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Curious layman

Antimatter explosion

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Just say I had enough anti-matter to get the US space shuttle (34 Mtr long), to the nearest star. 

If the shuttle was half way between the earth and the moon and was hit by something and exploded, would the explosion be big enough to wipe out the earth? What about the solar system? Would you be able to see it light years away?

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Posted (edited)
1 hour ago, Curious layman said:

If the shuttle was half way between the earth and the moon and was hit by something and exploded, would the explosion be big enough to wipe out the earth?

On the board of such spaceship there would have to be enough regular matter to annihilate together with antimatter. You're mentioning just antimatter or both matter and antimatter? Explosion would be in frame-of-reference of moving object, obeying inverse-square law, plus-minus some tolerance. Given this info, what would be amount of photons reaching Earth. [math]P = \frac{P_0}{4\pi r^2}[/math] where r = half distance between the Earth and the Moon.. ?

 

Edited by Sensei

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1 hour ago, Curious layman said:

Just say I had enough anti-matter to get the US space shuttle (34 Mtr long), to the nearest star. 

If the shuttle was half way between the earth and the moon and was hit by something and exploded, would the explosion be big enough to wipe out the earth? What about the solar system? Would you be able to see it light years away?

I don't think there is enough information to answer this. The amount of fuel isn't determined by the distance, but (roughly) by how quickly you want to get there; ie. how long you are accelerating for.

 

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13 minutes ago, Strange said:

I don't think there is enough information to answer this. The amount of fuel isn't determined by the distance, but (roughly) by how quickly you want to get there; ie. how long you are accelerating for.

 

I was thinking about when they talk about getting to the next star and stopping, at say about 10% c.

they say- you would need every part of your ship filled with anti matter, 

also ive heard that the amount of energy required would be more then is in the galaxy

surely then, even if you could theoretically do this, there would be absolutely no chance of you being allowed too. I mean that's a massive amount of energy all in one place, if it's equivalent to the amount of energy in the galaxy, wouldn't it just wipe out the solar system if it exploded?

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1 hour ago, Strange said:

I don't think there is enough information to answer this. The amount of fuel isn't determined by the distance, but (roughly) by how quickly you want to get there; ie. how long you are accelerating for.

..you probably misunderstood question.. which was "if spaceship accidentally exploded, could it (explosion) be dangerous to the Earth (which is 192,000 km away from it)"...

So, convert mass of antimatter+matter to energy and apply inverse-square law with the above distance, to receive irradiance on the Earth caused by explosion. Voila.

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7 minutes ago, Sensei said:

So, convert mass of antimatter+matter to energy and apply inverse-square law with the above distance, to receive irradiance on the Earth caused by explosion. Voila.

The mass is what appears missing. It was only specified as "enough anti-matter to get the US space shuttle (34 Mtr long), to the nearest star." If you know how to calculate the mass from that, then the question can be answered.

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36 minutes ago, Curious layman said:

I was thinking about when they talk about getting to the next star and stopping, at say about 10% c.

they say- you would need every part of your ship filled with anti matter, 

"They" do? 

Do "they" do a calculation to show this?

Quote

also ive heard that the amount of energy required would be more then is in the galaxy

Again, is there a calculation? (This is the third recent thread on interstellar travel where people seem to be allergic to any kind of rigorous analysis)

 

The KE of a moving object is 1/2 mv^2 if we are not in the relativistic regime. That energy needs to be dissipated at the end of the trip to come to orbit around the destination, so you will need a minimum of mv^2. No process will be 100% efficient converting to KE. The mass of the probe is not fixed, so that makes this a more complicated problem, but if the mass changes are small we can get a decent approximation.

The mass of the probe is M, and you need Mv^2 of energy. If your final speed is 0.1c then you have 0.01Mc^2 of kinetic energy. That's the minimum mass energy you need to convert to get there. But since you need to eject something to make your rocket work, there's a lot of energy that needs to go into whatever is providing you your thrust. 

The most efficient way (energetically) of doing this is to have a mass split in half, each moving at the final speed, after some explosion. That's doubles the energy required. But doesn't let you stop, unless you did something similar at the far end. So now you've got several percent of your mass being the antimatter, at a minimum. For a system that probably accelerates way to fast to be practical.

The problem gets worse if you try for a larger final speed (Not that this is simple, since it's orders of magnitude faster than we've accelerated a macroscopic object.) because now you're using up more mass, meaning the total mass has to be bigger, and you're accelerating the fuel you need for later on, which increases your fuel demand.

