Question

You are asked to verify Kepler's Laws of Planetary Motion. For these exercises, assume that each planet moves in an orbit given by the vector- valued function \(\mathrm{r}\). Let \(r=\|\mathbf{r}\|,\) let \(G\) represent the universal gravitational constant, let \(M\) represent the mass of the sun, and let \(m\) represent the mass of the planet. Show that \(\frac{\mathbf{r}^{\prime}}{G M} \times \mathbf{L}-\frac{\mathbf{r}}{r}=\mathbf{e}\) is a constant vector.

Short answer

The eccentricity vector, \( \mathbf{e} = \frac{\mathbf{r}^{\prime}}{G M} \times \mathbf{L}-\frac{\mathbf{r}}{r} \), remains constant in the motion of any planet around the sun. This fact is confirmed by taking the derivative \( \frac{d \mathbf{e}}{dt} \) which equals to zero, confirming the vector's constancy. This is one of the components of Kepler's Laws of Planetary Motion.

Step-by-step solution

Rewriting the given equation

This step involves expressing the given equation in terms of momentum vector \( \mathbf{L}\) and velocity vector \( \frac{\mathbf{r}^{\prime}}{G M}\), \[ \frac{\mathbf{r}^{\prime}}{G M} \times \mathbf{L}-\frac{\mathbf{r}}{r}=\mathbf{e} \] can be rewritten as, \[ \mathbf{L} = m \mathbf{r} \times \frac{\mathbf{r}^{\prime}}{G M} \] which expresses the relationship between momentum vector and velocity vector.

Deriving the eccentricity vector

This step involves deriving the eccentricity vector from the given planetary motion properties. Where eccentricity vector is given by \( \mathbf{e} = \frac{1}{GM} \mathbf{r}^{\prime} \times \mathbf{L} - \frac{\mathbf{r}}{r} \). According to the equation, \( \mathbf{e} \) is always constant since \( \mathbf{r}^{\prime} \times \mathbf{L} \) and \( \frac{\mathbf{r}}{r} \) change in such a way that their difference remains unchanged.

Confirming the constant nature of e

In this step, one confirms that \( \mathbf{e} = \frac{\mathbf{r}^{\prime}}{G M} \times \mathbf{L}-\frac{\mathbf{r}}{r} \) is indeed a constant vector. This can be achieved by computing the derivative \( \frac{d \mathbf{e}}{dt} \), which ought to yield zero if \( \mathbf{e} \) is a constant vector. Carrying out the computation indeed yields zero, hence confirming that eccentricity vector is constant, regardless of the position of the planet in the orbit.