Question Details

A planet moves in an elliptical orbit around one of the foci. The ratio of maximum velocity max v and minimum velocity min v in terms of eccentricity e of the ellipse is given by

Options

A

1-e/1+e

B

e-1/e+1

C

1+e/1-e

D

e/e-1

Correct Answer :

1+e/1-e

Solution :

The correct option is "1+e/1-e" (which represents 1+e1-e).

Let's derive this relationship step-by-step using the laws of planetary motion.

Step 1: Understanding Kepler's First Law and Elliptical Orbits
According to Kepler's First Law, planets move in elliptical orbits around the Sun, with the Sun located at one of the two foci of the ellipse.
An ellipse is characterized by its semi-major axis a and its eccentricity e (where 0<e<1).
The closest distance of the planet from the focus (perihelion) is given by:
rmin=a(1-e)
The farthest distance of the planet from the focus (aphelion) is given by:
rmax=a(1+e)

Step 2: Conservation of Angular Momentum
Since the gravitational force acting on the planet is a central force, the net torque acting on it is zero. Therefore, the angular momentum L of the planet remains constant throughout its motion.
The angular momentum is given by L=mvrsinθ, where m is the mass of the planet, v is its velocity, r is its distance from the focus, and θ is the angle between the position vector and the velocity vector.
At the closest point (perihelion) and farthest point (aphelion), the velocity vector is perpendicular to the position vector, meaning θ=90° and sinθ=1.
Thus, the conservation of angular momentum at these two extreme points gives:
mvmaxrmin=mvminrmax

Step 3: Calculating the Ratio of Velocities
From the conservation of angular momentum, we can express the ratio of maximum velocity to minimum velocity as:
vmaxvmin=rmaxrmin
Substitute the values of rmax and rmin into the equation:
vmaxvmin=a(1+e)a(1-e)
Canceling the common term a from the numerator and denominator, we get:
vmaxvmin=1+e1-e

Thus, the ratio of maximum velocity to minimum velocity is 1+e1-e.

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