While I was thinking about how tidal forces can make objects float at the surface of a planet orbiting a massive object like a black hole, the fact that any material on the Earth isn't held together by gravity only, but also by chemical bonds which give it its tensile strength came into my mind.

The scenario I thought of is as following:

1- A 10 kg metal sphere is tied to a 1 meter long thin rope and this rope is nailed into the surface of the Earth.

2- The tensile strength of the rope is 10 N. (So the metal sphere has to accelerate at 1 $m/s^2$ to break the rope)

3- The Earth is orbiting a black hole at its Roche radius, and so our metal sphere on the surface of the Earth is effectively weightless and floating, but it is still held by the rope.

Here is a simple picture to summarize:

enter image description here

Now, if we move the Earth to orbit the black hole even closer until the tidal forces on Earth due to the black hole become $\Delta a$ = 10.8 $m/s^2$ and so the metal sphere is pulled by the difference between $\Delta a$ and the Earth's gravitational acceleration (9.8 $m/s^2$) which is 1 $m/s^2$ towards the BH, will the rope break or not ?

In other words, if a is the gravitational acceleration towards the BH, in order to break the rope, which one do we need ? :

1- $a_{Point A}$ - $a_{Point B}$ = 1 $m/s^2$


2- $a_{Point A}$ - $a_{Point C}$ = 1 $m/s^2$

  • $\begingroup$ I think that the Earth orbiting the black hole at its Roche radius and then moving the Earth towards the black hole is fraught with danger? For the problem as stated I would worry about the rock holding the rope fracturing rather than the rope breaking. $\endgroup$
    – Farcher
    Jan 28, 2016 at 11:26

4 Answers 4


Between points A and C. The effect of having the Earth there is that it is providing a link between the center of mass of the Earth (C) and the test sphere (A). Effectively, there is an infinitely strong, rigid rod going from B to C, which is then connected to the rope. The tensions in that rod (the Earth) must match that of the rope.

Another way to think about it, is that if you removed the Earth entirely from the problem, and wanted to see if the rope was strong enough to resist the tidal forces --- then you would consider the difference in acceleration between points A and B.

  • $\begingroup$ You mean there is an infinitely strong rigid rod going from C to B, correct ? $\endgroup$ Jan 26, 2016 at 20:53
  • $\begingroup$ No. A to B --- the Earth is acting as the rigid rod, conveying the force to the rope. $\endgroup$ Jan 26, 2016 at 21:19
  • $\begingroup$ But A to B is the rope itself. You said the Earth is the rigid rod. This rod should be represented by C to B and not A to B. Isn't that what you mean ? $\endgroup$ Jan 26, 2016 at 21:45
  • $\begingroup$ @AbanobEbrahim Damn, yes, you're absolutely right --- sorry about that! $\endgroup$ Jan 26, 2016 at 21:49
  • $\begingroup$ What confuses me is whether Point B is actually accelerating with the acceleration of Point C or Point A. I mean, the rope is very short compared to the radius of the Earth that if Point B is accelerating, it will be doing so at roughly the same acceleration as Point A, which means that the entire rope is accelerating and won't break. So, is point B accelerating with the acceleration of Point C or Point A ? $\endgroup$ Jan 28, 2016 at 0:09

Between points A and C The earth can be considered as a wall and thus to break off the 'wall' a relative acceleration with respect to the wall is a must , If the tensile strength is 10N and mass 10 kg then the rope must move at a relative acceleration of 1m/s^2 or

Acceleration(a) - Acceleration(c) = 1


The acceleration of the points $B$ and $C$ will be same assuming that the Earth is not breaking apart, i.e., the acceleration of all points on the Earth will be same. If you think of the Earth as a very strong rod, only the Tension in the rod varies so that the entire rod accelerates with the same acceleration as that of the center of mass of the rod.

So, for the rope to break, we need -

${ a }_{ A } - { a }_{ B } = 1 m/s^2$


${ a }_{ A }-{ a }_{ C } = 1 m/s^2$

As ${ a }_{ B }={ a }_{ C }$


You can't add forces on different objects to get an acceleration of one object. All the forces should act on the center of the sphere to have effect on breaking the sphere. All the three forces: the tidal force $F_{T}$, the gravitational force $F_{g}$ and the Tension force because of the rope pulling the sphere $T$ act at the center of the sphere for the sphere to move intact without breaking apart. Mathematically, $$F_{T} - F_{g} - T = m_{s}a_{s}$$ where $m_{s}$ is mass of the sphere and $a_{s}$ is acceleration of the sphere towards the Blackhole.

$$\Rightarrow m_{s}a_{T} - m_{s}g - T =m_{s}a_{s}$$ where $a_{T}$ is the tidal acceleration on the sphere. which gives us, $$a_{T} - a_{s} = \frac{m_{s}g + T}{m_{s}}$$ $a_{T}$ and $a_{s}$ act at the center of the sphere. g is acceleration due to gravity and acts again at the center of the sphere not at points A, B and C.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.