When there is a large object in the bottom of the ocean of height h, the surface of the ocean rises by a height less than h. In fact, if the ocean depth >> h, then the surface will rise negligibly. (Assume the object is not massive enough to cause changes in the gravitational field.)

1. What fluid mechanics equation governs the rise in sea level due to the object in the bottom of the ocean?
2. Why is the rise in sea level much less than h?

  • $\begingroup$ Hi....Hint.....why does exactly the same effect happen when you put a brick into a small bucket of water? Think about the ratio of the density of water compared to the density of the brick...best of luck $\endgroup$ – user74893 Mar 20 '15 at 4:38
  • $\begingroup$ @irishphysics Density has much to do with it as long as the density of the object is greater than that of the sea water (so that the object isn't buoyant). The important thing is the object's volume. $\endgroup$ – Dai Mar 20 '15 at 5:26
  • $\begingroup$ The question is more than a little vague: what is the object's "largeness" relative to? The object may be large, but the oceans are vast... $\endgroup$ – Dai Mar 20 '15 at 5:27

It has to do with the volume of the object, the height is irrelevant. If I put a tower that reaches to the bottom of the ocean but has negligible width and thickness then the tower displaces almost no water. The water height raise will be determined by the displaced water. An equation for the water rise:

$h = \frac{V_{obj}}{Surface Area}$

Where the surface area is the area of the top of the ocean.

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