# What major event could disrupt planet Earth's orbit around the Sun?

What artificial disaster could disrupt the orbit of our planet? Not necessarily making it leave the Solar System, but maybe changing its trajectory around the sun. And by artificial, I mean that the cause can't be an asteroid or an object from outer space. Any ideas? I thought about a single, concentrated powerful atomic bomb explosion "pushing" the planet and disrupting its orbit, but I found that a little bit cliché.

• Read Worlds in Collision by Immanuel Velikovsky for some interesting insights. The amount of energy required may be more than you expect and take long enough for traditions of Jubilee cycles (49 years) to build up from repeated attempts to disrupt orbits. – KalleMP Dec 29 '16 at 18:22
• Change the orbit how much? Moving your pinkie finger up and down is enough to change Earth's orbit very, very, very slightly. – smithkm Dec 29 '16 at 19:31
• Any event confined to earth's gravity (Like a nuke) isn't going to move it in the slightest. In order to change the orbit, something MUST escape earth's gravity OR it must be an external gravity influence/change. The only man-made event that might do it without destroying nearly all life on earth would have to involve the moon. Splitting a huge chunk of the moon and ejecting it from earth's gravity would throw a lot of chaos into the system and could change the combined moon/earth orbit of the sun a bit. – Bill K Dec 29 '16 at 22:09
• @BillK If you prefer, shining a flashlight into space would avoid that the problem of the moving pinky arguably being part of the Earth. – smithkm Dec 29 '16 at 22:55
• Note that Velikovsky is not orbital-mechanics but made up ad-hoc motions. – JDługosz Dec 30 '16 at 1:01

# Lets find out how much energy that takes

The energy of an orbiting object is the sum of its kinetic and potential energy and can be expressed by $$E = \frac{1}{2}mv^2 - \frac{GMm}{r} = \frac{-GMm}{2a}$$ where $GM$ is the gravitational constant of the sun ($1.327\times10^{11} \text{km}^3\text{s}^{-2}$), $v$ is the orbital speed (variable for a non-circular orbit), $m$ is the mass of the Earth ($5.972\times10^{24}\text{ kg}$) and $a$ is the semi-major axis of the Earth ($1.496\times10^{8} \text{ km}$). Plug this all in and we get $$E = \frac{-1.327\times10^{11}\cdot5.972\times10^{24}}{2\cdot1.496\times10^{8}} = -2.649\times10^{30} \text{J}.$$

Note that you go up by 3 orders of magnitude to convert km to m to get Joules. Also, the energy is negative because the gravitational potential of the sun is greater than the Earth's kinetic energy. If it were not, then the Earth would be at or above escape velocity and flying out of the sun's orbit.

So lets calculate how much it would take to transfer to a different orbit. Instead of doing the Hohmann transfer calculation, we can just do the raw $\Delta E$ required. If we increment $a$ by 1, we get $a = 1.49600001\times10^{8}.$ Somewhere my high school calculus teacher is screaming about significant digits. Re-doing the energy calculation for this new, greater orbit, we get the same energy value, with rounding. But if we conserve all significant digits, we find that the difference between the two numbers is $1.77\times10^{22}$ joules. While this number is insignificant compared to the orbital energy of the Earth, it is very significant when compared to...well anything else that has energy.

For example, Tsar Bomba's blast released about $2.1\times10^{17}$ joules, which means that even if all of Tsar Bomba's energy were directed into changing the orbit of the Earth, it would take about 100000 of them to move the Earth in orbit by 1 measly kilometer. The Chicxulub impact was $4.20\times10^{23}$, so that is more like it; but still not enough energy to produce more than a few km of orbital deviance.

Conclusion: It will take a lot more explosive power than we have managed so far to make any sort of significant change in the Earth's orbit. Any explosion that is large enough to make a significant difference in the Earth's orbit will be an extinction event, to say the least.

