Orbital Dynamics

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Orbital dynamics is the study of the motion of objects in the presence of gravitational forces of other bodies, due to gravity. All bodies provide a gravitational pull on other bodies surrounding it.

The motion of one body about another body due to gravity, is known as an Orbit. Orbits are described by a body of theory called "Orbital Dynamics". Kepler discovered that in Newtonian Physics, which ignore Einstein's theories of relativity, orbits are elliptical in shape, at least for the simple case of one body orbiting another body without any influence from any third body. Kepler's Laws are three equations which describe elliptical orbits, and still hold true today.

Newton's Law of Gravity

The force exerted by two bodies on each other is given by

<math>F = \frac{Gm_{1}m_{2}}{r^{2}}</math>

where G is the Universal Gravitational Constant whose value is 6.67300 × 10-11 m3 kg-1 s-2, <math>m_1</math> is the mass of the first body, <math>m_2</math> is the mass of the second body, and r is the distance between them.

The centripetal force of an object that is held in rotation around a fixed point is expressed as

F = (m2 * V2) / r

where V is the linear velocity of the object, which in our case is the satellite. Notice that the velocity (and later the period) are independent of the mass of the satellite. Combining these two equations together and solving for the velocity of the satellite gives an orbital velocity of a circular orbit as

<math>V = \sqrt{\frac{m_{1}G}{r}} = \sqrt{\frac{\mu G}{r}} </math>

where <math>\mu = m_{1}G </math> is referred to as a gravitational parameter, for which tables can be found for the Sun and each of the planets.

The circumference of a circular orbit is <math>C = 2 \pi r </math>. The period is then equal to the circumference divided by the velocity, or

<math>P = \frac{C}{V} = 2 \pi \sqrt{\frac{r^3}{\mu}} </math>

Keplerian Orbits

A Keplerian orbit is a simplified orbit with the following assumptions

  1. There are only two bodies exerting a gravitational force.
  2. The bodies are homogeneous and spherical.
  3. No other forces act on the system.

These assumptions lead to an elliptical orbit. In fact, both bodies will orbit around a point that is located along the line connecting the center of mass of the two bodies. The location on this line is the center of mass <math>m_2/(m_1+m_2)</math>. For a man-made satellite orbiting the Earth, the center of mass of the combined system is essentially at the center of mass of the earth. For the Moon orbiting the Earth, the center of mass of the combined system is displaced 2949 miles from the center of mass of the Earth. It is 1014 miles below the surface of the Earth.

Simple Celestial Mechanics for Simple Problems

In the case in which the tabulated data available is surface gravity and radius, as is the case for some amateurs dealing with moons and planets, then surface orbital velocity can be found as the square root of the product of the radius and the surface gravity.

Vos = (gs * rs)0.5

To find circular orbital velocity at some altitude, first find the gravity at that altitude. Then the orbital velocity at that altitude is the square root of the product of gravity at altitude the radius of the orbit.

ga = gs * (rs2 / ra2)

Voa = (ga * ra)0.5

For hohmann transfer up or down between circular orbits, the delta v (change in velocity) needed to pass between tangently contacting orbits can be calculated considering the radii of circular orbits and the semimajor axis (Smj) of the transfer orbit to determine the periods of the various orbits, along with the semiminor axis (Smn) of the transfer orbit to determine its area. If we have delta r as the difference in orbital radius between orbit at the surface and orbit at a chosen altitude then we have the Smj of the transfer orbit between surface and chosen altitude as

Smj = rs + (0.5 * delta r)

Smn = (Smj2 - (0.5 delta r)2)0.5

The area of the transfer orbit is At.

At = Pi * Smj *Smn

The period of orbit at the surface (if orbit were possible at the surface without running into a hill) is Ps. The periods of the transfer orbit and the orbit at chosen altitude are Pt and Pa respectively.

Ps = (2 * Pi * rs) / Vos

Pt = Ps * (Smj / rs)1.5

Pa = Ps * (ra / rs)1.5

The area per unit time of an orbit is the area swept by a vector connecting the satellite with the gravitational center of the orbit. The area per unit time of the transfer orbit is Aptt.

Aptt = At / Pt

The area per unit time of the transfer orbit at the point of entry of the transfer orbit at the surface is Aptts.

Aptts = 0.5 * Vts * rs so: Vts = 2 * Aptts / rs

The area per unit time of the transfer orbit at the point of circularization of the orbit at altitude is Aptta.

Aptta = 0.5 * Vta * ra so: Vta = 2 * Aptta / ra


If we want to give a small rocket ship an impulsive acceleration with a mass driver on the moon so that it enters hohmann transfer to 30 km altitude and then circularize the orbit at altitude, we start by getting data. The radius of the moon is 1737.100 km. For the purpose of the example we keep three more significant digits than the source of the data justifies to help keep track of calculations in detecting error and to avoid loss of significance in calculation. The superfluous digits will be stripped away for the final answer. The surface gravity is 1.622000 m/sec2.

rs = 1737.100 km

ra = 1767.100 km

Vos = (gs * rs)0.5

Vos = (1.622000 m/sec2 * 1.737100 E6 m)0.5 = 1678.564 m/sec

g+30 = gs * (rs2 / r+302)

g+30 = (1.622000 m/sec2 * (1.737100 E6 m2 / 1.767100 E6 m2) = 1.567394 m/sec

Vo+30 = (g+30 * r+30)0.5

Vo+30 = (1.567394 m/sec 2 * 1.767100 E6 m)0.5 = 1664.254 m/sec

Smj = rs + (0.5 * delta r) = 1.752100 E6 m

Smn = (Smj2 - (0.5 delta r)2)0.5

Smn = ((1.752100 E6 m)2 - (1.5000 E4 m)2)0.5 = 1.752036 E6 m

Ps = (2 * Pi * rs) / Vos

Ps = (2 * Pi * 1.737100 E6 m) / 1678.564 m/sec = 6.502297 E3 sec

Pt = 6.502297 E3 sec * (1.752100 E6 m / 1.737100 E6 m)1.5 = 6.586700 E3 sec

At = Pi * Smj *Smn = Pi * 1.752100 E6 m * 1.752036 E6 m = 9.643880 E12 m2

Aptt = 9.643880 E12 m2 / 6.586700 E3 sec = 1.464144 m2 /sec

Vts = 2 * Aptts / rs = 1.685734 E3 m/sec = launch velocity

Vta = 2 * Aptta / ra = 1.657116 E3

Vo+30 - Vta = 7.138 m/sec = circularization impulse

Final Answer

Launch velocity = 1.6857 +or- 0.0001 E3 m/sec circularization impulse = 7.14 +or- 0.01 m/sec

This is a rough guide for what to expect of the type of launch technology. Actual launch velocity will be determined when people better know the full requirements of an industrialized Luna and its transportation systems.