Question

A body of mass m, which can be considered as a particle, is moving under the central force F(r) directed toward fixed point O. r is distance from a body to point O. Show that

Answer 1

\[\frac{ 1 }{ 2 }mv^2+\int\limits F(r)dr\] is constant during the motion.

Answer 2

mechanical energy is constant in these cases nd integration of f(r)dr gives P.E

Answer 3

how can I show that? It's not enough on exam just to write that comment

Answer 4

I wish I had the mathematical tools to give a through solid proof. All I can comment is that I don't think it's F(r)dr but it's
\[\int F(r)\:rd\theta\]

from hyperbolic flybys to this, your questions are hard

Answer 5

I have done something. I will write it and you can check it if it's not problem.

Answer 6

\[\vec F=-F(r)\ \vec e_r\]
\[V=-\int\limits F(r)dr\]
\[K=\frac{ 1 }{ 2 }mv^2\]
\[\frac{ 1 }{ 2 }mv^2+\int\limits F(r)dr=K-V\]
\[d \vec r=dr \vec e_r+dr \vec e_\theta\]
\[\vec F d \vec r=-F(r) \vec e_r (dr \vec e_r+dr \vec e_\theta)=-F(r)dr\]
I proved before that \[dK=-F(r)dr=-d \int F(r)dr\]
\[d(\frac{ 1 }{ 2 }mv^2+\int\limits F(r)dr)=d(K+\int\limits F(r)dr)=dK+d \int\limits F(r)dr=0\]

Answer 7

\[d \vec r=dr \vec e_r+r d\theta \vec e_\theta\]

Answer 8

Although I'm not sure why there is a minus sign, but after the correction, everything else looks fine.

Answer 9

minus is because \[\vec e_r\] is directed from O to the body and \[\vec F=\vec F(r)\] is directed from the body to the point O