# Lets find out how much momentum that takes

Now lets re-solve that first equation for $v$ to do some momentum conservation. The link helpfully boils this down to $$v = \sqrt{GM\left(\frac{2}{r}-\frac{1}{a}\right)}.$$ Assuming a circular orbit for simplicity so that $r = a$, I solve that as $v = 29.8 \text{ km/s}$ which is conveniently what Google thinks too.

The momentum of the Earth is then $mv = 1.779\times10^{29} \text{ kg}\cdot\text{m/s}$. So lets see what happens when we hit that with a Chicxulub. Lets say our impactor is going at a breezy 50 km/s, weighs 1e15 kg (a 5km radius, 2000 kg/m$^3$ sphere, approximately) and hits in such a way that all of its momentum is transfered to earth in the direction of Earth's motion. The added momentum is $mv = 5\times10^{19}$, so the new total for Earth is $1.7790000005\times10^{29}$. Resolving backwards through velocity and to semi-major axis of the orbit, we get an orbital change of about 80 meters.

Conclusion: Even if you were going to hit the Earth with something, you would destroy the crust, bore holes in the mantle and eradicate all life on it long before you move the Earth's orbit significantly.

# Extra final conclusion

You are not going to move the Earth any significant distance with a single energy or momentum event while allowing anyone on it to survive.

• Assuming you had that amount of energy, couldn't you apply it on the moon (or some other heavy body) and use the moon's gravity to lower earth's orbit (like a gigantic gravity assist maneuver)? Sure, the moon would be destroyed in the process, but people on earth might survive the event. – nikie Dec 29 '16 at 17:58
• What if you could spread the force out over a millenia or longer. – Dave Dec 29 '16 at 18:07
• @user17625 That is much more likely. I might do the math on that, but I suspect that the waste heat from whatever energy generation process we use on the Earth will cook the Earth uninhabitable before we get very far. In fact, I can calculate it right now. To increase the orbital energy of earth by 1%, we need about 1e28 J. If we assume 50% efficiency (which is high), that generates 1e28 J of waste heat. Since the Earth receives ~5.5e24 J annually from the sun, that waste heat is equivalent to 200 years of sunlight. I don't think we will like that, here on Earth. – kingledion Dec 29 '16 at 18:20
• @nikie I suppose I could try to work that out, but it seems like the destruction of the moon will make falling moon-bits a bigger problem than altered orbit for the next millenia or so. – kingledion Dec 29 '16 at 18:21
• I remember reading a sci-fi book where two planets would approach close enough for people to migrate via a water bridge. If they got that close I believe that tidal forces on the planet would have caused earthquakes that would have made both planets uninhabitable. – KalleMP Dec 29 '16 at 18:27

kingledion correctly points out that no single event can move the Earth without also basically destroying it, but that's for a single event. What if something on Earth was producing high thrust over a long period of time? Say a year?

# Polar Thruster?

Since the Earth is rotating, any long firing thruster on the surface will wind up cancelling itself out in the orbital plane. But we can move the Earth "vertically" relative to our orbital plane with a thruster placed at high latitudes, preferably at the North or South pole. Probably the South as there's solid land to work with.

Due to the Earth's axial tilt of 23.4°, a stationary thruster at the South Pole firing over multiple years will lose some of its efficiency due to the angle, but it isn't going to matter because the numbers will get absurd.

How powerful does our thruster have to be? Well, how fast do you want to change the Earth's orbit? What would you like the Earth's acceleration to be? What would it take to, say, move the Earth 10° "north" of it's current orbit over a year?

The distance the Earth needs to travel is, roughly, the base of an isosceles triangle whose long arms are 1.52e8 km long and whose vertex is 10°. That's 2 x 1.52e8 x sin(10°/2) or 2.64e7 km. 26 million km.

There's 3.15e7 seconds in a year, so we have that to move 2.64e7 km. How much acceleration do we need? We start with with d = vit + at2/2 and solve for a. v1 is 0 so it's just d = at2/2. Solving for a gives a = 2d/t2. If t = 3.15e7 seconds and d = 2.64e7 km we get an acceleration of just 5.32e-8 km/s2! A small acceleration over a long time adds up!

