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As I understand, an object can contract under its own gravitantional pull. Then as it does so, different parts of the object must move with some velocity. This in turn means that their relativistic mass will increase. Well we would have neglected this as the speed of the contracting matter might be small. But then the increase in mass leads to an even more increase in gravitating force. This leads to an even more contraction, according to Hook's law. This seem to mean that elastic objects will keep on contracting ad infinatum. Or in other words, relativity seems to predict that all objects will eventually become blackholes with time. Is this correct?

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In short: No.

In addition to gravity pulling things together, the electroweak force pushes things apart. The more you push things together, the more the atoms the things consist of will resist the pushing together, meaning that unless the gravitational pressure is sufficient, gravitating objects are stable (asterisk).

Essentially, your initial premise is false: An object reaches an equilibrium where the acceleration due to electrostatics cancels out the acceleration due to gravity, and the various parts of the object are stationary wrt. one another.

Edit: For everyday matter, these are the primary, although not only, forces in play. As you move into the realms of degenerate and exotic matter, things get more complicated, but similar considerations exist for e.g. neutron stars, preventing them from spontaneously collapsing into black holes as well.

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  • $\begingroup$ So the term k, appearing in Hook's law is not realy a constant but a function of preasure? $\endgroup$
    – user58594
    Commented Mar 2, 2015 at 15:01
  • $\begingroup$ @RoaringHarmless: Well, no. But Hooke's law relates to springs, not gravitational compression of massive bodies. I suppose the compression due to gravity could be expressed in terms of Hooke's law, but as the body is compressed, the properties of the model spring, and hence k, would change. It's just not a good model for this case. $\endgroup$ Commented Mar 2, 2015 at 15:08

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