# What would happen if two Earth size planets very slowly collided?

1. Sep 9, 2011

### blarznik

Hypothetically this pseudo-Earth suddenly appears and follows the same orbit as Earth, but moves slightly faster. If it eventually collides with Earth at snail pace what would the gravitational effects be like? (maybe ignore the moons)

Last edited: Sep 9, 2011
2. Sep 9, 2011

### CosmicEye

You mean like earth #2 chasing earth #1? well, it would be one hell of an unforgettable sight for a few weeks as the planet approaches. I assume E1 would slow down in it's orbit around the sun as E2 gains speed towards us. Can you imagine the sight as it passes the moons orbit? Scary but fascinating. I picture the planets from the movie Pitch Black, except we all die a feiry explosive death from a planet collision.

3. Sep 10, 2011

### Staff: Mentor

I believe the two planets would pull each other towards themselves and eventually both would be falling towards each other at 9.8 m/s^2 right before they hit.

4. Sep 10, 2011

### Pengwuino

That approximation fails magnificently when you start talking about entire earth-sized objects on a collision course :)

5. Sep 11, 2011

### cepheid

Staff Emeritus
Well, since g = GMearth/Rearth, and the distance between the centres of the two planets would be 2Rearth right before they hit each other, one would think that the acceleration at that instant would be g/2, no?

This is assuming that the planets have not only the same mass, but also the same size.

EDIT: Nevermind. At this point they really can't be considered to act as point masses and you have to take into account tidal effects and all that...

6. Sep 11, 2011

### Monsterleg

I had this big chart with a bunch of the stats of gravitational acceleration for both planets at certain distances and velocity changes and all that jazz took a while to do on my computer calculator. But alas I had most of it on an Excel chart and blam crashola program not responding, I was on a roll and hadn't saved in a while... My initial obsessive motivation has mostly drained out of me for now. Might pick it up again tomorrow.

I did make a few mental notes though. Two big factors came in to play: initial distance between Earth 1 and Earth 2, and time until impact. You have any idea what you want those to look like? Either one would do nicely I assume this is for a story. If not and it's just for your own curiosity I'll just throw something together saying they started like 630,729,000m apart moving at the same speed.

I think at point of impact it would be more like an acceleration of 39.2266 m/s2 because Earth 2 is accelerating at 19.6133 m/s2 due to its pull on Earth 1 (9.80665 m/s2) plus Earth 1's pull on it (9.80665 m/s2) and at the same time Earth 1 is decelerating at 19.6133 m/s2 due to the same factors. I could be wrong though.

But regardless the acceleration at point of impact isn't really what matters because you take into account the acceleration due to gravity across the vast distances of space between them and it continually builds up decelerating Earth 1 and accelerating Earth 2 until the impact and at that point it doesn't really matter how snail pace faster Earth 2 was moving or even if Earth 2 was moving slower initially. By the time they collide the impact due to their velocity differential would be quite something. This all of course depends on the initial distance between them.

I was mostly using the equation gh=go(re/(re+h))2 and the variations I derived from it would that equation be correct? Or would I have to double re because there are two Earths?

gh= Earth's gravitational acceleration on a object at a distance outside Earth's radius m/s2
go= Earth's standard gravitational acceleration of an object in m/s2 (usually 9.80665m/s2 but I used both 19.6133m/s2 and 39.2266m/s2 to get results due to 2 Earths being present)
re= Radius of Earth in meters
h= Distance in meters of object outside of Earth's radius

Oh and all my work is disregarding the whole orbital part I haven't really studied orbital acceleration and orbital factors very extensively yet. Also disregarding spin and the influence of the moons. That's just a bit much for me right now.

7. Sep 11, 2011

### Pengwuino

Yup it gets a bit crazy :D

8. Sep 11, 2011

### Staff: Mentor

Yep. I don't know the math to say anything else more correct though.

9. Sep 11, 2011

### Janus

Staff Emeritus
If you want it to hit the Earth, you don't want it appearing in the same orbit as the Earth.