Such an acceleration is so small only sensitive scientific equipment would notice. That might delay an investigation and allow a super villain to get away with firing a thruster at the South Pole for a year (though the people living there will probably notice). Might this work?!

No, sorry.

Now that we have the acceleration we need, how much force is that? F=ma. a is 5.32e-8 km/s2 and m is the mass of the Earth, about 6e24 kg. That's 3.2e17 kg km/s2 or 3.2e20 kg m/s2. kg m/s2 is a Newton.

How much force is 3.2e20 N? Oh dear, that's a lot. It's the force of 1e13 Saturn V rockets. It's roughly the same as the gravitational attraction between the Earth and the Moon.

What if it took 10 years? Looking at a = 2d/t2 we see that acceleration is the inverse square of time. If we increase time by 10, we decrease acceleration (and thus our force) by 100. Or for every order of magnitude increase in our thrust time, we get two orders of magnitude decrease in our thrust.

If over 1 year we need 1e13 Saturn V rockets, over 10 years we'd need 1e11 Saturn V rockets. That's still a lot of rockets. Over 100 years it's 1e9 rockets. Over 100,000 years it's 1e3. To move the Earth 10° "vertically" in its orbit we'd need 1,000 Saturn V rockets burning continuously for 100,000 years.

I think this illustrates why this isn't going to happen. Even though the needed acceleration is so small you wouldn't even feel it, the mass of the Earth is just too big, and the forces involved dwarf anything we can produce.

# Paint One Hemisphere White, One Black?

Sunlight exerts a force on the Earth pushing it away ever so slightly. Light that's reflected exerts twice the pressure as light that's absorbed. If the Earth absorbed all its sunlight, it would exert about 570 N.

If the northern hemisphere were to become a perfect absorber, and the southern hemisphere are perfect reflector the Earth would experience a continuous net thrust "north". How would we do this? Ask Sherman-Williams.

How much this would be is beyond my calculations, but not more than 570 N or about 10 Saturn V rockets. So it's still not going anywhere fast.

While this can have a measurable effect on small bodies like asteroids, known as the Yarkovsky effect (thanks @IwillnotexistIdonotexist) it does not work on the Earth. Why? That pesky square-cube law.

Radiation pressure depends on the surface area of a sphere. As a sphere gets bigger its surface area increases as the square of its radius, but its volume (and thus mass) increases as the cube! So as you make a body bigger the ratio of it's surface area to its mass drops. That means less radiation pressure, less thrust, per unit of mass.

• There's another very long-term effect that could alter Earth's orbit - a sudden change in the albedo of Earth's surface or its gradient sufficient to change its radiation pressure characteristic. See Yarkovsky effect; This is known to have a measurable effect on satellites. Also: "Possible asteroid deflection strategies include "painting" the surface of the asteroid or focusing solar radiation onto the asteroid to alter the intensity of the Yarkovsky effect and so alter the orbit of the asteroid away from a collision with Earth." – Iwillnotexist Idonotexist Dec 30 '16 at 3:27
• @IwillnotexistIdonotexist Great minds think alike! (see my edits) – Schwern Dec 30 '16 at 3:42
• "any long firing thruster on the surface will wind up cancelling itself out in the orbital plane" - couldn't it be fired every day (or every sidereal day, depending on what kind of long-term thrust you want) whenever it's facing within some angle of the direction you want to thrust in? You could even have an array of them. – Random832 Dec 30 '16 at 12:43
• @Random832 Yes, you could do that. However, the OP wanted a "artificial disaster" and they're normally not so controlled. Maybe something triggered by the sun? I drifted away from the disaster part myself in pursuit of making anything that works. – Schwern Dec 30 '16 at 18:45
• @Random832 The reason I liked the albedo idea is that if you had a relatively reflective planet, and suddenly an apocalyptic solar event "baked" cinder-black the facing portion of this planet, you could plausibly achieve that black/white hemisphere look that maximizes the Yarkovsky effect. – Iwillnotexist Idonotexist Jan 1 '17 at 21:11

As kingledion points out the possibility of a man-made force on earth itself is next to impossible. My alternative may be outside your rules of "Can't be an asteroid or object from outer space" but it depends on how willing you are to stretch this.

