What would happen if Buck Rogers threw his cell phone at the earth at 1/4 of the speed of light? Science cartoonist Randall Munroe maintains that a baseball traveling at .9c would at ground level would result in fusion, a plasma fireball and a colorful mushroom cloud. But what about an object traveling at .25c when it first hits the earth’s atmosphere?
The question is motivated by a recentproposal to shoot a nano-spacecraft at Alpha Centauri, the nearest star other than the sun. The 4.4 light year journey should take 20 years. Breakthrough Starshot has raised $100 million to study the proposal. Presumably the nano-craft wouldn’t spend too much time in the distant star system. I know space is big and empty, but I wondered anyway what would happen if it hit something.
Bonus question: What would be the size of the crater if it hit the moon?
.9 c to .25 c reduces the kinetic energy by about a factor of five. A baseball and an iPhone weigh about the same, but the a lot of the energy in Randall’s argument comes from fusion of the elements in the air, which I think would be greatly reduced by the lower speed and both reduced and spread out by entering the atmosphere from outside. So I’m thinking you’d get a spectacular flash, and some radiation on the ground.
An iPhone weighs about 0.13 kg. The relativistic kinetic energy equation is:
KE = mc[sup]2[/sup](1/sqrt(1-v[sup]2[/sup]/c[sup]2[/sup]) - 1)
Plugging the numbers in, we get 383.710[sup]12[/sup] J of energy. A kiloton of TNT is considered to be 4.18410[sup]12[/sup] J, so the iPhone is about equivalent to a 92 kiloton bomb. The bomb that destroyed Hiroshima was 15 kilotons, though we have other bombs that are much more powerful (for instance, the common W87 warhead is 300 kilotons, and the largest ever was 50 megatons).
That said, I agree with naita in that any planet with an atmosphere will not see much damage on the ground, as most of the energy will be dissipated in the upper atmosphere.
On airless bodies–I really have no idea. It doesn’t seem like the physics match that of typical impactors like asteroids, even if the kinetic energy is the same. How it transfers that energy to the body is important and hard to guess.
I think that, with either an ordinary asteroid or a relativistic phone, the energy is converted to heat more quickly than radiation or conduction can transfer the heat away from the impact point. And once you’ve got a whole lot of heat deposited really quickly, what happens past that point doesn’t depend on how the heat got there. In other words, it probably would be about the same effect as a same-energy asteroid.
Note that although the “iPhone” comparison has been going around in the news media, the technologists are actually talking about the “innards” of an iPhone, which are on the order of grams. So you can probably knock this down to by a factor of 2–4 at least.
You can punch in some representative numbers into the Impact Effect Calculator. I used an iron projectile with a radius of 1.5 cm, which gives it a mass of about 120 grams, and had it hit the planet at at 60,000 km/s, which is one-fifth the speed of light. If you do this, they contend that the projectile burns up in the atmosphere, and does not impact the surface.
Amusingly, in the accompanying technical paper (PDF) that describes the calculator, they specifically state that “the program uses the relativistic energy equation to accommodate the requests of several science fiction writers.”
Neat! The error is only about 3% at 0.25 c, so not a big deal in this case, but still worth pointing out.
I amused my coworkers at lunch the other day by doing this calculation in my head. Except that I screwed it up and forgot the square root, so I came up with 6%. They didn’t know the difference, though, and were mildly frightened… “did you just do that in your head?”
It’s easy, though. 0.25 squared is 0.0625 (everyone should know the low powers of two in their heads), which I rounded to 0.06. 1-0.06 is 0.94, and the square root of a number close to 1 is half of the “distance” to 1, so sqrt(0.94) is 0.97. And the inverse of a number close to 1 is the same distance on the other side of 1, so 1/0.97 is 1.03. So 3% error in this case (the real number is 3.279%).
You may well be right, but I can’t quite convince myself that the penetration distance isn’t a factor here. If the impactor can make it a significant distance below the surface before transmitting its energy to the surroundings, then you just have a below-ground explosion, which isn’t too exciting. I think there’s clearly some size where this becomes relevant–a single atom with that kinetic energy isn’t going to do a whole lot to the surface–but I don’t know what that size is, even within a few orders of magnitude.
Another comparison: the Chelyabinsk meteor liberated about 375*10[sup]12[/sup] J, which is amazingly similar to the iPhone at 0.25c. Intuitively you would expect the iPhone to give up that energy faster and higher in the atmosphere due to the lower mass and density. However I don’t know what the actual behavior would be, or whether that would be better or worse for those on the ground. Chelyabinsk meteor - Wikipedia
Probably the most impressive relativistic calculation I’ve ever done in my head is from Queen’s “In the Year of '39”: Assuming that the song is talking about relativistic time dilation, and assuming that the ship uses a constant-acceleration drive, what acceleration would be needed?
[spoiler]It’s about 800 gees.
First of all: The volunteers are “older but a year”, and they both leave and return in '39, to “a land that our grandchildren knew”, so 100 years on Earth. So the average time dilation factor is 100. They’re spending 1/4 year (ship’s time) accelerating out, 1/4 year decelerating to the world so newly born, 1/4 year accelerating back, and 1/4 year decelerating to Earth.
A gamma of 100 is easily in the “highly relativistic” regime, and in that regime, proper velocity is almost exactly equal to gamma. And if you use proper velocity, proper acceleration, and proper time, the same equations hold as for the Newtonian case. So their average proper speed is half their maximum proper speed, or the speed they reach in half the proper time it takes them to reach their maximum.
So it takes them 1/8 year to reach a proper speed of 100, their average gamma factor. If they continued accelerating forward for a full year, they’d reach a proper speed of 800. And g is about equal to 1/yr (a convenient coincidence), so that means their proper acceleration must be 800 gees.[/spoiler]
Not looking at your spoiler yet, but here’s my solution:[spoiler]The ships leaves and arrives the same year of the century, so the journey must have been a multiple of 100 years. Since the crewmember talks to the daughter of his love, I have to assume the journey was 100 years and not more.
The crewmember aged a year, so we have a gamma of 100. So (really only keeping track of the left side):
100 = 1/sqrt(1-v[sup]2[/sup])
Invert and square:
1/10000 = 1-v[sup]2[/sup]
Get rid of the 1-:
9999/10000 = 0.9999 = v[sup]2[/sup]
And then the square root trick gives:
v = 0.99995 c
Let’s see how I did:I solved the wrong problem! Oops (so what if I can’t read more than one sentence). But that 1 gee = 1/yr coincidence is fun; I’ll have to keep that one around.[/spoiler]