Could there be a planet so big that it is bigger than itself?
June 18, 2009 11:02 AM   Subscribe

Would it be possible to have a planet with significantly more surface area than the earth and similar gravity (at ground level) by decreasing its density?

I know that gravity decreases with the square of the distance, so does that mean that the earth is in the sweet spot for the size/density necessary to produce 1G? Or could that be scaled up to double or triple the surface area and still have a viable (livable) planet?
posted by blue_beetle to Science & Nature (5 answers total) 2 users marked this as a favorite
It is indeed possible. 'g' scales with the mass of the planet divided by the square of the radius, so if you double the radius and quadruple the mass - 'g' stays the same.

Incidentally, if you double the radius and quadruple the mass:
-You also quadruple the surface area
-The density is 1/4
posted by atrazine at 11:21 AM on June 18, 2009

In short, yes. The recently discovered TrES-4 is 70% larger than Jupiter, but has only 3/4th's of its mass: the planet has the density of cork.

TrES-4 is an oddity, but is not completely unique in this regard - Saturn, for example, is lighter than water, and would float in an appropriately sized cosmic bathtub.

However the habitability of such planets, at least for human beings, is very much in question. There would likely be little or nothing in the form of a solid or liquid surface in order to develop on.

There are some speculative treatments of the idea - The Integral Trees, by Larry Niven, and the concept of a Dyson Sphere.
posted by Bora Horza Gobuchul at 11:29 AM on June 18, 2009

Best answer: Actually the density is halved. (g = GM/r2, but M = 4πρr3/3 => g ~ ρr (for uniform density), so if g is constant ρ ~ 1/r)

Several planets in the solar system have a similar surface gravity to Earth even while differing radically in mass: Saturn's g is only about 14% greater than Earth's even though it has a mass 95 times that of Earth. Saturn is less dense than water.
posted by Electric Dragon at 11:38 AM on June 18, 2009

Best answer: You might also consider that the gravity field outside a hollow sphere of mass M is identical to the field outside a solid sphere of mass M. If we want constant g (same gravity as Earth) and constant ρ (same density as Earth, at least where there's material), then you can consider the two equations for a hollow sphere:
  • g = GM/ro2
  • ρ = 4 π (ro3 - ri3)/(3 M)
with ro and ri the inner and outer radii of the hollow sphere, respectively. (ri = 0 for Earth, or other solid sphere of uniform density.) Then we can apply a scaling law to a solid sphere for a factor c > 1 to make a hollow sphere that keeps g and ρ constant:
  • ro' = ro c
  • M' = M c2
  • ri' = ro c (1 - 1/c)1/3
With this, you can make hollow Earths of any size you want (neglecting material properties, non-uniform density, etc.).

Also, following up on Bora Horza Gobuchul's point, the Extrasolar Planets Encyclopaedia has a list of all known planets, and their masses and radii (if known), if you'd like to find more for yourself. You won't find many near the size of Earth, though (excepting maybe GJ 581e).
posted by Upton O'Good at 12:45 PM on June 18, 2009

Response by poster: One of the criteria would have to be that the planet should be dense enough on the surface to walk on (and construct buildings). Low density planets wouldn't necessarily be habitable.

Thanks for the tips everyone!
posted by blue_beetle at 11:41 AM on July 6, 2009

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