Small enough planet to notice rounding This is my first question on Physics StackExchange, so bear with me.
I am wondering. How dense would a planet need to be for a human to notice the fact that the planet is not flat, but round, by looking at the horizon, whilst maintaining a natural feel (by means of gravitational force).
Here on earth it is hard to notice the rounding of our planet, but could such a planet exist?
 A: The question isn't well defined, but I think (see details below) that it is $> 10^{7}$ kg/m$^3$. 
Horizon
If a planet is smooth, then the fact that there is a horizon tells you that it is not flat. We can obtain a general formula for horizon distance versus density for a planet with surface gravity the same as Earth.
The distance to the horizon as seen from something $h$ high (assuming a perfectly smooth, spherical planet) is
$$d = \left( [R+h]^2 - R^2 \right)^{1/2},$$
where $R$ is the planet radius. 
the mass of the planet is given in terms of surface gravity by
$$ M = \frac{g R^2}{G}$$
and the average density is
$$\rho =\frac{3M}{4\pi R^3}$$
Thus
$$d = \left( \left[\frac{3g}{4\pi G\rho} +h\right]^2 - \left[\frac{3g}{4\pi G\rho} \right]^2 \right)^{1/2}$$
If we assume that we are only talking non-absurd situations, where $h \ll R$, then this can be approximated as
$$d \simeq \left(\frac{3gh}{2\pi G\rho}\right)^{1/2} =11.8 \left(\frac{h}{2\ {\rm m}}\right)^{1/2} \left(\frac{\rho}{1000\ {\rm kg\ m}^{-3}}\right)^{-1/2}\ {\rm km}$$
Or turning this around, the required density for the horizon distance to be $d$ is
$$ \rho = \left(\frac{3gh}{2\pi G}\right) d^{-2}$$
e.g. If you, with $h \simeq 2$ m, wanted $d <100$ m, to make it obvious you were on a spherical planet, then $\rho > 1.4 \times 10^{7}$ kg/m$^3$. 
Curvature
So you could define a perceptibly spherical planet in terms of this, but planets aren't smooth. A better way to do this might be in terms of the visual curvature of the horizon, defined as
as
$$ \kappa = \left( \left[1 + \frac{h}{R}\right]^2 -1\right)^{1/2},$$
where $\kappa=1$ corresponds to an apparent circle of angular radius 45 degrees.
Assuming again that $h\ll R$, surface gravity is $g$ and you are observing the horizon from height $h$, then
$$\rho = \left(\frac{3g}{8\pi G h}\right) k^2$$
As a "perceptible" threshold, we could take $\kappa > 0.05$, which is the apparent curvature of the horizon as seen from a typical aeroplane flight on Earth (there is at least some evidence that this is about the right threshold, but it might need to be a bit larger). With $h= 2m$, this requires  $\rho > 2.2\times 10^{7}$ kg/m$^3$ and a planet with $R< 1.6$ km.