Here's why. As it begins to approach the Earth from behind, the Earth's gravity starts to pull forward on it. But a forward force acting on a orbiting object pushes it into a higher orbit. higher orbits are slower orbits.
Meanwhile, the planet is pulling backward on the Earth, causing it to fall into a lower faster orbit
So what happens is that before the planet reaches the Earth it will be pushed into an orbit that is higher and slower and the Earth drifts into a lower faster one and will begin to pull away from the planet.

Fast forward up to the point where the faster Earth is catching up to the slower planet. Now the Earth pulls back on the planet and the Planet pulls forward on the Earth. The Earth drifts into the slower orbit and the planet into the faster one. The planet pulls away until it begins to catch up with the Earth again and we are back where we started.

This is called a "horseshoe" orbit, from the the fact that the path of one planet looks like a horseshoe from the perspective of the other.

10. Sep 11, 2011

### Staff: Mentor

Can you elaborate on this a bit? If one object is slowed down and pulled into a "faster" orbit, does that object take a full revolution to get to the required speed for that orbit? And why wouldn't an object that is sped up in it's orbit be lost? How does it slow down to get into the slower orbit?

11. Sep 11, 2011

### cepheid

Staff Emeritus
I imagine not. As soon as it descends "deeper" into the sun's gravitational potential well, the potential energy it loses is turned into kinetic energy gained.

Same reason as above, only opposite. It expends energy climbing higher out of the sun's potential well.

12. Sep 11, 2011

### tony873004

13. Sep 11, 2011

### Staff: Mentor

So if it lost 100 m/s and fell towards the sun, then it gained that 100 m/s plus extra back? That makes sense. As does the opposite effect. Thanks!

14. Sep 13, 2011

### Android Cat

What happens when Earth1 and Earth2 pass within each others' http://en.wikipedia.org/wiki/Roche_limit" [Broken]?

Last edited by a moderator: May 5, 2017
15. Sep 13, 2011

### phinds

I don't know this stuff well so sure could be wrong, but my understanding is that the Roche limit has to do with loosely bound objects such as comets and would not apply to strongly bound objects such as a planet.

16. Sep 13, 2011

### Android Cat

There would still be huge tidal stresses, massive earthquakes, and anything not nailed down (like oceans, atmosphere, people, soil...) would shift in a big way.

It reminds me of science-fiction stories (Rocheworld, heh) by Dr. Robert L. Forward of two planets which were so close that ocean and atmosphere would occasionally spill back and forth.

17. Sep 13, 2011

### Janus

Staff Emeritus
It would apply to any object that was massive enough that its own gravity determined its shape. (essentially any body that has pulled itself into a sphere).

18. Sep 13, 2011

### phinds

Since I don't know this stuff well, I have no trouble believing I have a misconception about this and maybe you can clear it up for me. Here's my confusion:

(1) It seems to me that a loosely cohesive grouping of rocks that had brought itself into more or less of a sphere would be torn apart far further out from a planet than would an object that was one solid rock (USED to be many rocks but has brought itself into a sphere then cooled off enough to no longer have a liquid core).
(2) The Roche limit is based on the mass of the planet and is independent of the approaching object.

If both of these were true then only one of the incoming objects could be torn apart specifically at the Roche limit, so I assume I'm wrong about one or the other, possibly both.

Thanks

19. Sep 13, 2011

### Janus

Staff Emeritus
The Roche limit for a rigid body would be less for that of a loose collection, yes. But since no object is perfectly rigid, the true limit will be somewhere between these two extremes. Even if an object no longer has a liquid core, its shape can be determined by gravity and it will flex somewhat under tidal forces.
The Roche limit is actually related to the ratio of the body's densities. Each body has a Roche limit in terms of the other. The less dense of two objects will start to break up when the objects are further apart then the denser object will. It is even possible, the ratio is large enough, for the Roche limit of one object to be inside the other.
In pretty much every case, one object will be smaller or less dense than the other, and this object will be the one that breaks up.

For example, the Roche limit for the Moon with respect to the Earth is 9,496 km (rigid body limit) but for the Earth with respect to the Earth, it is only 1932 km. Since this is the center to center distance, and this is smaller than the Earth's radius, you wouls not see the Earth being torn apart by the Moon.

If however, both bodies were of the same mass and density, then the Roche limit would be the same for both.

20. Sep 13, 2011

### phinds

Very clear. Thank you.