0dd 1211MmrmMG mF= Then 21mm =. 2. A mass m is inside a uniform spherical shell of mass M and a mass M is outside the shell as shown in Figure 1. The magnitude of the total gravitational force on m is 22)(dRsMmGF+=. Solution: 222onononontotal)(0dRsGMm rGMmFFFFFFFtotalmMtotalmMmMmM+=+=rrrQrrr3. A certain neutron star has a radius of 10.0 km and a mass of 4.00 1030 kg, about twice the mass of the Sun. The magnitude of the acceleration of an 80.0 kg student foolish enough to be 100 km from the center of the neutron star is 2.671010 m/s2 . The ratio of the magnitude of this acceleration and g is 2. 72109 . If the student is in a circular orbit of radius 100 km about the neutron star, the orbital period is 3.8410-5 s . Solution: (a) The magnitude of the force is marGMmF=2So 210 2530112m/s1067. 2)101 (1041067. 6=rGMa M s m M R Fig.1 d (b) The ratio of the magnitude of this acceleration and g is 910 1072. 281. 91067. 2=ga(c) Since the acceleration is222 242 TrrTra= The orbital period is ) s (1012. 11041067. 6)101 (14. 34442 3011352322 =GMr arT4. The magnitude of the total gravitational field at the point P in Figure 2 is 2.3710-3 m/s2 ,the magnitude of the acceleration experienced by a 4.00 kg salt lick at point P is 2.3710-3 m/s2 , the magnitude of the total gravitational force on the salt lick if it is placed at P is 9.48 N . Solution: (a) The total gravitational field at the point P is )sin(cos 22jirGMirGMggg PEEPMM EMtotal+=+=rrrjijiijii)1012. 2()1007. 1 ()92. 038. 0(103 . 21092. 1) 1016. 41084. 3 1016. 4106 . 1()1016. 4(1098. 51067. 6 )106 . 1 (1036. 71067. 633348888282411282211+=+=+ +=The magnitude of the total gravitational field at the point P is 23m/s1037. 2=totalg (b) The acceleration has no relation with anything at some point. So the magnitude of the acceleration experienced by a 4.00 kg salt lick at point P is also equal to (a). 23m/s1037. 2=totalga (c)The magnitude of the total gravitational force on the salt lick if it is placed at P is N48. 937. 24= maF III. Give the Solutions of the Following Problems Moon Earth 7.361022kg5.981024kg1.60108m 4.16108m 90 P Fig.2 ij1. Several planets (the gas giants Jupiter, Saturn, Uranus, and Neptune) possess nearly circular surrounding rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous ring of mass M and radius R. (a) Find an expression for the gravitational force exerted by the ring on a particle of mass m located a distance x from the center of the ring a