Question

write down the constants of motion for central conservative forces​

Answer 1

Answer:

Your answer is

Explanation:

Conservation of Energy for a Particle in a Central Force Field. Since central forces are conservative forces, we know that total energy must be conserved. Now we derive expressions for the total energy of a particle of mass m in a central force field.

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Answer 2

Answer:

tronomy) at University of Victoria

I consider the two-dimensional motion of a particle of mass m under the influence of a conservative central force F(r), which can be either attractive or repulsive, but depends only on the radial coordinate r. Recalling the formula \ddot{r} - r \dot{ \theta }^2 \)for acceleration in polar coordinates (the second term being the centripetal acceleration), we see that the equation of motion is

m \ddot{r} - m r \dot{ \theta }^2 = F(r). \tag{21.2.1}\label{eq:21.2.1}

This describes, in polar coordinates, two-dimensional motion in a plane. But since there are no transverse forces, the angular momentum m^2 \dot{ \theta }^2 is constant and equal to L, say. Thus we can write Equation \ref{eq:21.2.1} as

m \ddot{r} = F(r) + \frac{L^2}{mr^3}. \tag{21.2.2}\label{eq:21.2.2}

This has reduced it to a one-dimensional equation; that is, we are describing, relative to a co-rotating frame, how the distance of the particle from the centre of attraction (or repulsion) varies with time. In this co-rotating frame it is as if the particle were subject not only to the force F(r), but also to an additional force \frac{L^2}{mr^3} . In other words the total force on the particle (referred to the co-rotating frame) is

F'(r) = F(r) + \frac{L^2}{mr^3}. \tag{21.2.3}\label{eq:21.2.3}

Now F(r), being a conservative force, can be written as minus the derivative of a potential energy function, F = - \frac{dV}{dr}. Likewise, \frac{L^2}{m^3} is minus the derivative of \frac{L^2}{2mr^2} . Thus, in the co-rotating frame, the motion of the particle can be described as constrained by the potential energy function V', where

V' = V + \frac{L^2}{2mr^2}. \tag{21.2.4}\label{eq:21.2.4}

This is the equivalent potential energy. If we divide both sides by the mass m of the orbiting particle, this becomes

\Phi' = \Phi + \frac{h^2}{2r^2}. \tag{21.2.5}\label{eq:21.2.5}

Here h is the angular momentum per unit mass of the orbiting particle, \Phi is the potential in the inertial frame, and \Phi ' is the equivalent potential in the corotating frame