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If Yavin Prime was fired upon, the scattered debris would have destroyed Yavin 4 as well.

Why did they wait to make a direct shot?

Valorum
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Zsmaster
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8 Answers8

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This was addressed in an "Ask the Jedi Council" feature on the StarWars.com website.

According to LucasFilm's (former) Head of Fan Relations/Director of Content Management Steve Sansweet, the Death Star was simply not powerful enough to destroy a gas giant.

Q. In Star Wars, why does the Death Star go around the planet Yavin to blow up the fourth moon when it can simply blow up Yavin first (as it did Alderaan) and then the fourth moon without wasting any time?

A. The Death Star's superlaser is very powerful, but it's not all powerful. Relatively speaking, a terrestrial world of rock and metal like Alderaan is easier to blow up than an immense gas giant like Yavin. The Death Star simply couldn't blow up Yavin, and had to circle the gas giant in order to get to the much smaller moon Yavin 4.

Starwars.com - Ask the Jedi Council

Community
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Valorum
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    This begs the question why it didn't shoot through the gas. – Captain Man Sep 06 '17 at 12:44
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    @Captain Man Presumambly the atmosphere of Yavin Prime would block or dissipate the laser – Landric Sep 06 '17 at 12:58
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    @Landric Even if it diffracted some of the beam away I imagine most would still hit Yavin 4. I assume because it's a moon it's smaller than Alderaan which I also assume means it's easier to blow up. I mean, I feel like Steve Sansweet is just making making something random up after the fact. I feel like a better explanation would be that the laser takes a long time to recharge (like 24 hours or something) so it's quicker to go around Yavin. I don't know if they ever said how long the laser takes to charge/reload or anything though. In any case I'd consider this answer more cannon than my guess. – Captain Man Sep 06 '17 at 13:15
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    @CaptainMan He's obviously just making stuff up, because fans ask all sorts of impertinent questions about irrelevant plot holes that don't actually matter to the story, even if they are "wrong." However, Stack Exchange only concerns itself with answers that have objective support, either from an official source (like Mr Sansweet) or derived from analysis of canon material; delving into what we each think would be better is just opinion-based hearsay. That being said, do you have any support for the recharging idea from the movie itself? If so, you may want to make an answer out of it. – Steve-O Sep 06 '17 at 13:32
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    Just because it's called a "gas giant" doesn't mean that there isn't a solid core under the gas. Even a small distance (compared to the overall diameter) under the "clouds" the "gas" is going to be too thick to send a laser through or even a solid projectile. – JPhi1618 Sep 06 '17 at 13:42
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    @CaptainMan It may be just a moon, but it was large enough to have Earth-like gravity, putting it in the same size class as Alderaan. – Joel Coehoorn Sep 06 '17 at 14:14
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    @CaptainMan You might want to look into what exactly a gas giant is. Gravity compresses the material in a gas giant to very high density closer to the core. It might have been possible to shoot through the outer layers of the "atmosphere" of Yavin, but not straight through. Even a shot partially through Yavin would have lost a lot of energy to the intervening gas. A gas giant is like a failed star, and you probably wouldn't suggest shooting through a star. – Todd Wilcox Sep 06 '17 at 14:38
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    Plus they might not be able to accurately target Yavin 4 without a direct line of sight, especially given all the error introduced by shooting through another planet to get there. – Brad Sep 06 '17 at 15:27
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If we begin by assuming that if they could have, they would have, it can be instructive.

How much firepower does it take to destroy a planet the size of Yavin?

We know of one planet destroyed by the Death Star - Alderaan - and we can say that Alderaan is an Earth-type terrestrial planet whereas Yavin is a Jupiter-type gas giant.

If firepower required is a function of planetary mass, then Jupiter, having 317 times the mass of Earth, would require 317 times the firepower to destroy than Earth would.

Likewise Yavin would require a similar magnitude increase in firepower to destroy.

So going back to our original assumption, we conclude that the Empire didn't destroy Yavin because the Death Star didn't have enough firepower to destroy a gas giant that large.

