Energy of particle when under a central force

[Saptarshi Sarkar]

Homework Statement

A particle moves in a circular orbit about the origin under action of a central force $\vec F = -k\hat r/r^3$. If the potential energy is zero at infinity, what is the total energy of the particle?

Relevant Equations

$E_{total} = E_{kinetic} + E_{potential}$

I calculated the potential energy of the particle as follows :

But I am not sure how to calculate the kinetic energy. I know that if it was a satellite orbiting a Earth, I could use $\frac {GMm} {r^2} = \frac {mv^2} {r}$ to calculate the velocity v and they I could calculate kinetic energy as $E_{kinetic} = \frac {mv^2} 2$.
But how do I calculate the velocity for the given force?

 

[Orodruin]

I know that if it was a satellite orbiting a Earth, I could use $\frac {GMm} {r^2} = \frac {mv^2} {r}$ to calculate the velocity v and they I could calculate kinetic energy as $E_{kinetic} = \frac {mv^2} 2$.
But how do I calculate the velocity for the given force?

Where did the left-hand side come from in the case of the gravitational force?

 

[Saptarshi Sarkar]

Where did the left-hand side come from in the case of the gravitational force?

The term is the gravitational force acting between Earth and the satellite.

I put $\frac k {r^3} = \frac {mv^2} {r}$ and found $E_{kinetic} = \frac k {2r^2}$. Is this correct? If this is correct, that would mean the total energy of the particle is 0. Is this possible?

 

[vela]

If this is correct, that would mean the total energy of the particle is 0. Is this possible?

Sure. Why not?

 

[Saptarshi Sarkar]

Sure. Why not?

Won't that mean that the particle is not bound anymore?

 

[vela]

It's right on the edge. Give it a little more energy, and it will go off to infinity. Take a little away, and it'll spiral in.

 

[gneill]

Hmm. A circular orbit should have a negative total mechanical energy, as should any elliptical orbit. The on-the-edge orbit is that of a parabola where the total mechanical energy is zero. Hyperbolic orbits have positive values for total mechanical energy.

I think there must be a sign issue with your PE. If you look at some common expressions for the specific mechanical energy for an orbit:

$\xi = \frac{v^2}{2} - \frac{\mu}{r}$
$\xi = - \frac{\mu}{2a}$
$\xi = \mu \frac{\left(e^2 - 1 \right)}{2p}$

where $a$ is the semi major axis, $e$ is the eccentricity, and $p$ is the semi latus rectum.

You should be able to tell from these expressions that bound orbits will have negative mechanical energy.

 

[vela]

Hmm. A circular orbit should have a negative total mechanical energy, as should any elliptical orbit. The on-the-edge orbit is that of a parabola where the total mechanical energy is zero. Hyperbolic orbits have positive values for total mechanical energy.

For an inverse-square force. The problem the OP is dealing with is an inverse-cube force.

 

[gneill]

For an inverse-square force. The problem the OP is dealing with is an inverse-cube force.

I thought the cube was an error, since in the original problem statement there was a vector r in the numerator. The "extra" r in the denominator accounts for forming a unit vector: $\frac{\vec{r}}{|r|}$

 

[vela]

The original problem statement has a unit vector in the numerator.

 

[gneill]

The original problem statement has a unit vector in the numerator.

Fair enough. I stand corrected.

 

[kuruman]

But how do I calculate the velocity for the given force?

For future reference, with such problems you can get the answer directly by considering the virial theorem.