# How was the torpedo that Luke Skywalker shot into the first death star able to make the distance down to the reactor?

How much fuel is in a proton torpedo? How far is the maximum effective range of said torpedo? How far into the interior of the first Death Star (DS-1) is the reactor?

Then there are the technical questions of:

• What is the time delay of the explosive?
• Are there any forces or gases pushing on the progress of the torpedos? Two torpedos—fired at the same hole—are going dozens of miles down the same tube.
• I can't be 100% sure, but I think it was Wookie magic.
– Daft
Dec 19, 2014 at 9:30
• The Force!!!!!! Dec 19, 2014 at 11:13
• "Use The Force Luke" Dec 19, 2014 at 14:03
• I'm not sure the missiles themselves travelled down the exhaust port — "A precise hit will start a chain reaction, which should destroy the station." I've always interrupted this as the missiles exploding once inside the port, and this explosion travelling all the way into the reactor due to bad design of thermal exhaust port allowing a chain reaction to take place. Dec 19, 2014 at 14:14
• @VicSmith yeah that was my impression as well, it wasn't that the torpedo made it to the reactor, its that it blew something up that started a chain reaction that resulted in the reactor blowing up. Dec 19, 2014 at 14:25

While I'm sure someone will come up with a more canonical answer, here's my approach from a pure physics perspective:

Basically, it boils down to Isaac Newton's Laws of Motion... and I immediately had to think of the soldier explaining the basics in regard to space combat as seen in Mass Effect 2 (dead link removed)

The most important part is this bit:

An object that is in motion will not change its velocity unless an external force acts upon it.

What this means is that the torpedo doesn't need enough fuel to burn through the whole (or more specific: half) Death Star.

The torpedo will only consume fuel in the following situations:

• Acceleration
• Deacceleration
• Turning (which - as mentioned in the comments - is just (de)acceleration again)

So in short, it will only use up its energy when it has to steer around or through some obstacle(s) (assuming there's no dense atmosphere inside the shaft).

As depicted in the movie, the shaft is pretty much straight and leads down right into the center/reactor. As such, there's probably not much the torpedo has to do, except that one turn on the surface and probably some additional corrections along the way.

In no way will it have to accelerate all the way down.

Note that sci-fi games, movies and settings in general will forego or simply violate (or forget about) Newton's laws to make things more interesting or fair, e.g. by limiting the effective range of weapon systems.

So is it impossible in a dogfight for a fighter to outrun a missile or torpedo till it burns out? Of course they can, even while considering Newton's laws. It makes it even more logical:

The dumbest thing you might do, is trying to burn aways fleeing from a torpedo or missile in a straight line. Even if Newton's laws are in effect, this will allow the pursuing weapon to follow you for the longest time. Instead, if you try to fly in radical and unpredictable patterns, turning around etc, the missile will have to steer and adjust its path, burning its ressources a lot faster, therefore making the escape a lot easier.

• This is correct, but how it fits in with Lucasian Physics, where space is a sound-carrying medium, and time and distance may be measured in each other's units, I have no idea.
– user8719
Dec 19, 2014 at 8:46
• @Darth Satan: General misunderstanding. You don't measure the time it takes, you measure the shortest distance based on your skill in navigation when using shortcuts through monster dens, ultra-dense asteroid fields, and while abusing interstellar superhighways. Dec 19, 2014 at 13:45
• Don't you be comin' round these here parts with yer fancy book-learnin'. (yes, I know, I was making a joke)
– user8719
Dec 19, 2014 at 14:08
• "Acceleration, Deacceleration, Turning" -- All of which are, in the end, acceleration. :P Dec 19, 2014 at 15:15
• what about air resistance? it is an exhaust pipe afterall Dec 19, 2014 at 22:53

The torpedo didn't travel to the reactor; it started a chain reaction that did. This was explained during the briefing before the Battle of Yavin:

A precise hit will start a chain reaction, which should destroy the station."

See Thermal exhaust port (Wookieepedia).

• While canon doesn't give details of the vulnerability, it's not hard to imagine that fabricating an entire exhaust pipe from material which is both physically strong and able to withstand exposure to the exhaust might be difficult. A pipe with an exhaust-resistant inner layer and a physically-strong outer layer could be much more practical to construct, but a sufficiently large rupture might expose a nearby portion of the outer layer to the exhaust, which would cause it to fail, thus exposing more of the outer layer to the exhaust, etc. Not every rupture would cause a chain reaction, but... Dec 19, 2014 at 22:08
• ...it's not hard to imagine that a large rupture could do so if it occurred deep enough that exhaust expanding through the rupture would start to experience turbulent flow before it was clear of the station. Depending upon the pressure and velocity of the exhaust, it may not have been necessary for the torpedo to go very deep in order to create such turbulence; whether the chain reaction could propagate indefinitely inward would depend on the construction of the station near the pipe. Dec 19, 2014 at 22:17
• In particular, there should be sufficient heat and back pressure in the area of the rupture to allow the exhaust there to melt more of the pipe, but not so much heat near the outer parts of the station to expand the outer opening enough to reduce the heat and back pressure to safe levels. The further the damage works its way down the pipe, the more the exhaust would be able to cool off before reaching the outer layers, so once the damage propagated a certain distance it would be able to propagate indefinitely. Dec 19, 2014 at 22:20
• @jpmc26 if it was fired from a phase cannon, then it is a CANONical answer. Har har har. Dec 19, 2014 at 22:50
• @jpmc26: Considering further hypotheticals, even a hit which didn't cause a chain reaction all the way to the reactor might have caused some problems, depending upon what other means of exhausting were available. If that exhaust port represented 90% of the station's exhaust capability, damaging it could force the station to cut back greatly on energy generation, which might leave it vulnerable to other attacks. That might have been too expensive to shoot, though, and tried audiences' patience. Having the whole Death Star go "boom" was cheap and effective. Dec 19, 2014 at 23:22

Well, you have to understand that there is some level of gravitational pull from the Death Star, be it artificial or actual. Also, you have to accept the possibility of magnetic shielding within the port itself, which would automatically act as a guide for the torpedo, all the way to the core. In that case, as seen in the movie, they only had to make a direct hit, and the torpedoes would be sucked in and down towards the core, just like you see in the movie. It could have been the shielding, not the force, that caused the 90 degree turn down into the port. It just so happened that to hit that small of a target was nearly impossible for someone without the Force as a guide. That's also why there was no concern about the port being horizontally oriented, because the target area was the effective range of the magnetic shielding, with a strength necessary to pull both torpedoes into the port and towards the core.

To address the chain reaction just inside the port, recall during the briefing the "pong" animation indicated the missile traveled to the center then caused the chain reaction. Minor point of sequence clarification.

• Something in the animation traveled to the center, but it wouldn't have had to be the missile. High-power electrical cables can suffer a chain-reaction called arc tracking where a spark travels down the cable; heat from the spark burns away the insulation and vaporizes the cable ahead of it, allowing the spark to travel down the cable. It's not hard to imagine that a similar phenomenon could occur with the pipe feeding the exhaust port; the animated blip moving through the station could have been a "fireball" burning its way down the pipe. Dec 19, 2014 at 22:03
• @supercat a bit of a stretch, but I can see where you're coming from. First rule of presentations - Keep it simple. Dec 20, 2014 at 2:47