In central force motion we change two body problem into single body. In the former case while we write the equation of force acting on one particle we write is as $m_1 r_1'' = F_{1_{ext}} + F_{12_{int}}$. Here $r_1$ is the position vector for particle $m_1$ from the origin, $r_2$ is the position vector for the second particle and $r$ is the distance between $m_1$ and $m_2$. Why is there no term of force in the form of $\frac{mv^2}{r}$ as we see in the centripetal force?
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$\begingroup$ Hi and welcome to the Physics SE! The equations become much easier to read, search and edit when mathjax is used. I've proposed an edit to your post this time, but you should use it yourself in future posts. $\endgroup$– StygCommented Jan 29, 2018 at 9:16
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$\begingroup$ Additionally, I would venture that your title does not reflect your question very well. Perhaps something that includes the term "centripetal force" or "centripetal acceleration" might suit better. I would propose an edit, but I'd rather not make substantive edits on someone else's post. $\endgroup$– StygCommented Jan 29, 2018 at 9:28
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It might be better to not think of $\frac{mv^2}{r}$ as the expression for centripetal force, and instead to think of $\frac{v^2}{r}$ as the acceleration of a particle moving in a circle of radius $r$ with constant speed $v$.
Thus, if it is known that a particle moves in a circular trajectory of given radius with given speed, it can be reasoned from Newton's second law that (the magnitude of) the total force acting on the particle must be $ma = m\frac{v^2}{r}$.
In the case of a particle in a potential of the form $V(r)$ (the single-body equivalent of the the two-body problem), we do not a priori know the particle trajectory, but we do know the force acting on the particle. We can then use Newton's second law to write the differential equation for $\vec{r}$ written in the question.
