**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

**A**= ½ [**r****x**(r + D r )] =**r**^**x**DrNow, the

*areal velocity*at P is by definition:

lim Dt

_{ ®0}(D

**A**/D t)

Then : d

**A**/ dt = ½ r

**x**r

**'**= ½ (

**r**

^{2 }dq /dt )

Or dA/dt =

**h/ 2**(Since: **h **=** **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 (d^{2}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 qNote 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 (d

^{2}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/ dtThe 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 (d**

^{2}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 d**A**/ 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 abAnd 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)**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? (*p ab )__Hint__: Take the area of an ellipse to be:
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