But it's nowhere near the energy available in the whole galaxy. You're not quite to the point if requiring half the ship to be antimatter, but that isn't far off, and for a more reasonable set of assumptions might be the case. (This assumes that you aren't getting propulsion form a source that isn't on board)

 

 

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Posted (edited)
14 minutes ago, Strange said:

The mass is what appears missing. It was only specified as "enough anti-matter to get the US space shuttle (34 Mtr long), to the nearest star." If you know how to calculate the mass from that, then the question can be answered. 

But space-shuttle size is 56.1 meters (not 34 meters long). What is "34 Mtr long" if not mass of antimatter (or antimatter+matter together)? It could mean 34 metric long tons ("34 Mtr long").

Edited by Sensei

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5 minutes ago, Sensei said:

But space-shuttle size is 56.1 meters (not 34 meters long).

That is the length of the space shuttle. The fact it is wrong, is not really relevant.

Quote

What is "34 Mtr long" if not mass of antimatter (or antimatter+matter together)? It could mean 34 metric long tons ("34 Mtr long").

Well, top marks for creative thinking. But it could be a the length of 34 Mass Transit Railways. Or 34 Mid Term Reviews. Or 34 Magnetic Tape Recorders.

Meanwhile, back in the real world where "34 metres long" means the length is 34 metres.... the mass is still unknown.

In the unlikely event that it is referring to 34 metric tons, that is equivalent to about 1021 joules.

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1 hour ago, swansont said:

"They" do? 

Do "they" do a calculation to show this?

Again, is there a calculation? (This is the third recent thread on interstellar travel where people seem to be allergic to any kind of rigorous analysis)

 

But it's nowhere near the energy available in the whole galaxy. You're not quite to the point if requiring half the ship to be antimatter, but that isn't far off, and for a more reasonable set of assumptions might be the case. (This assumes that you aren't getting propulsion form a source that isn't on board)

 

 

I really appreciate people like you replying me Swansont (and everyone else) but please remember, i work in factory, I have no idea how to do rigorous analysis on physics. Even Wikipedia is mostly incomprehensible. 

Note: this site has motivated me to rejoin the library again after years, so hopefully that will start to  change:)

The same amount of energy as in the galaxy was a reference to the Alcubierre drive, but Wikipedia says they think they've brought it down now.

Also, I thinks it's next to impossible for interstellar travel, this was something I thought was another reason why.

1 hour ago, Sensei said:

But space-shuttle size is 56.1 meters (not 34 meters long). What is "34 Mtr long" if not mass of antimatter (or antimatter+matter together)? It could mean 34 metric long tons ("34 Mtr long").

Your right. That's me not doing research. Another lesson for me.

that was just a reference to the size of the ship I was thinking about, my point being that even the energy required to get this to the nearest star (if all of it onboard) would be so massive it would be too dangerous to even think about.

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9 minutes ago, Curious layman said:

I really appreciate people like you replying me Swansont (and everyone else) but please remember, i work in factory, I have no idea how to do rigorous analysis on physics. Even Wikipedia is mostly incomprehensible. 

And there's nothing wrong with this; you're here trying to find things out, which is great.

But when you're repeating things you read or heard elsewhere, you should try and indicate the source.

9 minutes ago, Curious layman said:

 The same amount of energy as in the galaxy was a reference to the Alcubierre drive, but Wikipedia says they think they've brought it down now.

That's non-standard technology based in wishful thinking, and there is no guarantee that it's even possible to build and operate one, so it's not a proxy for a solution based on mainstream physics.  (and here's an example where knowing this up front would have made it easier to respond to the statement)

 

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18 hours ago, Sensei said:

But space-shuttle size is 56.1 meters (not 34 meters long). What is "34 Mtr long" if not mass of antimatter (or antimatter+matter together)? It could mean 34 metric long tons ("34 Mtr long").

Got it from this. Still not 34 though.

IMG_1298.thumb.JPG.2d19a1e129137a61eacb7759190e2786.JPG

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37.24 m is length of STS orbiter.

56.1 m is length of entire device with fuel tank attached (brown colored).

 

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On 7/12/2019 at 6:50 AM, Curious layman said:

that was just a reference to the size of the ship I was thinking about, my point being that even the energy required to get this to the nearest star (if all of it onboard) would be so massive it would be too dangerous to even think about.

As already pointed out, the energy required is dependent on the speed you want to get there and so there is no real lower limit other than that needed to reach Solar escape velocity. 

But let's, so the sake of argument, assume that we have learned how to make and store a enough antimatter to get to another star at 10% of c.  ( How much this would be depends on what kind of exhaust velocity our rocket can generate.)  This could be a fair fraction of the mass of the ship.*

The empty STS orbiter masses around 75,000 kg, so if we assume a 1 to 1 mass ratio,  this, annihilated with an equal amount of matter would created an explosion with 1.35e22 joules of energy, or over 3 million megatons.     A fairly significantly sized explosion to be sure, at least on a terrestrial scale.  But still just 1/28592 the amount of energy the sun produces every second.    Of course, at half the distance to the Moon, it would be a lot closer than the Sun (~1/800 the distance) and thus roughly 22 times as much energy would hit the Earth per square meter than does from the Sun every second,  much of it as hard radiation.

