# Module 3 -- Mechanical Energy of Orbits

### Learning Goals

• Write the equation of mechanical energy conservation for a Keplerian orbit.

## Conservation of Mechanical Energy in Keplerian Orbits

The only interaction involved in a Keplerian orbit is the internal and conservative gravitational interaction between the two bodies. Therefore, the conditions for conservation of mechanical energy are met.

## Form of the Mechanical Energy

Since the mass M is fixed at the axis of rotation, it has no kinetic energy. Thus, the mechanical energy of the system can be expressed:

$E = \frac{1}{2}mv^{2} - \frac{GMm}{r}$

where v is the speed of m.

## Total Mechanical Energy of Orbit

A simple expression for the total mechanical energy of the orbit can be obtained by evaluating the mechanical energy at the apoapsis and periapsis. At these extrema, the mechanical energy and angular momentum have a very simple relationship, which can be seen by multiplying the mechanical energy by r2:

$E r^{2}_{\rm ex} = \frac{1}{2}mv^{2}_{\rm ex}r^{2}_{\rm ex} - GMmr_{\rm ex} = \frac{L^{2}}{2m} - GMmr_{\rm ex}$

We showed in the previous module, however, that angular momentum is a constant of the motion. We can therefore solve the mechanical energy equation for the angular momentum at periapsis and apoapsis and set these equal to obtain:

$E r^{2}_{a} + GMmr_{a} = E r^{2}_{p} + GMmr_{p}$

Solving for E gives:

$E = \frac{GMm (r_{p} - r_{a})}{r^{2}_{a} - r^{2}_{p}} = -\frac{GMm}{r_{a}+r_{p}}$

Using the fact that the apoapsis distance plus the periapsis distance equals the full major axis distance, we can rewrite this expression to give an expression for the energy as a function of the semi-major axis distance a:

$E = -\frac{GMm}{2a}$

Note that the total energy of the bound orbit is negative.

## Comparison with Circular Case

A special case of the Keplerian Orbit Model is a perfectly circular orbit. Because the dynamics of the circular case are very simple, it is enlightening to examine the energy of a circular orbit.

For a circular orbit, the velocity can be determined using the Uniform Circular Motion model. The motion will necessarily be uniform, since the centripetal force provided by gravity will be constant if the radius of the motion is constant. Thus:

$F= ma = \frac{mv^{2}_{\rm circ}}{r} = \frac{GMm}{r^{2}}$

Rearranging the above expression and multiplying by 1/2 gives the kinetic energy of a circular orbit:

$\frac{1}{2} mv^{2}_{\rm circ} = \frac{GMm}{2r}$

Of course, since the radius is constant, the potential energy of the motion is simply:

$U_{\rm circ} = -\frac{GMm}{r}$

Thus, for a circular orbit, the kinetic energy is 1/2 the size of the potential energy. Adding this kinetic energy to the potential energy, remembering that the potential energy is negative, gives:

$E = + \frac{GMm}{2r} - \frac{GMm}{r} = - \frac{GMm}{2r}$

which is consistent with the more general expression derived above.

## Periapsis and Apoapsis

In the elliptical case, the kinetic energy will not be exactly half the size of the potential energy. You can see this because the total energy of an elliptical orbit with semi-major axis a is the same as that of a circular orbit of radius a, but the elliptial orbit will give a potential energy that varies throughout the motion. At the periapsis, the potential energy is larger in size than it would be in a circle of radius a, but at apoapsis the potential energy will be smaller in size than it would be in a circle. To see what this means for the ratio of kinetic energy to potential energy, consider the following example.

## Exploring Orbital Motion with PhET Interactive Simulations

Suppose that an object is in orbit with its velocity instantaneously perpendicular to the position vector. It is possible to use the result of the previous section to determine whether the object is at periapsis or apoapsis. The simulation below will allow you to experiment with this. At the beginning of the simulation the earth is set up to orbit the sun in its true orbit. Since that orbit is very nearly circular, the velocity is basically perpendicular to the position vector from the sun to the earth and it gives a kinetic energy that is 1/2 the potential energy. The simulation allows you to adjust the mass of the sun and the mass of the earth. Predict what will happen if you adjust these masses, and test your predictions.

Simulation courtesy PhET Interactive Simulations