Potential energy of a single point-mass in an upright object Does a point-mass on one end of an object sitting upright have gravitational potential energy?  
An example would be a balanced pencil.  Does a point-mass on the top end of the pencil have gravitational potential energy?
 A: Nice question. Elementary work done by force is $$ dW = F \,dx  $$
Substituting for a gravitational force and integrating both sides gives :
$$ W = \int G \frac {Mm}{x^2} dx $$
Now changing it into definite integral (work needed to move mass by gravitational field from $R$ to $r$ distance) and noticing that $W = E_{pot}$, gives :
$$ E_{pot} = \int_R^r G \frac {Mm}{x^2} dx$$
which is :
$$ E_{pot} = GMm \left(\frac 1R - \frac1r\right) $$
where $r$ is your point mass current distance from gravitational body center of mass (Earth COM this time) and $R$ is zero gravitational energy distance, i.e. distance from a gravitational body COM where you consider $E_{pot} = 0$.
So the final answer it depends how you choose zero energy point $R$, and it can be arbitrary point. In case :


*

*$R=r \,\,\,\,\,\,\to E_{pot} = 0\,\,\,\,\,$ (As in $E_{pot}=mgh$, when $h=0$)

*$R=\infty \,\,\,\to E_{pot} = -G\,M\,m\,r^{-1}$

*$R=0 \,\,\,\,\,\to E_{pot} = \infty \,\,\,\,\,\, \text{with any}\,\,r$ (Almost never used because of singularity)


So choose your zero point wisely.
EDIT
As @looksquirrel101 correctly pointed out, singularity at point $R=0$ can be removed by other means.
A: Yes, gravitational potential energy ( assuming it's only affected by the gravitational field of the earth) of a point mass exists at any place in the Universe. As on the top of a pencil. 
It is given by the formula :
$U(r)$ = $\frac{- GMm}{r}$
A: 
The potential energy is:
$$U=-mgL $$
With:
$$mg=\dfrac{mMG}{\left( R+L\right)^2}\approx\dfrac{mMG}{R^{2}}$$
Thus $g=\frac{MG}{R^2}$
And
$$U=-mMG\left( \dfrac{L}{R^{2}}\right) $$
