Saturday, November 10, 2012

Odds of jumping into a star

In an attempt to be more like Dwight Schrute I just finished watching the new Battlestar Galactica.  Overall, it was a good show.  I did, however, have a problem with it.  Its particular answer to faster than light (FTL) travel was a jump drive.  The practical effect was that coordinates could be entered and then the ship would be more or less instantly transported to those coordinates, without covering the area between.  The coordinates were relative to the current location, so they had to be calculated for each jump at the time.  At several points during the series, in order to escape an enemy, an unexpected jump was needed immediately.  Each time a jump to anywhere was ordered, and the response was that it could land them into a star.  The jump was performed anyway and they beat the odds.

The problem is space is almost entirely empty space (hence the name).  I knew the odds of actually jumping into something would be remote, and decided to calculate them.

Before I start, a quick note: I seem to recall a similar objection in Star Wars.  However, in Star Wars FTL travel was via hyperdrive, which did seem to cover the area between the start and end point.  This would greatly increase the odds of actually hitting something.  I'm assuming the jumps in BSG don't cover the area in between, which I feel is justified by the name, dialogue, and jumps that were made surrounded by matter.

Volume of the Milky Way:
The series is purposly coy about if it involves our Earth, but without spoilers it's safe to say it takes place in a galaxy very similar to the Milky Way.  Wikipedia tells me that the Milky Way is a disk 100,000 light years across, and about 1000 light years thick on average.
`V = \pi \cdot r^2 \cdot h`
`V = \pi \cdot (50,000" ly") ^2 \cdot 1000" ly" = 7,854,000,000" ly"^3`

Volume of stars:
Estimates for stars in the Milky Way are 300-400 billion.  I'll round that up to 500 billion.  I'll also assume stars are Sun like.  While there is a lot of stars larger and smaller than the Sun, it's an alright average.
`V = 4/3 \cdot \pi \cdot r^3 \cdot n`
`V = 4/3 \cdot \pi \cdot (1.47 \times 10^-7" ly")^3 \cdot 500 \times 10^9 = 6.69 \times 10^-9" ly"^3`

As you can see, the stars occupy a very tiny volume of the Milky Way.  The odds of a random point being inside a star are about 1 in 1,174,000,000,000,000,000.

Larger stars:
Some of you may be protesting that there is other matter in the galaxy besides stars.  However, this stuff is as rare relative to stars as stars are relative to the galaxy as a whole.  Still, it could be argued that jumping in close to a star would be a problem.  So let's increase the size of the average star we are using to the size of the orbit of Mercury.  Jumping in at the distance of Mercury from the Sun shouldn't pose a threat for a ship in BSG.  This also more than allows for all the other non-star matter.

`V = 4/3 \cdot \pi \cdot (7.38 \times 10^-6" ly")^3 \cdot 500 \times 10^9 = 8.42 \times 10^-4" ly"^3`

Still a pretty small chance at about 1 in 9,330,000,000,000.

Entire star systems:
Let's go ahead and say that the entire star system is off limits to a jump.  Pluto orbits at a max of 50 AU.  The Voyager probes are just shy of 100 AU and are currently at the heliopause, considered the edge of the solar system.  Using 100 AU radius spheres gives:
`V = 4/3 \cdot \pi \cdot (1.59 \times 10^-3" ly")^3 \cdot 500 \times 10^9 = 8348" ly"^3`

 Which gives odds of about 1 in 940,876.

So there is only about a 1 in a million chance of jumping into a star system, which itself is almost entirely empty.

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