## Tuesday, May 18, 2021

### Looking Again At Kepler's 2nd Law

Analytical diagram to determine Kepler's 2nd law

Kepler's 2nd law of planetary motion, i.e.  a planet completes equal areas of its orbit in equal intervals of time, is perhaps the most important of the 3 laws.  In assessing it the diagram shown is instructive.  In the first instance we have for the change in area, D A :

D =    ½  [r   x (r +  r )] =   r^  x   Dr

Now, the areal velocity at P  is by definition:

lim Dt ®0  (D A /D t)

Then :  dA/ dt =  ½   r   x  r'  =     ½  (  2   dq  /dt )

Or   dA/dt =  h/ 2

(Since:  =  r 2   dq  /dt )

We can also use a "torque" model for radial motion,  in concert with the polar coordinates of a point mass, to further examine an orbit of radius r under gravitational attraction to the Sun,  viz:

a r   =   dr 2/ dt 2   -  r (dq  /dt ) 2

Similarly:

a q  =  r (d2 q  /dt 2)   + 2 dr/dt ((dq  /dt )

In polar coordinates we are also interested in the angular momentum, e.g.

L = mr 2   dq  /dt   =   r p q  =  r p sin q

Note that the angular momentum for a planet orbiting the Sun is constant! The force components in polar coordinates are:

F r    =  m[dr 2/ dt 2   -  r (dq  /dt ) 2 ]

F q    =  m  [r (d2 q  /dt 2)   + 2 dr/dt ((dq  /dt ) ]

If we take:  d/dt (r p q )  we get:

r  d p q /dt   =   r F q    =   dL/ dt

The rate of change of angular momentum which is also called the torque of the force F about the point 0, is.

t   =   dL/ dt

The change in angular momentum is then:

dL/ dt = r F q    =

m  [r (d2 q  /dt 2)   + 2 dr/dt ((dq /dt ) ]

=   d/ dt  m r 2 (dq  /dt)

Note the angular momentum per unit mass m is just:

h  =  L/ m  =  r 2   dq  /dt   =  r p q   /  m

This  quantity is exactly the h used in the Kepler 2nd law, and a constant of the motion.  In effect, with a bit of further working:

r 2   dq  /dt   =  2 dA/ dt  =   h  or dA/dt =  h/ 2

The preceding can be integrated to obtain:

A  =  ½  h t +   c

Then the area swept out  by the radius vector in time  t   -  t 0 ,  according to Kepler’s 2nd law would be:

A  =   ½  h (t   -  t 0)

Given a period of revolution P,   i.e. for elliptic motion about a fixed origin, when t = t + P, and P > 0, the area swept out is:

A  +  p ab  =    ½  h (t  + P   -  t 0)

b  =  a (1   -  e 2 ) 1/2

Subtracting the first equation above from the second, i.e.

½  h (t  + P   -  t 0)  -   A  +  p  ab

And solving for P, one arrives at:

P  =   2 p  ab h -1

An alternative approach uses:

dq =    h dt/  r 2

Then integrate to obtain:

qq   +    ò t  0      h dt/  r 2 (t)

Suggested Problem:

An asteroid moves in an elliptic orbit around the Sun. The lengths of the major and minor axes are 2a and 2b, respectively. If the asteroid’s velocity at the point of closest approach (where it crosses the major axis) is   v o   then how much time is needed for the object to make one complete orbit?  (Hint:  Take the area of an ellipse to be:   p ab  )