Sunday, July 29, 2012

Star Trek: Orbits

Star Trek the Next Generation (1987-1994) (Picture)

During Star Trek episodes it is all too common to hear Captain Picard say "standard orbit, Mr. LaForge." This puts the ship close enough to the planet to keep it moving in circular motion around it.

Synchronous orbit is also sometimes requested. What is the distinction between the two, and what physics is required to keep a ship in orbit?

Orbits from a classical mechanics perspective

From the perspective of classical mechanics, a ship in standard orbit is an easy problem. There is a simple force balance equation that needs to be solved, namely:

GMm/r2 = mv2/r

This simplifies to:

GM = v2r

where G is the gravitational constant 6.673 x 10-11 m3kg-1s-2, M is the mass of the planet, v is the linear velocity of the ship around the planet, and r is the distance between the center of the planet and the ship. Note: the velocity and distance from the planet required for the ship to remain in orbit is independent of the mass of the ship.

What a satellite looks like in orbit around the earth. In my equations ms = m and me = M (Source)

Since G and M are known, it is up to Mr. LaForge to choose a proper velocity and distance from the planet. He has infinitely many choices because he has two free parameters. Let's look at a quick example of the Enterprise in orbit around the earth:

G =  6.673 x 10-11 m3kg-1s-2
M = 5.97 x 1024 kg

Geordi chooses to place the ship about 10 km from the surface of the earth. Since the radius of the earth is about 6378 km that means:

r = 6388 km = 6.388 x 106 m

By then solving for v, the Enterprise must be traveling with a linear velocity of

v = (GM/r)1/2 = 7898 m/s

in order to remain in orbit around earth.
Synchronous orbit

Sometimes Picard needs to be in "synchronous" orbit around a planet. This means that the ship remains above the same point on the planet while in orbit. This introduces another constrain into the equation we used above:

GM = v2r

The constraint is that the angular velocity ω of the ship must be the same as the angular velocity of the planet. The angular velocity of earth, for example is one revolution per day. This corresponds to 2 pi radians per 24 hours. There is a useful relation between linear and angular velocity: v = rω. Reformatting our equation we get:

GM = ω2r3

Since the angular velocity ω is that of the earth, there is only one unknown: r. We can solve this example for earth. On earth the angular velocity turns out to be:

ω = 7.27 x 10-5 radians/s

Then solving for r we get:

r = (GM/ω2)1/3 = 4.22 x 107 m = 3.58 x 107 m from the surface of the earth

The corresponding linear velocity v of the ship would be

v =  rω = 3071 m/s

In this way, using only the mass of the planet he is trying to orbit, Mr. LaForge can choose a velocity and distance from the surface of the planet that will allow the ship to orbit the planet.


Satellites traveling around earth employ the same physics to keep them in orbit.

A satellite in orbit around earth. (Source)

There are even geosynchronous satellites. What do you think they do? They orbit the earth and remain above the same point on the earth's surface, like the Enterprise does when it is in synchronous orbit around a planet.

1 comment:

  1. it appears that neither the relative velocity nor the angle of the orbit to the planet poles has any effect on the orbit calculations.