Homework 24

Homework 24 - Solution for Long-answer Homework 24...

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Unformatted text preview: Solution for Long-answer Homework 24 Gravitation Problems Solution to Long-answer Homework Problem 24.1() Problem: We often use U = mg y to approximate the change in gravitational potential energy of an object close to the surface of the earth, with g e = 9 . 8 m s 2 . Similarly, we can approximate the change in gravitational potential energy of an object close to the surface of the sun as U s = mg s y where m is the mass of the object, g s is the gravitational acceleration at the surface of the sun, and y is the height above the surface. (a)Calculate g s at the surface of the sun. (b)Will the gravitational potential approximation given above for the sun be accurate over a larger or smaller range of values of y than that for the earth? Justify your answer (do the math). Solution to Part (a) The gravitational force between two objects is given by F g =- Gm 1 m 2 r 2 1 , 2 (the negative sign means the force is attractive) and by Newton II, we know that F = ma So taking m 1 = M s and m 2 = m , and a = g s at the surface of the sun ( r 1 , 2 = R s ), we have mg s = GM s m R 2 s g s = GM s R 2 s = parenleftBig 6 . 673 10 11 Nm 2 kg 2 parenrightBig ( 2 . 10 30 kg ) (7 . 10 8 m) 2 = 272 . 4 m s 2 Grading Key: Part (a) 2 Points Solution to Part (b) Gravitational potential is given by U ( vector r ) =- GMm r So the amount that gravitational potential changes over a (radial) distance r is U = U f- U i =- GMm parenleftbigg 1 r f- 1 r i parenrightbigg = GMm parenleftbigg 1 r i- 1 r f parenrightbigg We are working this out for a distance r above the surface of the sun, so if we take r i = R s and r f = R s + r , we have U = GMm parenleftbigg 1 R s- 1 R s + r parenrightbigg 1 Now we have to get a little creative (or notice that this derivation is in the book). Taking the second term in parentheses, we can divide both the numerator and denominator by R s , giving 1 R s + r = 1 R s 1 + r R s = 1 R s parenleftbigg 1 + r R s parenrightbigg 1 Now the binomial theorem tells us that for small x , we have (1 + x ) n 1 + nx So as long as r is very small relative to R s , we have 1 R s + r = 1 R s parenleftbigg 1 + r R s parenrightbigg 1 1 R s parenleftbigg 1- r R s parenrightbigg And we can approximate U as U GMm bracketleftbigg 1 R s- 1 R s parenleftbigg 1- r R s parenrightbiggbracketrightbigg = GMm r R 2 s And so U m parenleftbigg GM R 2 s parenrightbigg ( r ) = mg s r All this is wonderful, but the question asked us whether the approximation...
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This note was uploaded on 05/04/2008 for the course PHYS 2054 taught by Professor Stewart during the Spring '08 term at Arkansas.

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Homework 24 - Solution for Long-answer Homework 24...

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