Question
Mon February 14, 2011 By: Arun Iyer

# Can you pls expain Keplers Laws and their derivation in a simple way ?

Mon February 14, 2011
Dear student,

o derive Kepler's first law, define:


where the constant



has the dimension of length. Then



and



Differentiation with respect to time is transformed into differentiation with respect to angle:



Differentiate



twice:




Substitute into the radial equation of motion



and get



Divide by the right hand side to get a simple non homogenous differntial equations for the orbit of the planet:



An obvious solution to this equation is the circular orbit





These solutions are



where  and  are arbitrary constants of integration. So the result is



Choosing the axis of the  coordinate system such that , and inserting , gives:




If  this is the equation of an ellipse and illustrates Kepler's first law.

Only the tangential acceleration equation is needed to derive Kepler's second law.

The magnitude of the specific angular momentum



is a constant of motion , even if both the distance , and the angular speed , and the tangential velocity , vary, because



where the expression in the last parentheses vanishes due to the tangential acceleration equation.

The area swept out from time t1 to time t2,



depends only on the duration t2 Ã¢ÂˆÂ’ t1. This is Kepler's second law.



which is Kepler's third law for the special case.

In the general case of elliptical orbits, the derivation is more complicated.

The area of the planetary orbit ellipse is



The areal speed of the radius vector sweeping the orbit area is



where



The period of the orbit is



satisfying



implying Kepler's third law


Hope this helps.
Team
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