Question

A comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being about 0.620 AU and its greatest distance from the sun being 35.5 AU (1 AU = the Earth-Sun distance). If the comet's speed at closest approach is 54.0 km/s, what is its speed when it is farthest from the Sun? (The gravitational force exerted by the Sun on the comet is parallel to the moment arm, so exerts no torque. Therefore, angular momentum of the comet around the Sun is conserved.)

Answer 1

Given :-

Vclosed = 54 km/sec

Rclosed = 0.620 A.U

Rfar = 35.5 A.U

from coservation of angular momentum

Lclose = Lfar

(MVR)close = (MVR)far

Since the problem neglected any change in the comet's mass, we can focus on the velocity and radius

(VR)close = (VR)far

Vfar = Vclose(Rclose / Rfar)

Vfar = 54 km/sec*(0.620 A.U / 35.5 A.U)

Vfar = 0.943 km/sec