Can a Celestial body have a mountain rise all the way to Synchronous Orbit? On Earth, the tallest mountain is Mauna Loa. In the solar system, to date, the tallest mountain is Mons Olympus. Both fall far short of GSO.
Mars gravity is 0.376 of Earth gravity. If Mauna Loa were on Mars it would have an altitude of ~80000 ft. If Mons Olympus were on Earth it would similarly be no more than ~27000 ft. Yet these are almost comparable altitudes. 
Is it possible for a mountain to rise as high as GSO? Are all known tallest peaks on planets in the Solar System similarly limited?
 A: Is it possible for an object in the solar system to have a bulge large enough to create weightlessness on the top of that bulge? 
Absolutely. Heck, you don't even need a bulge. A perfectly sperical object of mass $M$ and radius $R$ revolving around its axis in a period $t$ would have zero gravitational acceleration at its equator if the gravitational acceleration matches the centripetal acceleration:
$$\frac{G M}{R^2} = \frac{(2 \pi R)^2}{R t^2}$$
Defining the density $\rho = \frac{4 \pi M}{3 R^3}$, this equatorial weightlessness condition can be written:
$$G \rho t^2 = 3 \pi$$
A spinning spherical grain / boulder / asteroid with density $\rho = 6 \ 10^3 kg/m^3$ revolving around its axis roughly every 1.5 hrs ($5 \ 10^3 \ s$) would do.
Note that the size of the object doesn't enter the picture. The rotational speed and mass density is all that matters. For non-spherical (disc-shaped) objects the above equation still gives a good approximation if you substitute for $\rho$ the object's mass divided by the volume of the smallest sphere enclosing the object.
