How much damage would a space probe cause if it can get about 4 lightyears away in 10 years and doesn't have any brakes when it arrives? Could it cause a global extinction event on the planet it was sent to? Do actual concept probes think about this?

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    $\begingroup$ If it was a desolate planet with no atmosphere and was being visited by a single alien explorer and it landed on him, then technically yes :) $\endgroup$ – Wossname Aug 26 '16 at 7:12
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    $\begingroup$ Ignoring the minor gravity difference, the impact energy can't exceed the kinetic energy we put in the probe. And at the moment, exactly that is our biggest limitation. $\endgroup$ – MSalters Aug 26 '16 at 7:59
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    $\begingroup$ The proposal I think you're referring to is not to crash the probe into the planet, but to have it "fly by". The trajectory would be carefully planned and probably wouldn't get within more than a few AU of the planet. $\endgroup$ – zwol Aug 26 '16 at 12:57

Not enough to cause a global extinction.

I could plug things into equations, but I'm lazy. Plugging relativistic kinetic energy of a 100kg mass traveling at 0.4c gives $8 \times 10^{17}$ Joules. The page on the impact that is hypothesized to kill the dinosaurs, the Chicxulub impact, says that the energy released in this event was $4\times 10^{23}$ Joules.

A quick google search shows that the juno probe is about $3000$kg, or about thirty times heavier than the mass I plugged in. This gives that your probe would release 0.00006 times the amount of energy as in the Chicxulub impact. This is a lot, but not in the grand scheme of things!

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    $\begingroup$ Beat me by 20 seconds... and happened to use the same made-up mass, and the same extinction event! What are the odds? $\endgroup$ – Cort Ammon Aug 26 '16 at 6:15
  • $\begingroup$ @CortAmmon hah! That's pretty good :) $\endgroup$ – user12029 Aug 26 '16 at 6:16
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    $\begingroup$ I'm bad at physics; however, wouldn't some(probably large?) portion of the probe burn in the atmosphere, reducing the mass, and also slow down as a result, leaving even less energy for impact? $\endgroup$ – MatthewRock Aug 26 '16 at 9:51
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    $\begingroup$ @MatthewRock: Well, whether that energy goes into the air or the earth it's bad news for whatever lives on that immediate part of the planet. It's "only" 200 Megaton TNT equivalent, 4 times the largest nuclear weapon ever detonated. So you might merely be burned to death and flattened by the shock wave if it goes off a hundred miles over your head, rather than heated to plasma if it's right next to you on the ground, but it's a bad day either way. $\endgroup$ – Steve Jessop Aug 26 '16 at 11:31
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    $\begingroup$ One thing that hasn't been considered here is that if there's an atmosphere involved, the relativistic nature of the impact means that the shock is likely sufficient to induce nuclear fusion. This means you're not just transferring kinetic energy, you're also releasing energy from the fusion reactions. Normally we talk about shock heating due to extreme compression during reentry. At 0.4c, you're talking more like particle accelerator impacts with the atoms in the atmosphere, with all the fun things that happen in those collisions. $\endgroup$ – Dan Bryant Aug 26 '16 at 14:37

Let's assume it travels at a constant speed of .4c (since we don't know anything about its acceleration) and its mass is 100kg. At those speeds, the relativistic effects are minor compared to the fact that I just totally made the mass up out of thin air, so we can do a classical solution.

Using $E=1/2mv^2$ we find that the energy is $1/2 \times 100 \times 119916983^2 = 7.19004143 \times 10^{17} ~\mathrm J$. Using my favorite page on the internet, Orders of Magnitude (Energy), we see that that's on par with detonating the Tsar Bomba, the largest nuclear weapon ever built. It's 6 orders of magnitude off from the energy released by the Chicxulub meteor event that is believed to have caused the extinction of the dinosaurs.

Thus there's little risk of an extinction event. That and hitting a planet by accident 4 light years away without trying is pretty hard!

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    $\begingroup$ To save others the bother of looking this up: ​ The energy calculated by this answer is between 3 and 4 times the Tsar Bomba's yield. ​ ​ ​ ​ $\endgroup$ – user21968 Aug 26 '16 at 8:57

Calculating the damage done due to an impact is an imprecise business, but we could use the KT extinction event (that finished off the dinosaurs) as a benchmark. This impact had an energy of around $4.2\times 10^{23}$ joules. The question is then how fast our probe would have to be going to deliver this much energy.

