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. Prove that \(\mathbf{r} \cdot \mathbf{r}^{\prime}=r \frac{d r}{d t}\)

Short answer

The proof of Kepler's law in this case means using mathematical formulation and vector calculus to validate the relationship \(\mathbf{r} \cdot \mathbf{r}^{\prime}=r \frac{d r}{d t}\). Understanding the vector mathematics and its derivative is vital in confirming this relationship.

Step-by-step solution

Understanding the parameters

The first key to this proof is to understand the vector-valued function given by \(\mathrm{r}\), that represents the position of the planet in its orbit. The planet’s position at any time can be represented by a radius vector. \(r\) is the magnitude of \(\mathbf{r}\), which is calculated with the Pythagorean theorem. \(G\) is the universal gravitational constant, \(M\) is the mass of the sun, and \(m\) represents the mass of the planet.

Deriving \(r\)

The magnitude of \(\mathbf{r}\) is \(r\), which is the distance between the planet and the sun. We know from calculus that the derivative of the magnitude of a vector is equal to the vector divided by its magnitude, all multiplied by the derivative of the vector. Therefore, \(\frac{d}{dt}(\mathbf{r}) = \mathbf{r} \cdot \frac{d}{dt}(\mathbf{r}) / r\). Let's replace \(\frac{d}{dt}(\mathbf{r})\) with \(\mathbf{r}^{\prime}\) to keep notation consistent. Therefore, we have \(\mathbf{r} \cdot \mathbf{r}^{\prime} = r \cdot \frac{d}{dt}(\mathbf{r})\).

Calculating \(\mathbf{r} \cdot \mathbf{r}^{\prime}\)

When we take the derivative of the magnitude of a vector, we get the vector divided by its magnitude multiplied by the derivative of the vector. Therefore, when we dot multiply the vector \(\mathbf{r}\) with its derivative \(\mathbf{r}^{\prime}\), we get \(r \cdot \frac{d}{dt}(\mathbf{r})\). The dot product gives the magnitude of a vector when multiplied with the cosine of the angle between the vectors. Since there is no angle between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) (they are in the same direction), the cosine value is 1 and we get \(r \cdot \frac{d}{dt}(\mathbf{r})\), which is the given equation \(\mathbf{r} \cdot \mathbf{r}^{\prime}=r \frac{d r}{d t}\). This verifies the given equation and that's the proof.