Analytical Mech Homework Solutions 73

Analytical Mech Homework Solutions 73 - But the total...

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But the total orbital energy is 2 k E a =− So 2 2 k Ea a δ = Since planetary orbits are nearly circular ~ k V a and ~ 2 k T a Thus, a ET a and Ta = We obtain 2 av = 6.20 (a) 0 1 τ = VV d t () k Vr r From equation 6.5.4, lr 2 θ = ± 2 d dt r l = or 2 rd dt l = 2 00 kr Vdt d l τπ ∫∫ From equation 6.5.18a … 2 1 1c o s a r ε = + 2 2 1 o s ka d Vdt l + From equation 6.6.4 … 2 2 2 1 a l π 2 2 0 1 21 c o kd V a εθ s + 2 2 0 2 o s 1 d = + , 2 1 < k V a (b) This problem is an example of the virial theorem which, for a bounded, periodic system, relates the time average of the quantity
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