orbit, its eccentricity, the perihelion distance, the comet's speed at that point, and
the ... 2. mV 2. To calculate the angular momentum, we may use either h = r2 ˙.
Worked Example The Orbit of a Comet A comet of mass m is travelling through space under the gravitational influence of Sun. Initially it is a long way away, travelling at speed V along a line which, if Sun exerted no attraction, would pass a distance b from the Sun. Find the shape of orbit, its eccentricity, the perihelion distance, the comet’s speed at that point, and angle through which the comet’s trajectory is deflected.
the the the the
This is a standard orbit question, and the usual formulae apply, in particular h2/(GM ) r= 1 + e cos θ and
E=
1 2m
h2 r˙ + 2 r 2
−
GM m r
G2 M 2 m 2 = (e − 1). 2h2 The quantity α ≡ GM/(V 2 b) will turn out to be important. The initial potential energy is zero, so the comet’s total energy is E = 21 mV 2 . To calculate the angular momentum, we may use either h = r2 θ˙ or the equivalent formula h = |r × r˙ |; in this case the latter is more useful, as it reduces to the speed of the comet multiplied by the perpendicular distance from the Sun to its velocity vector, i.e., h = V b. The eccentricity e satisfies e2 − 1 = i.e., e=
2h2 E V 4 b2 = , G2 M 2 m G2 M 2 √
1 + α−2 > 1.
The orbit is therefore hyperbolic (as is obvious from the fact that E > 0).
Perihelion r = rp occurs when θ = 0 (because then 1 + e cos θ is a maximum, so r is a minimum). Hence h2/(GM ) rp = 1+e V 2 b2/(GM ) √ = 1 + 1 + α−2 √ 1 + α−2 −1 1 − =α b 1 − (1 + α−2 ) √ = 1 + α2 − α b. At perihelion, the speed of the comet is just rθ˙ (since the other component of the velocity, r, ˙ is zero there). Hence vp =
h V =√ rp 1 + α2 − α √ = 1 + α2 + α V.
Finally, note from the diagram that the angle φ through which the comet is deflected is given by φ = 2θa − π where θa is the incoming polar angle (an asymptote of the hyperbola). We have r = ∞ at θ = θa , so that cos θa = −1/e. Hence cos 21 (φ + π) = −
1 e
or equivalently, 1 . e The expression for φ looks neatest using the following trick: sin( 12 φ) =
cot2 ( 12 φ) = cosec2 ( 12 φ) − 1 = e2 − 1 = α−2 , i.e., φ = 2 tan−1 α.
