Topic: Engineering Mechanics
Pair of differential equations that describes motion of planet about sun using first two laws of Kepler is given as:
- A. $2\dot{r}\dot{\theta}+r\dot{\theta}=0$ and $\dot{r}+r\dot{\theta}^2(r\frac{a}{b^2}-1)=0$
- B. $2\dot{r}\dot{\theta}+r\dot{\theta}=0$ and $\dot{r}+r\dot{\theta}^2(r\frac{a}{b^2}+1)=0$
- C. $2\dot{r}\dot{\theta}-r\dot{\theta}=0$ and $\dot{r}-r\dot{\theta}^2(r\frac{a}{b^2}-1)=0$
- D. $2\dot{r}\dot{\theta}-r\dot{\theta}=0$ and $\dot{r}-r\dot{\theta}^2(r\frac{a}{b^2}+1)=0$
Answer: A. $2\dot{r}\dot{\theta}+r\dot{\theta}=0$ and $\dot{r}+r\dot{\theta}^2(r\frac{a}{b^2}-1)=0$
Explanation: Note: The question paper has typographical errors denoting the second derivatives (using single dot instead of double dots). Kepler's second law (conservation of angular momentum) mathematically dictates $\frac{d}{dt}(r^2 \dot{\theta}) = 0$, leading to $2r\dot{r}\dot{\theta} + r^2\ddot{\theta} = 0$, dividing by $r$ provides $2\dot{r}\dot{\theta} + r\ddot{\theta} = 0$. For the radial equation resulting from gravitational inverse-square law, $\ddot{r} - r\dot{\theta}^2 = -\frac{\mu}{r^2}$. Utilizing properties of an ellipse $l = b^2/a$ and $\mu = h^2/l$, substituting into the radial equation matches the form $\ddot{r} + r\dot{\theta}^2(r\frac{a}{b^2}-1)=0$.