Explanation Given: The equation of the ellipse is r = p ( 1 + e cos θ ) Where r is the radial distance e is the eccentricity θ is the angle a is the semimajor axis r a p + r p e r = 2 a Where r a p is the aphelion r p e r is the perhelion E = 1 2 m v 2 − G M m ‖ r ‖ = C Where v = ‖ r ' ‖ C is a constant Results from previous solutions r p e r = a ( 1 − e ) r a p = a ( 1 + e ) And ( 1 − e ) v p e r = ( 1 + e ) v a p Calculation: (a) It is given that E = 1 2 m v 2 − G M m ‖ r ‖ = C The energy is constant. So the energy at the perihelion and the aphelion must be equal to the constant E p e r = E a p = C So the energies at perihelion and the aphelion are E p e r = 1 2 m v p e r 2 − G M m r p e r = C E a p = 1 2 m v a p 2 − G M m r a p = C Equating the energies E p e r = 1 2 m v p e r 2 − G M m r p e r = E a p = 1 2 m v a p 2 − G M m r a p 1 2 v p e r 2 − G M r p e r = 1 2 v a p 2 − G M r a p 1 2 v p e r 2 − 1 2 v a p 2 = G M r p e r − G M r a p v p e r 2 − v a p 2 = 2 G M ( 1 r p e r − 1 r a p ) (b) Since r p e r = a ( 1 − e ) r a p = a ( 1 + e ) 1 r p e r − 1 r a p = 1 a ( 1 − e ) −

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Chapter 14.6, Problem 19E

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Perhilion velocity of an ellipse

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Chapter 14 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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Perhilion velocity of an ellipse

Solution Summary: The author explains that the equation of the ellipse is r = p ( 1 + e cos ).

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Perhilion velocity of an ellipse

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Chapter 14 Solutions