For a relativistic projectile the total energy is given by:

$$ E^2 = \frac{m^2v^2c^2}{1 - v^2/c^2} + m^2c^4 $$

and the kinetic energy is just this energy minus the rest mass energy $mc^2$.

Let's assume our probe has a mass of a ton ($10^3$ kg), which is around the mass of the Voyager probes. In that case to match the energy of the Chicxulub impact would need a speed of around $0.99999998c$. You're suggesting a speed of $0.4c$, and at that speed the kinetic energy would be around $8\times 10^{18}$ joules, which is a factor 50000 lower than the Chicxulub impact energy. Even so I wouldn't like to be standing underneath it when it hit.

But there are a couple of points that need to be made.

Firstly space is big - really big - and it's mostly empty. The chance of a probe hitting anything is so ridiculously small that no-one is going to take it seriously.

Secondly NASA does care what happens to its probes and this is part of the mission planning. For example the Cassini probe is going to be deliberately crashed into Saturn where the heat of reentry will destroy it (and any Earth life forms that might be hiding on it).

  • $\begingroup$ any thoughts on the possibility of the impact triggering fission or fusion of the surface material of the planet? it seems like global dispersal of radioactive fallout could encourage an extinction event. $\endgroup$ – james turner Aug 26 '16 at 16:32
  • $\begingroup$ @jamesturner: 'Fraid not, fun though it might be. For example see Can Jupiter be ignited?. Our probe would just make a large bang. $\endgroup$ – John Rennie Aug 26 '16 at 16:35

First of all, even without an artificial braking system, a spaceship can slow down using tidal forces of the planets/astronomical bodies it passes by, just as it can speed up using gravitational assist. So, it will not necessarily reach cataclysmic speeds simply by travelling for a long time through space.

Now, let's estimate the energy that will be released if the space probe New Horizons crashed into a planet like earth. New Horizon has a gross weight of about 400 kg. Its last recorded speed (the world record currently) is 16.26 km/s = 16260 m/s. Let's assume it crashes with this speed. And all of its kinetic energy is converted into heat. The heat released will be about 53,000,000,000 J. A ton of TNT produces about 4.184 $*$ $10^9$ J of energy. The energy released in the crash event is about 10 times this energy. That's about enough to blow a large enough tunnel in an average mountain. The planet won't even notice.

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    $\begingroup$ When your going at 0.4c gravity assists are not going to make a significant difference. $\endgroup$ – Taemyr Aug 26 '16 at 20:30

The space probe won't do any damage as it will be vaporized in the exosphere. When the probe enters the interplanetary medium at a distance of the order of 100 AU from the star, it will collide with ions at a density of the order of a 5 ions per $\text{cm}^3$. The collisions with protons happen at a speed of 0.4 c, therefore at an energy of the order of 100 MeV, so this is quite similar to being exposed to intense alpha radiation. The temperature of the heat shield will be raised to about 600 K. The heat shield may tolerate temperatures up to about 6000 K before it will vaporize. This temperature will be reached when the density is $10^4$ times larger, so 50,000 ions per $\text{cm}^3$. But at 1000 km altitude above Earth you'll typically already have $\sim 10^{6}$ atoms per $\text{cm}^3$.

So, as the probe approaches an Earth like planet, the temperature of its heat shield gradually rises due to the increasingly intense bombardment of the ions and atoms it scoops up. At 10,000 km the shield will be red hot, 0.02 seconds later when it is at 4000 km it will most likely have vaporized, exposing the interior to temperatures of the order of 6000 K. The entire probe then simply vaporizes completely before even reaching 1000 km distance.

The atoms the probe consist of will keep on moving into the atmosphere until they collide with atoms. Because of the disintegration, each atom is now "on its own" while in a solid only the atoms at the surface would suffer collisions. All the atoms will have collided with atoms from the upper atmosphere already at a height of a few hundred kilometers. In these high energy collisions, the atoms get ionized and will then experience the strong Lorentz force due to the Earth's magnetic field. These processes will cause the energy of the probe to get dispersed well above 100 km altitude.


protected by Qmechanic Sep 1 '16 at 1:52

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