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Homework 24

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

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Solution for Long-answer Homework 24 Gravitation Problems Solution to Long-answer Homework Problem 24.1() Problem: We often use Δ U = mg Δ 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 Δ U s = mg s Δ y where m is the mass of the object, g s is the gravitational acceleration at the surface of the sun, and Δ 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 Δ 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 . 0 × 10 30 kg ) (7 . 0 × 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 Δ r is Δ 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 Δ r above the surface of the sun, so if we take r i = R s and r f = R s r , we have Δ U = GMm parenleftbigg 1 R s - 1 R s + Δ r parenrightbigg 1

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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 + Δ r = 1 R s 1 + Δ r R s = 1 R s parenleftbigg 1 + Δ 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 Δ r is very small relative to R s , we have 1 R s + Δ r = 1 R s parenleftbigg 1 + Δ r R s parenrightbigg 1 1 R s parenleftbigg 1 - Δ r R s parenrightbigg And we can approximate Δ U as Δ U GMm bracketleftbigg 1 R s - 1 R s parenleftbigg 1 - Δ r R s parenrightbiggbracketrightbigg = GMm Δ r R 2 s And so Δ U m parenleftbigg GM R 2 s parenrightbigg r ) = mg s Δ r All this is wonderful, but the question asked us whether the approximation Δ U = mg s Δ r
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Homework 24 - Solution for Long-answer Homework 24...

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