However, there is no reason why such a launch would be made from from the Earth, or anywhere near it.    By the time we got anywhere near making a journey to the stars, we will have conquered our own solar system.    The manufacture of the large stores of antimatter needed would be made and stored at some distant point of the system, maybe on a moon of one of the outer gas giants.   At such a distance that an accident would pose little risk to Earth. 

There are a lot of hurdles to be crossed before we could ever even think of such a venture. (At present, we are only capable of making very small amounts of antimatter and store it for short times, at a cost of billions of dollars per gram), And is quite possible that the difficulties and costs involved would make it impractical.

 

* if we assume a standard rocket, we have to carry not only the antimatter, but also matter with which to combine it with,  This extra mass requires more antimatter, which requires more matter for the reaction...   The total adds up pretty fast.   There is is a possible way to drive this down.  A few decades ago,  a idea was introduced known as the Bussard ramjet.   The design was based on the idea of "scooping up" interstellar hydrogen as you traveled and using it in a fusion drive to propel the craft.  initially is was argued that such a ramjet could reach near light speed because they didn't need to carry their own fuel.  Later studies showed this to be overly optimistic, and the whole concept might not be practical.

However, a modification of this idea might work with an antimatter drive.  With hydrogen fusion, only a small percentage of the mass is converted to energy, with antimatter-matter conversion, up to 100% of the mass is converted to energy.  So maybe you would only need to carry the antimatter, and scoop up the matter for the reaction as you go along.  Not needing to carry the reaction mass would drive down the amount of antimatter you would initially need.  Still an immense undertaking, but maybe just not quite as immense.

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My guesstimation is that it would take about 22 tons of matter + 22 tons of antimatter at the halfway-to-moon distance to make some serous damage on earth. This is based on total irradiation energy equivalent to the energy released in 1000 megaton explosion... 1000 megaton seem quite a lot, but when distributed over the whole earth's disc, I don't think it would be deadly (but I guess some significant damage will be done).

The Space Shuttle had considerably more mass than this, so its explosion (supposing half matter half antimatter composition) would cause serous damage on Earth surface. Possibly global-scale fires.

(It would not event touch the solar system - the sun converts 4 million tons of mater into radiation every second)

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13 hours ago, Danijel Gorupec said:

My guesstimation is...

Except that my guesstimation was wrong by factor 4. The actual result should be more like 85+85 tons... Which is more than was the mass of the full shuttle orbiter... So I guess it is all ok then.

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Ok these replies are great but I didn't explain myself properly, what I was trying to do was get an image in my head of how much energy it would be. It's like when people compare something to Olympic sized swimming pools, or Hiroshima bombs, it makes it's easier for everyday people to understand. I can get a good scale of the space shuttle in my head, that's why I chose that. The antimatter and location of ship was irrelevant, The explosion bit was because you can tell me all the numbers you want, but the truth is they don't really help me, I need an image.

13 hours ago, Danijel Gorupec said:

My guesstimation is that it would take about 22 tons of matter + 22 tons of antimatter at the halfway-to-moon distance to make some serous damage on earth. This is based on total irradiation energy equivalent to the energy released in 1000 megaton explosion... 1000 megaton seem quite a lot, but when distributed over the whole earth's disc, I don't think it would be deadly (but I guess some significant damage will be done).

The Space Shuttle had considerably more mass than this, so its explosion (supposing half matter half antimatter composition) would cause serous damage on Earth surface. Possibly global-scale fires.

(It would not event touch the solar system - the sun converts 4 million tons of mater into radiation every second)

This helps a lot. Thank you.

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1 hour ago, Curious layman said:

Ok these replies are great but I didn't explain myself properly, what I was trying to do was get an image in my head of how much energy it would be. It's like when people compare something to Olympic sized swimming pools, or Hiroshima bombs, it makes it's easier for everyday people to understand. I can get a good scale of the space shuttle in my head, that's why I chose that. The antimatter and location of ship was irrelevant, The explosion bit was because you can tell me all the numbers you want, but the truth is they don't really help me, I need an image.

I don’t know if it will help visualise it or not, but Wikipedia has a series of pages with “order of magnitude” comparisons. For example: https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy)

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The energy released from the collision will be both from the kinetic energy of the antimatter object and the matter object and from their mass since they will annihiliate with each other so it will be very big  . 

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