Lecture 8 -Proof of Conservation of Momentum and II Law
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Transcript of Lecture 8 -Proof of Conservation of Momentum and II Law
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Proof of Kepler’s II lawDerivation for conservation of
momentum and Energy
Lecture - 7
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Conservation of momentum
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Angular momentum of body m2 relative to m1 is the
moment of m2’s relative linear momentum
--- velocity of m2 relative to m1.
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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
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Equation of motion is Using this, Second term
h - specific relative angular momentum is constant
First term
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and
Hence, path of m2 around m1 lies in a single plane.
Cross product
= constant.
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We can rewrite the equation
That is
Angular momentum depends only on the transverse (perpendicular) component of the relative velocityNOT the radial component
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Proof of Kepler’s II law
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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.
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Derivation of Energy Law
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Derivation of Energy Law
Relative linear momentum per unit mass is just the relative velocity
Equation of motion
Taking DOT product with
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LHS is
RHS is
Let us use the relations
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This equation reduces to
LHS
RHS
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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
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= - 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