Kepler's 2nd law and potential energy relation

[karnten07]

Homework Statement

A particle moves under the influence of a central potential V(r). Show that Kepler's second law, that the radius vector from the force centre to the particle sweeps out area at a constant rate, is true whatever the form of V, as long as it is central. Derive the law in the form
.
dA/dT = 0.5 r^2 (theta)

Where (r,theta) are polar coordinates in the orbital plane with origin at the force centre. There is meant to be a dot above theta!

Homework Equations

The Attempt at a Solution

All i have at the moment (after trawling the net, and searching in books) is a statement and a derivation for the above equation given. But i don't know how to link it to the potential energy.

I get the equation given as equal to L/2m and the fact that it is a constant. The statement i have is:

It is true no matter what the forms of the force and potential energy functions are because it is entirely a result of the conservation of angular momentum.

Any thoughts on how to include the potential energy in a more quantitative way??

 

[anathema]

Dear forum post author,

Thank you for your question. I would be happy to help you understand how Kepler's second law can be derived using the central potential V(r).

First, let's start with the definition of angular momentum, L, which is given by L = mvr, where m is the mass of the particle, v is its velocity, and r is the distance from the force center. Since the potential is central, the force is always directed towards the force center, and therefore the particle's velocity is always perpendicular to the radius vector r.

Next, we can express the rate of change of angular momentum as dL/dt = mvdv/dt = mva, where a is the acceleration of the particle. Since the force is always directed towards the force center, the acceleration can be written as a = v^2/r. Substituting this into our equation, we get dL/dt = mv^2/r.

Now, let's consider the area swept out by the radius vector r in a small time interval dt. This area is given by dA = (1/2)r^2d(theta), where d(theta) is the change in angle swept out by the radius vector. To find the rate of change of this area, we divide both sides by dt, giving us dA/dt = (1/2)r^2d(theta)/dt. But d(theta)/dt is simply the angular velocity, omega, which is given by v/r. Therefore, we can rewrite our equation as dA/dt = (1/2)rv, which is equal to (1/2)L/m, as we showed earlier.

Since angular momentum is conserved for a central potential, we can say that (1/2)L/m is a constant, and therefore dA/dt is also a constant. This proves Kepler's second law, which states that the radius vector sweeps out area at a constant rate.

I hope this helps to clarify how the potential energy is linked to Kepler's second law. If you have any further questions, please don't hesitate to ask. Happy studying!

 

[ogb p]

Thank you for your question. Kepler's second law states that the radius vector from the force center to a particle in orbit sweeps out area at a constant rate, regardless of the shape of the potential energy function. This can be mathematically expressed as dA/dt = 0.5r^2(theta), where (r,theta) are polar coordinates in the orbital plane with the origin at the force center.

To understand the relationship between Kepler's second law and potential energy, we need to consider the conservation of energy in orbital motion. In a central force field, the total energy of a particle in orbit is given by the sum of its kinetic energy and potential energy:

E = 1/2mv^2 + V(r)

where m is the mass of the particle, v is its velocity, and r is the distance from the force center. Since the force is central, the angular momentum of the particle is conserved, given by L = mr^2(theta). This means that as the particle moves in its orbit, its speed and distance from the force center may change, but its angular momentum remains constant.

Now, let's consider a small change in the position of the particle, from r to r+dr. This change in position corresponds to a change in the angle theta, from theta to theta+dtheta. The area swept out by this change in position is given by dA = 1/2r^2dtheta. Using the conservation of angular momentum, we can rewrite this as dA = L/2m(r^2)dtheta.

Substituting this into the equation for conservation of energy, we get:

E = 1/2mv^2 + V(r) = 1/2m(r^2)(dtheta/dt)^2 + V(r)

Since the total energy E is constant, we can rearrange this equation to get:

dtheta/dt = (2m/E)^0.5 (r^2)^-0.5

Substituting this into our equation for dA, we get:

dA/dt = 1/2r^2(dtheta/dt) = (1/2)(2m/E)^0.5 (r^2)^0.5 = (1/2)r^2(theta)

This is the same equation we started with, showing that the rate of change of area swept out by the particle is constant, regardless of