Victim of Circumstance
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    Something something merge your accounts you could have loads of rep and privileges. Come back to us, that meta post was years ago now. We want you back. – Edlothiad Sep 06 '17 at 07:14
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    Actually, gravitational binding energy is proportional to the square of mass. That's the smallest amount of energy you'd need to break apart a planet (and prevent it from reassembling again later), ignoring all the extra electromagnetic forces that also hold the planet together (even in a gas giant, those aren't negligible). Jupiter more like 2000 times the gravitational binding energy of Earth (radius also plays a role, and the proportionality applies perfectly only for isotropic spheres, which Jupiter isn't), rather than just the 300 times you'd expect. – Luaan Sep 06 '17 at 11:12
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    Now, we do see Alderaan explode extremely violently (at least some fragments appear to move at low relativistic velocities), so there's certainly some energy left over; and it's also noted that the main reason a weapon like the Death Star is necessary is that even a sizeable fleet of star destroyers is largely ineffective against even low-grade planetary defence shields, so it can be assumed that the power of the weapon mainly has to do with defeating any existing and future planetary defences with one stroke - with the destruction of the planet itself being just an unfortunate side-effect. – Luaan Sep 06 '17 at 11:14
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    But if you think that this is the firepower necessary to destroy a shielded world, it seems outright ridiculous to expect that the same kind of firepower would easily destroy something three orders of magnitude more massive; not to mention that we have little idea how the weapon would interact with the gas giant. And even if it had enough power to disrupt Yavin enough to destroy its moon (which had a planetary shield, unlike the gas giant), it seems extremely unlikely that the Death Star wouldn't be affected - it would be far more vulnerable than the moon. – Luaan Sep 06 '17 at 11:19
  • @Luaan I added an answer which is an attempt to expand on your comment (and this answer), I would be interested to see what you think of it. – Charles Sep 07 '17 at 04:42
  • @Joseph old drama, no longer exists, don't worry. – Edlothiad Sep 07 '17 at 13:01
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Yavin 4 was being rapidly evacuated at the time of the Death Star's attack.
Debris from Yavin would (likely) strike the moon, but how quickly?

Charles generously did some of the math on the size of Yavin and such, and we could use Kepler's Third Law if we chose, but conveniently, someone has done much of the rest of the math for us.

In short summary, "Yavin IV probably orbits Yavin at about 500,000 km radius". The fastest-moving debris from any man-made explosion was as a result of the Pascal-B nuclear test, a 2,000lb steel plate, which ended up traveling at approximately 66 km/s. Assuming the debris travels at a similar speed, and assuming it would even sufficiently damage Yavin 4, it would end up giving the Rebels a comfortable 126 minutes to evacuate from the moment it was fired.

When the Death Star begins to orbit Yavin, according to the announcing voice, "The moon with the Rebel base will be in range in thirty minutes." By destroying Yavin 4, not Yavin (and as seen in other answers, the ability to do that is not certain) the Empire reduces the amount of time the Rebels have to evacuate by more than an hour.

Daniel B
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  • If this maths holds up this is a great hidden gem! – Edlothiad Sep 07 '17 at 06:15
  • The entire site is a gem, but particularly that page. And to forestall complaints, I realize this answer doesn't factor in the possibility that debris off of planet destruction travels much faster than 66 km/s, but I've been unable to find any remotely substantial data on how fast the debris actually travelled, so this seems a fair estimate. Frame-by-frame analysis proved untenable. – Daniel B Sep 07 '17 at 06:34
  • Very nice! I got a somewhat higher radius from the orbital period of Yavin Prime and Kepler's Third, but it's good to see that the two methods give roughly similar numbers. Figuring the same density of Jupiter I calculate an escape velocity of 84 km/s, so any fragments from destruction must leave at an average velocity at least this high. This drops the time to 99 minutes, but your point remains. :) – Charles Sep 07 '17 at 13:38
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    Calculation: sqrt(2 * G * 4/3 * Pi * (97176 km)^3 * 1.33 g/cm^3 / (97176 km)) = 83.4 km/s. Feel free to tweak the radius and density. – Charles Sep 07 '17 at 13:41
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    Yavin was not being evacuated at the time of the assault... it actually took awhile after the battle for the rebels to escape. – Skooba Sep 07 '17 at 17:30
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The Empire was particularly proud of the Death Star and they were power-blinded. They want to see rebels die quickly in one shot. Destroying Yavin would have taken time to destroy Yavin 4.

Plus, the Death Star was near to Yavin and destroying this planet would have caused debris fly to Death Star and damage it. The Death Star didn't have defense shield then. Therefore, it wasn't wise to destroy Yavin. However, this is just speculation.

Plus, if they have destroyed Yavin, you couldn't see the planning of attack and the final battle around Death Star and the movie would have been 30 minutes shorter, wouldn't it? ;-)

Not
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  • They could just use their Hyperdrives to get out of the area? – Edlothiad Sep 06 '17 at 07:11
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    They could, but they would require a guy to press the button when they fire. However, that's a possibility, but they were overconfident about Death Star and wanted to end rebels in one shot. Remember Tarkin saying Evacuate? In our moment of triumph? – Not Sep 06 '17 at 07:15
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    I agree with your answer, except that little bit. Also that last note about "it would shorten the plot" is generally not well received and discouraged – Edlothiad Sep 06 '17 at 07:17
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    I added this just as humor and I added that the second part is just speculation. – Not Sep 06 '17 at 07:20
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The question is, can they destroy it?

Gravitational binding energy U varies as

U ∝ m^2/r ∝ ρ^2r^5 ∝ ρ^{1/3}m^{5/3}

in the uniform density simplification, where ρ is the density, m the mass, and r the radius. Yavin Prime has a diameter 39% larger than Jupiter, but apparently its mass isn't known. Fortunately, we're living in the future! We have a database of exoplanets, and we can search for planets the same size as Yavin Prime. In fact, COROT-5b is almost the exact size of Yavin Prime: 187,000 to 201,000 km compared to Yavin Prime's 198,500 km. It's quite puffy: less than half Jupiter's mass despite being larger. But is this realistic for Yavin Prime?

