Problem Set 8

# Problem Set 8 - 7.13 Intersecting Orbits The two masses...

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Unformatted text preview: 7.13) Intersecting Orbits The two masses will orbit about their common center of mass, with the mass m always twice as far from the center of mass as the mass 2 m .The minimal eccentricity for the two orbits to intersect will have the closest approach for the small mass equal the furthest approach for the larger mass.-3-2.5-2-1.5-1-0.5 0.5 1 1.5-2.5-2-1.5-1-0.5 0.5 1 1.5 2 2.5 m 2m M Figure 1: Orbits for mass ratio 1 / 2. Wherever the masses are in their orbits, the center of mass lies in the same position on the line connecting them. Note that M represents the center of mass, and not an actual particle. The dashed orbit helps to give an idea of how far an eccentricity of 1 / 3 is from a perfect circle. Therefore we have 2 r min = r max Using equation 7.25, we see r ∝ 1 1 + epsilon1 cos θ → r max (1- epsilon1 ) = r min (1 + epsilon1 ) Using our relation above, we see 2 r min (1- epsilon1 ) = r min (1 + epsilon1 ) → 1 = 3 epsilon1 So we see epsilon1 = 1 / 3. 7.18) Circle to Parabola Going from a circle to a parabola requires increasing the energy from its value at the bottom of the well in the effective potential up to zero. For an orbit at radius r , the potential energy is- GMm r and the velocity can be found from GMm r 2 = m v 2 r → v = radicalbigg GM r so the kinetic energy is 1 2 mv 2 = GMm 2 r 1 and the total energy is- GMm 2 r So the increase in energy must be GMm 2 r and therefore the kinetic energy must double. This means the speed must increase by a factor ofand therefore the kinetic energy must double....
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## This note was uploaded on 01/22/2012 for the course PHYS 3201 taught by Professor Deheer during the Spring '10 term at Georgia Tech.

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Problem Set 8 - 7.13 Intersecting Orbits The two masses...

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