Orbital Energies, Kepler's Laws and Other Relationships

Orbital Energies, Kepler's Laws and Other Relationships

Kepler's Laws

Kepler's Three Laws can be used to describe the motion of the Planets:

1. Planets move in orbits that are ellipses
2. The planets move such that the line between the Sun and the Planet sweeps out the same area in the same area in the same time no matter where in the orbit.
3. The square of the period of the orbit of a planet is proportional to the mean distance from the Sun cubed.

The above rules were deduced empirically from the motions of the planet in the early 17th century, before Newton deduced the law of gravity and his laws of motion. When Newton's laws are applied to the planets, Kepler's laws can be derived with certain refinements.

Description of Orbits

Movement of a planet or satellite in an orbit can be described with the above rules and some simple plane geometry. The parameters of the ellipse in terms of eccentricity will be used but it also helpful understand the simple method of drawing an ellipse.

Drawing an Ellipse Two pins, a length of string, a sheet of paper and a pencil are used to draw ellipses. The eccentricity of the ellipse is set by the spacing of the pins relative to the length of string stretched between the pins. The eccentricity = (distance between the pins)/(length of string between the pins) This basic property of the ellipse can be used to determine relationships between certain parameters of the elliptical orbit. The length of the string equals twice the semi-major axis and the distance between the pins is twice the distance, c = e a. When the pencil is at the semi-major axis, the right triangle formed by the axes and the string allows one to write: a2 = b2 + c2 So...b = a ( 1 - e2)1/2 You can specify the shape of the ellipse that you wish to draw by its eccentricity, e or by its semi-minor axis, b. The size is determined by the semi-major axis. The ellipse in the drawing has an eccentricity = 0.8 so the width of the ellipse (twice the semi-minor axis) is ( 1- 0.8x0.8)1/2 = 0.6 of the length.

Energy of an Orbit

The Total energy of an object in orbit is the sum of kinetic energy (KE) and gravitational potential energy (PE).
KE = 1/2 mv²
PE = - GMm/r
r = the distance of the orbiting body from the central object and
v = the velocity of the orbiting body
E = 1/2 mv² - GMm/r

The semi-major axis is directly related to the total energy of the orbit: E = - GM/2a

Semi-major Axis and Total Energy The relationship between these two can easily be derived for a circular orbit and also works for elliptical and hyperbolic orbits. As we see in the diagram: a = F/m = GM/r2 = v2/r. In the case of a circle e = 0 and r = a. So v2 = GM/a and thus: E = 1/2 m(GM/a) - m(GM)/a = m (GM)/(2a) E/m = GM/(2a)

Incidently, the concept of circular velocity is useful in describing elliptical orbits. v_c² = GM/a The energy also provides an ｅxpression for the velocity in orbit
E/m = GM/(2a) = 1/2 v² - GM/r
and hence v² = GM/a[2/r - 1/a]
but written in terms of the circular velocity
v² = v_c²[2/r - 1/a]

---

Keplers 2nd Law

We have already discussed Kepler's 1st Law without giving it its name. Kepler discovered first that planets move in elliptical orbits about the Sun. The 2nd Law of Kepler describes the relative velocity of the objects in their elliptical orbits. He discovered that the line from the Sun to the planet swept out equal areas in equal times. At first this does not seem very helpful but if we use a little geometery then we can use it quantitatively. The diagram on the right illustrates the law.

The area of the shaded segment from A to B equal the area from segment C to D. Any body in the orbit around the Sun (o) will travel from A to B in the same time that it travels from C to D. The rate of sweeping our area by the line between Sun and orbiting object is called the Areal Velocity, A . In one period, P, of the orbit the line sweeps out the area of the ellipse so we can calculate this velocity from
A = (area of ellipse)/(period of ellipse) = (p a b) / P
A = p(1 - e²)^1/2 a²/P

Look at the diagram again; as an orbiting object goes from a to b the area swept out is approximately the area of the triangle o-a-b. That area is equal to the isoceles triangle o-a'-b' . The area of the later triangle can be calculate easily; that area is one half the base (length a'-c-b') times the height (length o-c).

The height of the triangle is just the radius, r, of the orbit at point c and the base of the triangle is the velocity perpendicular, v_, to the radius line at that same point times the time of transit from a to b. So the rate of sweeping out area in the triangle at c is:
A = v_r/2
There are only two points in the orbit where the perpedicular velocity equals the orbit velocity and that is a perihelion and aphelion. As a result we can relate the speed in orbit at these two poins most easily.
r_p = a (1 - e) and r_a = a (1 + e)
So...
v_ar_a/2 = v_pr_p/2 = (p a b) / P
and
v_a = v_p( 1 + e)/(1 - e)
With a little more derivation (using Kepler's 3rd Law) we can show that
v_p = v_c[( 1 + e)/( 1 - e )]^1/2
and
v_a = v_c[( 1 - e)/( 1 + e )]^1/2

---

Kepler's 3rd Law - Relationship between Period and Semi-major Axis

This law was derived empirically by Kepler. He found that if the period of the planet was given in years and the semi-major axis was given in Astronomical Units (AU) then
P² = a³

It is easily derived for a circular orbit and the result applies to elliptical orbits when the radius of the circle is replaced by the semi-major axis of the ellipse. The period of an object in a circular orbit where r = a is
P = 2pa/v
and hence since v = (GM/a)^1/2
P = 2pa^3/2/(GM)^1/2
This relationship is in metric units. If we transform to AU and years then we get 2p/(GM)^1/2 = 1yr/ AU^3/2

Newton refined Kepler's 3rd law using center of mass motion. When this is considered then the mass, M is not just the mass of the central body (the Sun for the planets) but the sum of the masses of both the 'central' and 'orbiting' object. In the case of the solar system, Kepler was not too far off because the mass of the Sun is more than a thousand times the masses of all the planets and their mass add only a small amount. So the correct relationship is:
P = 2pa^3/2/(G(M+m))^1/2