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So let's say that an asteroid the twice the size of Earth passed by going relatively slow say in between the Moon and Earth.

Among so many other things (and perhaps you can touch on some of the biggest 'other' things), would anything on the asteroid's side of Earth be essentially 'sucked' towards the asteroid. That is if I am standing outside and it passes would I lift off the ground and float towards it?

Conversely, would things on the opposite side experience two times the standard weight? That is if I am standing on the other side, would I feel heavier as the asteroid passes?

I realise there would be many other devastating effects but I'm most interested in this one...

The possible answer I suspect is that objects on Earth will not 'feel' anything as the Earth as a whole will move towards the asteroid and not every little thing inside Earth...

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  • $\begingroup$ The 1951 movie When Worlds Collide includes some reasonably accurate special effects footage of the sort of things that would happen if the earth were approached that closely by an earth mass planet. Your 2X the mass scenario is even nastier. imdb.com/title/tt0044207 $\endgroup$ – Wayfaring Stranger Jan 6 '15 at 22:16
  • $\begingroup$ I realize there would be many other devastating effects The worst one being that the Earth is pulled out of its orbit, along with the moon. Earth's new orbit may have extreme weather so no life could survive. It would also be extremely likely that the moon will collide with the Earth, ending all life. $\endgroup$ – LDC3 Jan 7 '15 at 1:26
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    $\begingroup$ Something that big would be a planet, not an asteroid. $\endgroup$ – Keith Thompson Jan 7 '15 at 16:56
  • $\begingroup$ The effect on Earth's orbit would depend on the trajectory of the visitor. A quick fly-by wouldn't impart enough momentum to the Earth to substantially change our orbit, but a slower pass would (but implies that the visitor was already on a similar orbit to Earth's, raising many questions). A collision with the moon would also be extremely unlikely, though disruption of the moon's orbit and consequently of tides would be possible. $\endgroup$ – Russell Borogove Jan 7 '15 at 22:38
  • $\begingroup$ So, I'm not sure this deserves a whole NEW question, but if that rogue planet were to scream through the solar system, say at 500,000mph, or just under 7.5 times the speed at which the Earth revolves around the sun. Let's say it still threads the needle and goes between the Earth and the moon. Would this be fast enough to get between the Earth and the moon without causing either the Earth or moon to hit the rogue planet? And if so, wouldn't this still cause the moon to collide with Earth? $\endgroup$ – Jimmy G. Jan 3 '16 at 2:34
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In a close approach, Earth would feel tidal forces: the near side to the approach would be affected by the vistor's gravity more than the far side, because the force of gravity decreases with distance from a mass. If you were standing on the far side, though, your effective weight wouldn't increase -- it would decrease! This is because the visitor would be pulling the center of the Earth more strongly than it would be pulling on you, again because of the gravitational falloff with distance.

In a very close approach, closer than the Roche limit, the tidal forces would overcome the Earth's gravity completely, and Earth would begin to break apart. So it matters very much how close an approach we're talking about. Assuming the visitor has the same density as the Earth, the critical distance is about 10 Earth radii, or 64000km -- about a sixth of the distance from the Earth to the moon.

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The Earth wouldn't simply split in half, it would be pulled towards the asteroid as a whole. To simplify matters, let's assume that the mass is equal to twice that of Earth (size alone is not enough, you must also consider density). You can then use Newton's law of gravitational attraction for the calculations:

$$F=G \frac{m_1m_2}{r^2}$$

$G$ is a constant, $r$ is half the distance between the moon and Earth, $m_1$ is the mass of Earth, $m_2$ is twice that. The gravitational force would be about $1.3 \times 10^{29} \text{ N}$. In comparison, The gravitational attraction between Earth and the moon is about $2.0 \times 10^{26} \text{ N}$. Which means that the attraction Earth-meteorite will be 2000 times larger than Earth-Moon. The Earth will speed rapidly toward the asteroid and crash into it. During that flight Earth indeed is to be expected to have gravity fluctuations and probably lose major parts of its atmosphere. However it is unlikely that Earth, being a telluric planet (made of rock and metal) would subdue major deformation. After all, Jupiter's moons are round, even though they are in much more extreme conditions.

You would however need to consider the velocity of the meteorite for any more precise estimations. The speed and the trajectory count. An object that passes faster would tear the Earth out of its orbit but it would continue moving, whilst if you just place that meteorite between Earth and Moon, Earth (and Moon) would simply fall on a perpendicular trajectory.

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