If you consider kingledion's point about changing orbits then we see that these depend on mass and radius. If the disaster were to involve a planet further out than earth and, depending on what rules you want your universe to have/stretch, less massive than earth then it may be more possible.

If any orbiting object loses its transverse momentum (along the path of the orbit) it will fall in towards the host star (and it is further out so it can fall past earth on the way). So lets look at the possibilities here.

Pluto at its lowest momentum : $$M_{pluto} = 1.3 \times 10^{22} kg , V_{pluto,min} = 3710 ms^{-1}$$ Gives: $$P_{min} = M_{pluto}V_{pluto,min} = 4.8\times 10^{25}kgms^{-1}$$ So you would need to hit it head on with something very massive to stop it entirely ($\frac{P_{min}}{c} = 1.6\times 10^{17}kg$ where $c$ is the speed of light). However you needn't use Pluto and you needn't stop the object entirely. I don't know much about the future you're creating, perhaps you have some death star like ship which can travel at the speed of light to get from system to system. That would do it I believe (see below).

If you had a large ship (perhaps a mining ship to make the excuse of why the ship was so heavy) crash into Pluto or some other fairly slow moving object then this could cause the effect you need. Perhaps a ship is rushed off badly repaired, the trajectory is badly calculated or some such and leads to it coming into the solar system head on with something.

However it happens this method means we can have the huge impact needed but have it occur on another celestial object and so not wipe out those on earth. It does still require a huge impact though.

• Based on the death star being 160km across, a volume of $V = \frac{2\pi 160^{3}}{3} = 8.6\times 10^{6} km ^{3}$, lets say that is mostly empty space and 10% some material like steel with a density of $\rho = 7850 kg m^{-3}$ so our death star has a mass of $\rho V = 6.7\times 10^{19}kg$ obviously this is all just speculative.
• In MathJax you are formatting units as variables. That's wrong. Try \mathrm or similar. – JDługosz Apr 27 '17 at 20:08

An "artificial" disaster simply needs to be "being-made" (similar to man-made, but not species specific). In order to produce the energy required by kingledion's answer, there are two strong possibilities:

One - an artificially produced, but sufficiently massive black hole or wormhole with gravitational qualities.

Two - a planetoid-ship or other powered/movable object of sufficient mass to interact with the Earth (possibly using a slingshot effect) to shift the Earth's orbit.

• Tacking the word "artificial" in front of a planetary or stellar body seems like a cop out. The question could benefit from a tech level. – Schwern Dec 30 '16 at 3:21
• @Schwern I think the question is rather unclear about what it means by "artificial," so I don't see a problem with this answer in principle. – jpmc26 Dec 30 '16 at 7:22
• @Schwern I refer you to the Dahak series. As no tech level was specified, even theoretical artificial events qualify for the question. Anything you can slap an engine or controllable motive force onto where said object is able to withstand the stress of controlled and variable acceleration pretty much counts. Should I bring up certain anime with mecha so large they use galaxies for bullets? ^^ I thought I was being fairly restrained in my answer. =P – nijineko Dec 30 '16 at 17:55

Something that can push the earth out of orbit is if large parts of it's mass was converted into energy. Nuclear power that humans can produce isn't going to do it but artificial miniature black holes are theoretically possible from high energy events subatomic collision events. They can possibly eat small part of the planet and then starve itself if no more matter is added. The black hole will then slowly release energy due to Hawking Radiation until they explode in a gamma ray burst.

It doesn't actually take much since all mass in this situation is converted into energy from E= MC^2, it will only take about 10^11 KG of matter to converted into energy to make the 10^22 KJ of energy from @kingledion's answer. This is less than the mass of a small mountain.

A event like this will likely take a long time for the black hole to gain and then lose the mass and might leave the side of the Earth that faces the black hole scorched due to energy released from the gamma ray explosion.

## protected by Community♦Jul 14 '17 at 12:06

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).