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Unformatted text preview: Solution for Longanswer Homework 24 Gravitation Problems Solution to Longanswer Homework Problem 24.1(Calculating little g ) Problem: Mars has a mass of 6 . 421 × 10 23 kg , and radius 3 . 4 × 10 6 m . (a)Calculate the gravitational acceleration “g”, at the surface of Mars. (b)Will the gravitational potential approximation given above for Mars 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 Newton’s Law of Universal Gravitation, F g = Gm 1 m 2 R 2 1 , 2 Using Newton’s Second Law, we can solve for the acceleration, F g = m 2 a = Gm 1 m 2 R 2 1 , 2 a = Gm 1 R 2 1 , 2 We give this acceleration the special symbol “ g ”. Remember that R in this context is the distance between the centers of mass of the objects, so it is the radius of the planet, in this case. g mars = GM mars ( R M ) 2 = (6 . 673 × 10 11 Nmkg 2 )(6 . 421 × 10 23 kg) (3 . 4 × 10 6 m) 2 = 3 . 7 m s 2 Grading Key: Part (a) 5 Points 2 point(s) : Correctly recognizes ‘ g ’ as acceleration due to gravity at the surface of Mars. 2 point(s) : plugs in the numbers for Mars correcty (1 each radius and mass) 1 point(s) : correct answer with units 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 1 We are working this out for a distance Δ r above the surface of Mars, so if we take r i = R M and r f = R M +Δ r , we have Δ U = GMm parenleftbigg 1 R M 1 R M + Δ r parenrightbigg 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 M , giving 1 R M + Δ r = 1 R M 1 + Δ r R M = 1 R M parenleftbigg 1 + Δ r R M 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 M , we have 1 R M + Δ r = 1 R M parenleftbigg 1 + Δ r R M parenrightbigg 1 ≈ 1 R M parenleftbigg 1 Δ r R M parenrightbigg And we can approximate Δ U as Δ U ≈ GMm bracketleftbigg 1 R M 1 R M parenleftbigg 1 Δ r R M parenrightbiggbracketrightbigg = GMm Δ r R 2 M And so Δ U ≈ m parenleftbigg GM R 2 M parenrightbigg (Δ r ) = mg mars Δ r All this is wonderful, but the question asked us whether the approximation Δ U = mg mars Δ r would be accurate over a larger or smaller range of values than the similar expression for gravitational potential energy change near the surface of the Earth. The key to the approximation is the binomial expansion, which is accurate only when x is small. Since R M is smaller than R e , Δ r R M will remain small over a smaller range of values of Δ r than Δ r R e will, so the approximation for gravitational potential energy change for Mars is accurate...
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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.
 Spring '08
 Stewart
 Physics, Acceleration, Mass, Work

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