Physics 11 Fall 2012 Practice Problems 7 - Solutions

Transcription

1 Physics 11 Fall 2012 Practice Problems 7 - s 1. Why is the gravitational potential energy of two masses negative? Note that saying because that s what the equation gives is not an explanation! The gravitational potential energy represents the energy of the system, which you can think of as the amount of work that it took to put the system together. Because the masses attract each other gravitationally, then they naturally fall together. This means that the work was done for us, and so the system has negative energy. This is also the amount of energy that we d need to add to the system to break it apart. Since we d need to do work on the system to break it apart (ending up with zero total energy), we had to start with a negative energy. The same thing occurs with oppositely-charged particles like protons and electrons, which also have negative electrostatic potential energy. 1

2 2. One of the greatest discoveries in astronomy in the past decade is the detection of planets outside the solar system. Since 1996, more than 100 planets have been detected orbiting stars other than the Sun. While the planets themselves cannot be seen directly, telescopes can detect the small periodic motion of the star as the star and planet orbit around their common center of mass. (This is measured using the Doppler effect.) Both the period of this motion and the variation in speed of the star over the course of time can be determined observationally. The mass of the star is found from its observed luminance and from the theory of stellar structure. Iota Draconis is the eighth brightest star in the constellation Draco. Observations show that a planet, with an orbital period 1.50 y, is orbiting this star. The mass of Iota Draconis is 1.05M Sun. (a) Estimate the size (in AU) of the semimajor axis of this planet s orbit. (b) The radial speed of the star is observed to vary by 592 m/s. Use conservation of momentum to find the mass of the planet. Assume the orbit is circular, we are observing the orbit edge-on, and no other planets orbit Iota Draconis. Express the mass as a multiple of the mass of Jupiter. (a) We can determine the semimajor axis using Kepler s law, which says that T 2 = 4π 2 GM a3, where T is the period (in years), M is the mass of the star, and a is the size of the semimajor axis (in AU). So, solving for a gives a 3 = GMT 2 4π 2. Now, we can write M = 1.05M Sun, and T = 1.5T Earth, so a 3 = 1.05(1.5) 2 C = 1.05(1.5) 2, since a 3 = GMT 2 = 1 is just the orbital period of the Earth, one year! 4π 2 So, we just find a = (1.05(1.5 2 )) 1/3 = 1.33 AU. (b) The star and it s planet are orbiting about their common center of mass. We can treat this like an elastic collision problem in which the momentum is conserved. In this case the momentum of each body is equal mv = MV, where m is the mass of the planet with velocity v, and M is the mass of the star with velocity V. If the velocity is found to change by a certain amount in the elastic collision, then the velocity of the star is half that change by conservation of momentum (it starts off heading one way, then it heads the other after the collision). Solving for the mass of the planet is m = V M. The velocity of the planet is just v = 2πa/T, so v m = V T 2πa M. Now, plugging in the different values, and recalling that the mass of Jupiter is m J = kg, gives m = ( ) ( ) = kg = 12.5m 2π( J. ) 2

3 3. Some people think that the shuttle astronauts are weightless because they are beyond the pull of Earth s gravity. In fact, this is completely untrue. (a) What is the magnitude of the gravitational field in the vicinity of a shuttle orbit? A shuttle orbit is about 400 km above the ground. (b) Given the answer in Part (a), explain why shuttle astronauts suffer from adverse biological effects such as muscle atrophy even though the are not actually weightless. (a) The shuttle orbits at a distance of r = R E +h from the center of the Earth, where R E is the radius of the Earth, and h is the height above the surface. If the height is h = 400 km, then r = = 6800 km, or meters. At this point, the gravitational field has a magnitude g = GM E r 2 = ( ) 2 = 8.6 m/s 2, which is still almost 90% of the acceleration at the surface of the Earth. (b) Remember that the weight that we feel is due to the normal force. The astronauts in orbit are in constant free-fall, and don t feel their weight. So, without the compensating normal force to fight against the muscles begin to weaken, not needing to do as much anymore. 3

