Halley\'s comet moves about the sun in an elliptical orbit,with its closest appr
ID: 1726160 • Letter: H
Question
Halley's comet moves about the sun in an elliptical orbit,with its closest approach to the sun being 0.652 AU and itsgreatest distance being 29.9 AU (1 AU= the earth-sundistance) if the comet's speed at closest approach is 44.2 km/s what isits speed when it is farthest from the sun? you may assume that its angular momentum about the sun isconserved answer in units of km/s. i don't get it, any help is greatly appreciated! thanks. Halley's comet moves about the sun in an elliptical orbit,with its closest approach to the sun being 0.652 AU and itsgreatest distance being 29.9 AU (1 AU= the earth-sundistance) if the comet's speed at closest approach is 44.2 km/s what isits speed when it is farthest from the sun? you may assume that its angular momentum about the sun isconserved answer in units of km/s. i don't get it, any help is greatly appreciated! thanks.Explanation / Answer
The hint is that angular momentum is conserved. Angularmomentum is defined as (r x p) or (r x vm); where r, p, and v arevectors. The magnitude of the cross product isr*v*m*sin. It just so happens that the angle betweenthe position vector and the velocity vector is 90 degrees atclosest approach (perihelion) and greatest distance(aphelion). That means we can use the product r*v*m, becausesine is 1 at 90 degrees. Since angular momentum is conserved,we can write: r1*v1*m =r2*v2*m. The m's cancel, so we have thesimple equation: (0.652 AU)*(44.2 km/s) = (29.9AU)*v2. Now solve for v2.
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