Several years ago a student asked why so many things in the solar system were round. He noted that many objects in the solar system, although not all, are round. The standard answer, which he knew, is that the mutual gravitational attraction of the molecules pulls them into the shape that gets them as close to each other as possible: a sphere. This argument works fine for fluid bodies such as the Sun or Jupiter, but it isn't so simple for a solid object—we have all seen rocks that are not round. There is still a gravitational attraction acting between the rock's molecules, butfor small rocks that force does not overcome the strength of the bonds holding those molecules in their relative positions. Since the strength of the gravitational force grows with the size of the object, a large enough rock will have a strong enough gravitational attraction to force a deformation into a round shape. But how large is that? A simple model gives an answer to this question. There is also renewed interest in this topic as a result of the new definition of a planet approved by the International Astronomical Union, which says in part, “A ‘planet’ is a celestial body that… has sufficient mass for its self‐gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape.”1 What size object is large enough to satisfy this criterion? Where does Pluto fall regarding this question?

1.
International Astronomical Union, www.iau.org, press release, 24 August 2006.
2.
Liffman, Kurt et al., “Stress in Sandpiles,” Second International Conference on CFD in Minerals and Process Industries, CSIRO, Melbourne, Australia (1999).
3.
nssdc.gsfc.nasa.gov/planetary/. This is NASA's web page for physical data on the solar system.
4.
The value used for the compressive strength of rocks came from a slideshow available at www.usouthal.edu/geography/allison/GY403/GY403_Lecture7_DynamicAnalysis.pdf. Another site with a table showing the compressive strength for a variety of common materials (but not rocks) is www.efunda.com.
5.
Students may be concerned that since much of the mass of Mauna Kea is below water there will be a significant buoyant force reducing the pressure at the base. Buoyant forces are due to the pressure of the surrounding fluid pushing up on the bottom of an object being greater than the fluid's pressure pushing down on the top. Since there is no water pushing up on the bottom, the buoyant force does not enter into the calculation. The water above the flanks of Mauna Kea does still push down, adding to the pressure at the base, but this additional weight is small compared to the weight of the mountain itself.
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