Physics 18 Spring 2011 Homework 9 - Solutions Wednesday March 16, 2011
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1 Physics 18 Sping 2011 Homewok 9 - s Wednesday Mach 16, 2011 Make sue you name is on you homewok, and please box you final answe. Because we will be giving patial cedit, be sue to attempt all the poblems, even if you don t finish them. The homewok is due at the beginning of class on Monday, Mach 30th. Because the solutions will be posted immediately afte class, no late homewoks can be accepted! You ae welcome to ask questions duing the discussion session o duing office hous. 1. Why is the gavitational potential enegy of two masses negative? Note that saying because that s what the equation gives is not an explanation! The gavitational potential enegy epesents the enegy of the system, which you can think of as the amount of wok that it took to put the system togethe. Because the masses attact each othe gavitationally, then they natually fall togethe. This means that the wok was done fo us, and so the system has negative enegy. This is also the amount of enegy that we d need to add to the system to beak it apat. Since we d need to do wok on the system to beak it apat (ending up with zeo total enegy), we had to stat with a negative enegy. The same thing occus with oppositely-chaged paticles like potons and electons, which also have negative electostatic potential enegy. 1
2 2. Some people think that the shuttle astonauts ae weightless because they ae beyond the pull of Eath s gavity. In fact, this is completely untue. (a) What is the magnitude of the gavitational field in the vicinity of a shuttle obit? A shuttle obit is about 400 km above the gound. (b) Given the answe in Pat (a), explain why shuttle astonauts suffe fom advese biological effects such as muscle atophy even though the ae not actually weightless. (a) The shuttle obits at a distance of = R E +h fom the cente of the Eath, whee R E is the adius of the Eath, and h is the height above the suface. If the height is h = 400 km, then = = 6800 km, o metes. At this point, the gavitational field has a magnitude g = GM E 2 = ( ) 2 = 8.6 m/s 2, which is still almost 90% of the acceleation at the suface of the Eath. (b) Remembe that the weight that we feel is due to the nomal foce. The astonauts in obit ae in constant fee-fall, and don t feel thei weight. So, without the compensating nomal foce to fight against the muscles begin to weaken, not needing to do as much anymoe. 2
3 3. Calculate the mass of Eath fom the peiod of the moon, T = 27.3 d; its mean obital adius m = m; and the known value of G. The moon is held to the Eath by the gavitational foce, F G = G mmoonm Eath, whee 2 is the obital adius. Now, because the obit is (petty much) cicula, then the net foce on the moon is the centipetal foce, F cent = mmoonv2. But, fo an object moving in a cicle, then v = 2π/T, and so F cent = m moon ( 2π T 2 ) 2 = 4π2 T 2 m moon. Because the obit is stable this foce is equal to the gavitational foce. Setting the two foces equal gives 4π 2 T m moon = G m moonm Eath. 2 2 Solving fo the mass of the Eath gives M Eath = 4π2 T 2 G 3. The peiod is T = = seconds, and so M Eath = 4π2 T 2 G 3 = 4π 2 ( ) 2 ( ) ( ) 3 = kg. 3
4 4. (a) If we take the potential enegy of a 100 kg object and Eath to be zeo when the two ae sepaated by an infinite distance, what is the potential enegy when the object is at the suface of Eath? (b) Find the potential enegy of the same object at a height above Eath s suface equal to Eath s adius. (c) Find the escape speed fo a body pojected fom this height. (a) The gavitational potential enegy of two objects of masses m and M, sepaated by a distance is P E gav = G mm, which sets the potential enegy to zeo when the objects ae infinitely fa apat ( ). So, if the object is on the suface of the Eath, then = R E, and M = M E, and R E. Now, we can ewite this ecalling that fo Eath, g = GM E. Thus, RE 2 R E = m GM E R RE 2 E = mgr E. Now, if the adius of the Eath is 6400 km, o metes, then P E g = mgr E = (100)(9.8)( ) = J. (b) The potential enegy at a distance of = R E + R E = 2R E will just be 2R E = 1 2 mgr E = J. (c) To detemine the escape velocity fom this distance we just ecall 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 smalle than the escape velocity fom the suface of the eath (v esc 11 km/s), as we should expect. 4
5 5. Black holes ae objects whose gavitational field is so stong that not even light can escape. One way of thinking about this is to conside a spheical object whose density is so lage that the escape speed at its suface is geate than the speed of light, c. If a sta s adius is smalle than a value called the Schwazschild adius R S, then the sta will be a black hole, that is, light oiginating fom its suface cannot escape. (a) Fo a nonotating black hole, the Schwazschild adius depends only upon the mass of the black hole. Show that it is elated to that mass M by R S = (2GM) /c 2. (b) Calculate the value of the Schwazschild adius fo a black hole whose mass is ten sola masses. (a) The escape velocity of a spheical mass M with adius R is given by 2GM v esc =. If we want light to be unable to escape fom the suface, then the escape velocity has to be geate than, o equal to, the speed of light. Thus, setting the escape velocity v esc = c and solving fo the adius gives R S = 2GM c 2. (b) The mass of the sun, M sun = kg. So, we can detemine the Schwazschild adius fo 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 lage city. 5
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