Question

Write theθLagrange equation for a particle moving in a plane ifV=V(r) (that

is, a central force). Use theθequation to show that:

(a) The angular momentum r×mvis constant.

(b) The vector r sweeps out equal areas in equal times (Kepler’s second law).

Short answer

Part a and b are verified.

Step-by-step solution

Definition of Kepler's Second law

Kepler's second law of planetary motion describes the speed of a planet traveling in an elliptical orbit around the Sun. It states that a line between the Sun and the planet sweeps equal areas at equal times.

Given Parameters

It is given that the Lagrange equation for a particle moving in a plane if

V=V(r)(Central force).

Show that the angular momentum r × mv is constant

Now Lagrangian for a particle moving in a plane is

$$\begin{gathered} L=\frac{1}{2}m{v}^2-V\left(r\right) \\ v=r\hat{r}+r\hat{\theta }......\left(1\right) \\ L=\frac{1}{2}m\left(r^2+r^2{\theta }^2\right)-V\left(r\right)......\left(2\right) \end{gathered}$$

Whereandare unit vectors in r and θ direction respectively.

Now applying the Euler-Lagrange equation to (2) in the θ direction yields as

$\begin{array}{rcl}\frac{\partial L}{\partial \theta }& =& m{r}^2\theta \\ \frac{\partial L}{\partial \theta }& =& 0......\left(3\right)\\ \frac{\partial L}{△\theta }& =& \frac{\partial L}{\partial \theta }\\ m{r}^2\theta & =& \mathrm{co}ns\tan t.....\left(4\right)\\ & & \end{array}$

Now to show that equation (3) is the z component of angular momentum,

$\begin{array}{rcl}{L}_z& =& m{\left(r\times v\right)}_z......\left(5\right)\\ r\times v& =& r\hat{r}\times \left(r\hat{r}+r\theta \hat{\theta }\right)\\ r\times v& =& rr\times \hat{r}+r^2\theta \hat{r}\times \hat{\theta }\\ \hat{r}\times \hat{r}& =& 0\\ \hat{r}\times \hat{\theta }& =& \hat{z}\\ {\left(r\times v\right)}_z& =& r^2\theta \\ L_z& =& m{r}^2\theta \end{array}$

By comparing equation (5) with (3) it shows that the angular momentum is constant.

Show that the vector r sweeps out equal areas in equal times (Kepler’s second law)

A figure below show that the differential area and displacement are related by

$$\begin{gathered} dA=\frac{1}{2}{\left(r\times dr\right)}_z \\ dA=\frac{1}{2}{\left(r\times v\right)}_{z}dt......\left(10\right) \\ A=\frac{1}{2}{\left(r\times v\right)}_z \end{gathered}$$

Angular momentum is constant therefore is constant. By transitive property

Hence, this is Kepler's second law.