Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.59 A.U. and its greatest distance being 35 A.U. (1 A.U. = Earth-Sun distance). If the comet's speed at closest approach is 62 km/s, what is its speed when it is farthest from the Sun? You may neglect any change in the comet's mass and assume that its angular momentum about the Sun is conserved.
-
Conservation of angular momentum:
mR²ω = constant
At perihelion and aphelion, ω = V/R
Assuming constant mass and the comet at aphelion and perihelion:
(RV)a = (RV)p
35Va = 0.59*62
Va ≈ 1.05 km/s
The balance of centrifugal and gravitational forces is only valid for circular orbits. The orbital velocity in elliptical orbits is continuously changing. The unbalance in those forces produces that change in orbital velocity.
mR²ω = constant
At perihelion and aphelion, ω = V/R
Assuming constant mass and the comet at aphelion and perihelion:
(RV)a = (RV)p
35Va = 0.59*62
Va ≈ 1.05 km/s
The balance of centrifugal and gravitational forces is only valid for circular orbits. The orbital velocity in elliptical orbits is continuously changing. The unbalance in those forces produces that change in orbital velocity.
-
Hello Jessica, we have to rely on the fact that the gravitational force of attraction between comet and sun is balanced by centrifugal force. Moreover, greater the distance slower the speed.
So GMm/r^2 = mV^2/r
As G, M remain constant and m get cancelled, V^2 * r = constant.
Hence Vf / Vn = ./ (rn/rf)
So Vf = Vn * ./ rn/rf
Vf = 62*./ 0.59/35 = 8.05 km/s
So GMm/r^2 = mV^2/r
As G, M remain constant and m get cancelled, V^2 * r = constant.
Hence Vf / Vn = ./ (rn/rf)
So Vf = Vn * ./ rn/rf
Vf = 62*./ 0.59/35 = 8.05 km/s