Yavin Prime's orbital period is 4818 days. If we knew the mass of Yavin, we could use this to determine its distance to Yavin Prime. But Yavin itself is class K, so between 0.45 and 0.8 times the mass of our sun. Kepler's Third Law gives a distance (semimajor axis) of 640 million to 775 million km (lighter = closer) for those figures. By comparison, CoRoT-5b orbits at a mere 7.4 million km, receiving 10,000 times more energy from its star by being 100 times closer. (Actually a bit more, since CoRoT-5 is a bit brighter than Yavin.)

So it's no surprise CoRoT-5 is cooked up to an enormous size. But certainly there's no reason to think Yavin is less dense than CoRoT-5, and it's probably closer to Jupiter which is a similar distance (778 million km) or even a bit denser. So in the likely case it will have gravitational binding energy 1^2 * 1.39^5 = 5.2 times that of Jupiter*, and in the 'worst case' where Yavin Prime is puffy like COROT-5b the GBE is (.217/1.33)^2 * 1.39^5 = 14% that of Jupiter's*. More reasonable would be if Yavin Prime is more like, say, KOI 1421.01 (Uehara et al., "Transiting Planet Candidates Beyond the Snow Line"); with its density Yavin Prime would have a GBE almost three times that of Jupiter's.

But as Jupiter's GBE is thousands of times larger than that of Earth's (the uniform density approximation, which should really not be trusted to scale between a gas giant and a terrestrial planet, gives a factor of 9000). So 2-4 orders of magnitude more than the most we've seen the Death Star produce, after a blast large enough to knock them out for some time as they recharged. So I don't think it's unreasonable to suggest that the Empire didn't destroy Yavin Prime because it was simply beyond their power.

* Using here the uniform density approximation, but only to scale between planets so it's not quite so bad.

Charles
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  • Remember, though, that's only to make it go to escape velocity. That's the minimum to make it break the GBE. When the death star destroys Alderan, in a few seconds, the plume is 2-3 times the radius of the planet. This means it's being accelerated at orders of magnitude times the escape velocity! – corsiKa Sep 08 '17 at 15:16
  • @corsiKa Agreed -- though it looks like only a small fraction of the planet's mass is going that fast, and presumably when at the upper end of its destructive capability there's no leftover energy to accelerate much of anything beyond escape velocity. – Charles Sep 08 '17 at 18:09
  • You can get the mass from. https://starwars.fandom.com/wiki/Yavin_Prime/Legends – blademan9999 May 05 '23 at 07:52
  • @blademan9999 Ah... it lists surface gravity at 4 times standard, so 49.81m/s^2/G(198,500 km/2)^2 = 5.8e27 kg. That gives a density of 1.4, just a bit more than Jupiter, so even higher GBE. – Charles May 05 '23 at 20:36
  • That would give it GBE ~6.7 times that of Jupiter. – Charles May 05 '23 at 20:41
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My opinion is: What would their reason be?

I mean, how does Grand Moff Tarkin feel about the rebels evacuating the moon or their last ditch attack attempt?

Evacuate? In our moment of triumph? I think you overestimate their chances.

I'd say he's pretty relaxed and confident, all things considered.

He could probably rush it a little, shooting through the outer atmosphere of Yavin Prime when the moon was only just coming to view, but why rush?

As far as he's concerned, victory is assured. That means no need to rush it, no need to focus on little details that, in a heated or uncertain fight, could tip the scales. None of that. He's slowly, deliberately closing in for the kill. This is his moment of glory and he's going to savor every second of it.

mario mario
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Is it safe to assume that Yavin 4 would have been quickly "destroyed" if Yavin were "destroyed?" I'm not sure that it is. I mean, eventually, yes. Yavin 4 would probably end up falling into the local star or whatever. There would certainly be a lot of terrible repercussions for it's ecosystem.

But the question is whether the Rebels would all have been sufficiently killed before they could flee. Killing rebels is what the empire if there for. If Yavin blows up, the local area probably becomes hazardous for navigation, especially for something as large as the Death Star. So, it might be hard to get in close and ensure the rebels are wiped out.

Some pieces of Yavin could, of course, destroy all the rebels on Yavin 4 or perhaps the moon itself. But, they also might not.

Mon Mothma is down there. Leia Organa is down there. The Empire wants to kill those high value targets. What kind of demonstration is it of Imperial power to destroy a gas giant but allow the leaders of the Rebellion to escape. Those would be some mixed signals.

Remember, the Imperials don't know that the Death Star is going to get blown up in five minutes. They are supremely confident. Waiting 5 minutes to line up the death blow makes perfect sense to them.

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I think the sudden removal of gravity from the destruction of Yavin Prime could send the Death Star into an unpredictable trajectory.

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