Proof of Kepler’s II lawDerivation for conservation of

momentum and Energy

Lecture - 7

Conservation of momentum

Angular momentum of body m2 relative to m1 is the

moment of m2’s relative linear momentum

--- velocity of m2 relative to m1.

Divide this equation by m2

Let h=H2/1 / m2, so that,

h - relative angular momentum of m2 / unit mass,

- specific relative angular momentum.

Units of h are km2 / s

Differentiating, h

Equation of motion is Using this, Second term

h - specific relative angular momentum is constant

First term

and

Hence, path of m2 around m1 lies in a single plane.

Cross product

= constant.

We can rewrite the equation

That is

Angular momentum depends only on the transverse (perpendicular) component of the relative velocityNOT the radial component

Proof of Kepler’s II law

Differential area dA swept out

by the relative position vector r

during time interval dt.

But angular momentum

So = Constant - Proves Kepler’s II law

Equal areas are swept out in equal times.

Derivation of Energy Law

Derivation of Energy Law

Relative linear momentum per unit mass is just the relative velocity

Equation of motion

Taking DOT product with

LHS is

RHS is

Let us use the relations

This equation reduces to

LHS

RHS

Let us consider periapsis pointh = r p v p

h 2 = r p 2 v p 2

h 2 μ ε = ------ ------ 2r2

p r pa (1 – e 2) = h 2 / μ, r p = a ( 1 – e)

Using the above, ε = - μ / 2a

vp2 = h2 / rp

2

= - μ / 2a

This proves Vis Viva equation – Conservation of energy

For circle and ellipse, ‘a’ is positive, ε is negative For parabola, ‘a’ is infinite, ε is zero For hyperbola, ‘a’ is negative, ε is positive

By definition of the above equation, at ‘r infinite, PE is zero