long duration orbits

[Gunhand1961]

How would I set up a long duration cometary orbit with an object travelling at near light speed, or at least at a very high velocity?

[buggs_moran]

How would I set up a long duration cometary orbit with an object travelling at near light speed, or at least at a very high velocity?

Easiest way...not realistic at all.
1) Find a comet you like in the comets.ssc file in the extras directory.
2) Copy into a new text file.
3) Change the period (in earth years to speed up)
4) Save as a new ssc file

example

Code: Select all

# was Ikeya-Zhang
"this is a test" "Sol"
{
   Class "comet"
   Texture "asteroid.jpg"
   Mesh "roughsphere.cms"

   EllipticalOrbit
   {
   Period              1.00                  # 50000 AU/yr is 3/4c I think....                  
   SemiMajorAxis     25000.0               # Oort cloud rough distance 50000AU
   Eccentricity      0.999991
   Inclination       28.1206
   AscendingNode     93.3718
   ArgOfPericenter   34.6666
   MeanAnomaly       0
   }

   # These are all made up (copied from Halley)
   Radius           7.5
   RotationPeriod 170

   Albedo 0.04
}


You should read these for some info...
http://www.lns.cornell.edu/~seb/celestia/transforming_ephemeris.html#1.0
http://www.nineplanets.org/kboc.html

[GlobeMaker]

The question was "How would I set up a long duration cometary
orbit with an object travelling at near light speed, or at least at a
very high velocity"?

For a realistic model, please consider a comet going at 0.5 c. That is close
to the speed of light, but not so fast that relativistic effects need to be
considered. The gravity of the Sun is too weak to hold this comet in an orbit.
But the comet could orbit a black hole.

The Sun has weak gravity, as measured from the Earth. It takes a year to
orbit the Sun, but only 16 minutes for the comet to go from the Earth to
the Sun. How far can the comet be moved by the Sun's gravity in
16 minutes? An orbit is just free fall.

"a long duration cometary orbit" may have a period of 100 years to
orbit the Sun. In 100 years, the comet will travel less than 50 light
years from the Sun. There are thousands of stars
within a 50 light year radius of the Sun. The
comet would be attracted to some of those stars while it loops back
toward the Sun. That interference from other stars would probably
prevent the comet from orbiting the Sun.

A black hole has enough gravity to capture the comet. The orbit can be
outside of the event horizon. The orbit can have a short period, but if
it has "a long duration cometary orbit" it would travel many light years
from the black hole. You can put the black hole between galaxies,
so the comet's orbit is not interfered with by other stars. You can
calculate the rest, based on library book-learning.

[GlobeMaker]

More notes on the orbit of a very fast comet.

Assume the comet is going almost as fast as light and it orbits an isolated star with the same properties as Sol. If the orbit is a circle, we can estimate the radius of the orbit by the following method. (See chapter XII of the book by Tullio Levi-Civita "The Absolute Differential Calculus". The chapter title is The Gravitational Equations and General Relativity).

Section 10 discusses the famous experiment where the Sun's gravity bends a light ray from a distant star. We can approximate the fast comet by a light ray. The ray came close to Sol, and its direction was bent by 1.7 arc seconds. This agreed with a prediction by Einstein, and the book shows the calculation. This calculation also shows how to vary the distance between the ray and the Sun's surface to calculate other angles of deviation for the ray's path. These formuli are now presented for the case of a comet moving near c, in a circular orbit around a star.

c = speed of light (300,000,000 m per second) = 3E+8m/second
R = radius of Sol (600,000,000 meters) 6E+8
r = radius of comet's orbit
d = angle of bending of direction of comet or photon
m = mass of Sun = 2 E+30 kg
G = gravitational constant 6.7E-11 (m^3)/(kg*s^2)

d = 1.7 arcsec (R/r)

Set d = 180 degrees so the comet returns in its orbit

180 degrees = (1.7/3600) degrees * R/r
r = (1.7/3600) degrees * R/180

r = 1.7*400,000 / (180*3600) miles = 1 mile

(black hole must be smaller than this orbit.
______________________________

To get independent confirmation using a second method see

http://zebu.uoregon.edu/~soper/Sun/mass.html

* the period P of a planet's orbit.
* the distance r from the planet to the Sun.
* a constant G measured in laboratory experiments.
* the mass M of the Sun.

M = 4 pi^2 * r^3 / (G*P^2)
Using this relation with P and r for the Earth gives the mass of the Sun.

M ~ 2 x 10^30 kg.
Find the radius r when velocity is v = c

P = circumference / velocity

circumference = 2 pi r

P = 2 pi r / c

2 E+30 kg = 4 pi^2 * r^3 / (G* (2 pi r / c) ^2)

2E+30 = (4 pi pi r r r c c ) / (G 4 pi pi r r)

cancel some factors in numerator and denominator:

r = 2E+30 * G / c^2

r = 2E+30 * 6.7E-11 / 9E+16

r = 13E+19 / 9E+16 = 1.4E3 meters = 1 mile !

The comet cannot orbit the sun at nearly the speed of light.
It must orbit a tiny black hole. The duration of its period will not be
long. The black hole with the mass of the Sun must have a
radius less than one mile and the period will be a small fraction
of one second. Sorry, can't be done with a long period. Try to
set your goals a little lower, please.