4 4. Calculate the mass of Earth from the period of the moon, T = 27.3 d; its mean orbital radius r m = m; and the known value of G. The moon is held to the Earth by the gravitational force, F G = G mmoonm Earth, where r 2 r is the orbital radius. Now, because the orbit is (pretty much) circular, then the net force on the moon is the centripetal force, F cent = mmoonv2. But, for an object moving r in a circle, then v = 2πr/T, and so F cent = m moon r ( 2πr T 2 ) 2 = 4π2 T 2 m moonr. Because the orbit is stable this force is equal to the gravitational force. Setting the two forces equal gives 4π 2 T m moonr = G m moonm Earth. 2 r 2 Solving for the mass of the Earth gives M Earth = 4π2 T 2 G r3. The period is T = = seconds, and so M Earth = 4π2 T 2 G r3 = 4π 2 ( ) 2 ( ) ( ) 3 = kg. 4

5 5. The Principle of Equivalence states that the free-fall acceleration of any object in a gravitational field is independent of the mass of the object. This can be deduced from the law of universal gravitation, but how well does it hold up experimentally? The Roll-Krotkov-Dicke experiment performed in the 1960s indicates that the free-fall acceleration is independent of mass to at least 1 part in Suppose two objects are simultaneously released from rest in a uniform gravitational field. Also, suppose one of the object falls with a constant acceleration of exactly 9.81 m/s 2, while the other falls with a constant acceleration that is greater than 9.81 m/s 2 by one part in How far will the first object have fallen when the second object has fallen 1.00 mm farther than the first object has? Note that this estimate provides only an upper bound on the difference in the accelerations; most physicists believe that there is no difference in the accelerations. The first object has an acceleration g = 9.81 m/s 2, while the other has a different acceleration, g + g = 9.81 ( ) m/s 2, where g = m/s 2. Now, if released from rest, the first object will fall a distance d 1 = 1 2 gt2, while the second object will have fallen a distance d 2 = 1 2 (g + g)t2 = 1 2 gt gt2. Now, the difference in distance that the two objects will have fallen is d = d 2 d 1 = 1 2 gt2. Now, solving d 1 = 1 2 gt2 for the time of fall gives t = 2d 1 /g. Plugging this in for d gives d = 1 ( ) 2 g 2d1 = g g g d 1. Now, solving for the distance, recalling that g = g, we find d 1 = g g d = 1012 d. So, if the first object has traveled an extra distance d = 10 3 m, then the distance that it s fallen will have been d 1 = d = = 10 9 m. So, the two objects will have to fall a distance of a billion meters to get out of step by 1 millimeter! 5

6 6. (a) If we take the potential energy of a 100 kg object and Earth to be zero when the two are separated by an infinite distance, what is the potential energy when the object is at the surface of Earth? (b) Find the potential energy of the same object at a height above Earth s surface equal to Earth s radius. (c) Find the escape speed for a body projected from this height. (a) The gravitational potential energy of two objects of masses m and M, separated by a distance r is P E grav = G mm r, which sets the potential energy to zero when the objects are infinitely far apart (r ). So, if the object is on the surface of the Earth, then r = R E, and M = M E, and P E g = G mm E R E. Now, we can rewrite this recalling that for Earth, g = GM E. Thus, RE 2 P E g = G mm E R E = m GM E R RE 2 E = mgr E. Now, if the radius of the Earth is 6400 km, or meters, then P E g = mgr E = (100)(9.8)( ) = J. (b) The potential energy at a distance of r = R E + R E = 2R E will just be P E g = G mm E 2R E = 1 2 mgr E = J. (c) To determine the escape velocity from this distance we just recall the usual method setting KE + P E = 0, so 1 2 mv2 = 1 2 mgr E, giving v esc = gr E = (9.8)( ) = 8 km/s, which is smaller than the escape velocity from the surface of the earth (v esc 11 km/s), as we should expect. 6

