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The definition of the Roche limit is: "The distance within which a celestial body, held together only by its own gravity, will disintegrate due to a second celestial body's tidal forces exceeding the first body's gravitational self-attraction". I can't say that I understand that perfectly, but let me try to explain why I am confused.

It is stated here that the Roche limit for the Sun-Earth system is 556,397 kilometers, and ignoring the fact that this distance is actually inside the Sun, I understand that this is the distance at which a test object placed on the surface of the Earth will be pulled equally between the Earth and the Sun, and decreasing this distance would lead that test object to start accelerating towards the Sun.

The other thing I thought of, and which is why I am confused, is the gravitational acceleration. I have calculated through $a=GM/r^2$ that an object on the surface of the Earth will have the same gravitational acceleration of 9.8 $m/s^2$ towards the Sun if the Earth is 3,680,000 kilometers away. Which is more than 6 times the Roche limit.

Now in both cases, it means, or at least how I understand it, that an object on the surface of the Earth is pulled by the same amount of force towards the Earth and the Sun, and getting closer or further will determine towards whom the object will accelerate.

So, why aren't the two numbers the same if they mean the same thing ?

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  • $\begingroup$ How much pull does the Earth have on the object if the Earth is 3,680,000 kilometers away? $\endgroup$ – immibis Jun 1 '16 at 4:07
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The Roche limit is a tidal effect. It is due to the fact the Sun pulls more strongly on the near side of the Earth than on the far side.

Suppose the Earth is at some distance $d$ from the Sun, by which we mean that the distance from the centre of the Sun to the centre of the Earth is $d$. The acceleration due to the Sun's gravity is then just given by Newton's law:

$$ a_{centre} = \frac{GM}{d^2} $$

But the near surface of the Earth is closer to the Sun than the centre of the Earth. The difference in the distance to the Sun is obviously just the radius of the Earth $r$. That means the acceleration of the surface due to the Sun's gravity is:

$$ a_{surface} = \frac{GM}{(d - r)^2} $$

The difference is $a_{surface} - a_{centre}$:

$$ \Delta a = \frac{GM}{(d - r)^2} - \frac{GM}{d^2} \tag{1} $$

Normally objects at the surface of the Earth experience the usual gravitational acceleration of 9.81 m/s$^2$. However as the Earth gets closer to the Sun this acceleration is reduced by $\Delta a$. When $\Delta a$ reaches 9.81 m/s$^2$ objects at the surface are effectively weightless and can float away.

So to calculate the Roche limit we simple set $\Delta a$ in equation (1) to 9.81 m/s$^2$ and solve for $d$, and the result comes out at about $d$ = 559,000 km. This is slightly different to the figure you got from Wikipedia because the calculation I've described is actually only an approximation and there are a few corrections needed for a really accurate result.

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  • $\begingroup$ Doesn't the same happen when the Earth is 3,680,000 kilometers away from the Sun ? At this distance, the gravitational pull of the Sun for an object on the surface of the Earth will be as strong as the pull from the Earth, and so the object will be weightless too. So what's the difference ? $\endgroup$ – Abanob Ebrahim Sep 24 '15 at 16:37
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    $\begingroup$ @AbanobEbrahim: No. At 3,684,000 km the acceleration due to the Sun's gravity at the Earth's surface is indeed about 9.81 m/s$^2$, but the acceleration due to the Sun's gravity at the centre of the Earth is about 9.78 m/s$^2$. So the difference is only about 0.03 m/s$^2$. The effect is to reduce the acceleration felt by objects at the surface by 0.03 m/s$^2$ (i.e. reduce their weight by 0.03m Newtons), which isn't going to make much difference. $\endgroup$ – John Rennie Sep 24 '15 at 16:45
  • $\begingroup$ It makes sense now. I didn't take the acceleration due to the Sun's gravity at the center of the Earth into consideration, and that is why I thought they meant the same thing. Thank you. $\endgroup$ – Abanob Ebrahim Sep 24 '15 at 16:59
  • $\begingroup$ One last thing, if $\Delta a$ is equal to 19.6 $m/s^2$, would this mean that an object on the surface of the Earth will be accelerating away from the surface at 9.8 $m/s^2$ ? $\endgroup$ – Abanob Ebrahim Sep 24 '15 at 18:02
  • $\begingroup$ @AbanobEbrahim: yes, though of course the Earth will have fallen to bits by the time it gets that close. $\endgroup$ – John Rennie Sep 24 '15 at 18:03

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