Applying Advanced Theoretical Mechanics To Circular And Planetary Motion (Part 2)

Application to Planetary Motion:

We begin with the polar coordinates of a point mass in an orbit of radius r under gravitational attraction to the Sun:

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

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

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.

 

t   =   dL/ dt

The diagram below  shows how torque is derived:

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 2^nd 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 2^nd 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:

q =  q ₀    +  ò ^t  ₀      h dt/  r ² (t) 

Problems:

1)  Prove the relations below:

i) dA/dt =  h/ 2

ii) P  =   2 p  ab h ^-1

And show:   P  =   2 p Ö (a ³ ) /  Öm

Where:   m  =  G (m1 + m2)

2) Because of the Earth’s rotation a plumb bob does not hang precisely in the direction of Earth’s gravitational attraction, i.e. in the direction of its weigh mg.  Given the radius of the Earth = 6.4  x 10 ⁶ m  and your latitude is 45 degrees, find the angle by which the bob deviates from the direction of g. (Hint: Sketch a diagram with forces, angle shown to assist in solution.)

3) 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  )