7 7. A point particle of mass m is on the x axis at x = L and an identical point particle is on the y axis at y = L. (a) What is the direction of the gravitational field at the origin? (b) What is the magnitude of this field? (a) The mass m at x = L creates a gravitational field g x = Gm L2 î at the origin, while the mass at y = L creates a field g y = Gm L2 ĵ at the origin. The net gravitational field at the origin is just the sum of the two fields, g = g x + g y, or g = G m ) (î + ĵ. L 2 (b) The magnitude of the field is g = g = g 2 x + g 2 y = G m L = 2Gm L 2. 7

8 8. A straight, smooth tunnel is dug through a uniform spherical planet of mass density ρ 0. The tunnel passes though the center of the planet and is perpendicular to the planet s axis of rotation, which is fixed in space. The planet rotates with a constant angular speed ω, so objects in the tunnel have no apparent weight. Find the required angular speed of the planet ω. The object falls due to the gravitational force, F G = G mm, where M is the planetary r 2 mass beneath the object (recall that only the mass between an object and the center of the planet makes any difference, gravitationally). Let s figure out how much mass this is. Because the density is constant, then the total mass inside a radius r is M in = ρ 0 V in = ρ 0 4π 3 r3. So, the net gravitational force on the object is F G = G mm in = 4πGmρ 0 r. r 2 3 At some radius, the gravitational force is balanced by the centripetal force, F cent = mv2, r causing it to seem to be weightless. Since v = rω, we can write F cent = mrω 2. Setting the two forces equal gives 4πGmρ 0 r = mrω 2. 3 Notice that the mass and radius of the object cancels out. Solving for the angular speed gives 4πGρ0 ω =. 3 8

9 9. Black holes are objects whose gravitational field is so strong that not even light can escape. One way of thinking about this is to consider a spherical object whose density is so large that the escape speed at its surface is greater than the speed of light, c. If a star s radius is smaller than a value called the Schwarzschild radius R S, then the star will be a black hole, that is, light originating from its surface cannot escape. (a) For a nonrotating black hole, the Schwarzschild radius depends only upon the mass of the black hole. Show that it is related to that mass M by R S = (2GM) /c 2. (b) Calculate the value of the Schwarzschild radius for a black hole whose mass is ten solar masses. (a) The escape velocity of a spherical mass M with radius R is given by 2GM v esc =. r If we want light to be unable to escape from the surface, then the escape velocity has to be greater than, or equal to, the speed of light. Thus, setting the escape velocity v esc = c and solving for the radius gives R S = 2GM c 2. (b) The mass of the sun, M sun = kg. So, we can determine the Schwarzschild radius for this black hole, R S = 2GM c 2 = ( ) 2 = 29.6 km. So, the mass of ten suns is squashed into a ball about the size of a large city. 9

10 10. It is believed that there is a supermassive black hole at the center of our galaxy. One datum that leads to this conclusion is the important recent observation of stellar motion in the vicinity of the galactic center. If one such star moves in an elliptical orbit with a period of 15.2 years and has semimajor axis of 5.5 light days (the distance light travels in 5.5 days), what is the mass around which the star moves in its Keplerian orbit? Kepler s law says that the square of the period, T, of an orbit is proportional to the cube of it s semimajor axis, a, so T 2 = 4π2 GM a3. So, all we need to do is to solve for the mass, M = 4π2 GT 2 a3. Now, we can just plug in numbers. One year has about π 10 7 seconds, while one day has = seconds. So, one light day is = meters. So, M = 4π2 GT 2 a3 = 4π (15.2 π 10 7 ) 2 ( ) 3 = kg. Our sun has a mass of about kg, and so this star has a mass of nearly four million suns! This is clearly a black hole. 10