Gravity in other than 3 spatial dimensions and stable orbits I have heard from here that stable orbits (ones that require a large amount of force to push it significantly out of it's elliptical path) can only exist in a three spatial dimensions because gravity would operate differently in a two or four dimensional space. Why is this?
 A: Specifically what that is referring to is the 'inverse-square law', nature of the gravitational force, i.e. the force of gravity is inversely proportional to the square of the distance:  
$F_g \propto \frac{1}{d^2}$.
If you expand this concept to that of general power-law forces (e.g. when you're thinking about the virial theorem), you can write:
$F \propto d^a$,
Stable orbits are only possible for a few, special values of the exponent '$a$'---in particular, and more specifically 'closed1', stable orbits only occur for $a = -2$ (the inverse-square law) and $a = 1$ (Hooke's law).  This is called 'Bertrand's Theorem'.
Now, what does that have to do with spatial dimensions?  Well, it turns out that in a more accurate description of gravity (in particular, general relativity) the exponent of the power-law ends up being one-less than the dimension of the space.  For example, if space were 2-dimensional, then the force would look like $F \propto \frac{1}{d}$, and there would be no closed orbits.
Note also that $a<-3$ (and thus 4 or more spatial dimensions) is unconditionally unstable, as per @nervxxx's answer below. 

1: A 'closed' orbit is one in which the particle returns to its previous position in phase space (i.e. its orbit repeats itself). 
A: I'll try to answer it by considering radial deviations from a circular orbit. First we have to assume two things about our n-dimensional universe: Newton's second law still holds, that is,
for a particle's position vector in n-dimensions $\vec{x} = (x_1, x_2, \cdots x_n)$, 
\begin{align}
m \ddot{\vec{x}} = \vec{F},
\end{align}
where $\vec{F}$ is some n-dimensional force,
and also that the law of gravity is given by Gauss' law:
\begin{align}
\nabla \cdot \vec{g} = -4\pi G\rho,
\end{align}
where $\vec{g}$ is the gravitational force field. (See wikipedia for more information).
The solution to that pde is
\begin{align}
\vec{g} \sim 
= - r^{1-n} \hat{e_r},
\end{align}
for $n \geq 2$. (For $n = 1$ the motion is on a line and because it's always attractive the 'orbit' will still remain an 'orbit')
Since the motion will always be constrained to move in the 2-plane spanned by the initial radial vector $\vec{r}_0$ and the initial velocity vector $\vec{v}_0$, it is easiest to analyze the motion in cylindrical coordinates. That is, Newton's second law becomes
\begin{align}
m(\ddot{r} - \dot{\theta}^2r)&=F_r \\
m(r\ddot{\theta}+2\dot{r}\dot{\theta}) &= F_\theta \\
m \ddot{x_3} &= F_{x_3} \\
m \ddot{x_4} &= F_{x_4} \\
&\cdots \\
m \ddot{x_n} &= F_{x_n},
\end{align} 
where $x_1$ and $x_2$ are coordinates of the plane spanned by $\vec{v}_0$ and $\vec{r}_0$. Here $r$ really means $\sqrt{x_1^2 + x_2^2}$, but it turns out that because the motion is just 2-D i.e. $x_3 = x_4 = \cdots x_n = 0$, we can say $r = \sqrt{x_1^2 + \cdots + x_n^2}$.
Now we make use of the fact that gravity is always radial, so $F_\theta = 0$ and we can combine the first two equations to get
\begin{align}
\ddot{r} - \frac{L^2}{r^3} = F_r = f(r),
\end{align}
where $L$ is a constant of motion (in 3D this is the angular momentum).
For a circular orbit  at $r = r_c$, $\ddot{r} = 0$, so we are left with
\begin{align}
-\frac{L^2}{r^3} = f(r).
\end{align}
Consider small deviations from $r_c$: $x = r-r_c$. Plugging this into newton's law and expanding to first order, one gets
\begin{align}
\ddot{x} + \left[-3f(r_c)/r_c-f'(r_c) \right]x = 0.
\end{align}
This is a simple harmonic equation if the stuff in the parenthesis is positive. So we obtain a stability condition
\begin{align}
\left[-3f(r_c)/r_c-f'(r_c) \right] > 0.
\end{align}
Let's check this on a radial force $f(r) = -kr^d$. The stability condition gives
\begin{align}
-k r_c^d -\frac{kd}{3}r_c^d < 0,
\end{align}
which implies $d > -3$. So if the force law goes as $r^d$ where $d > -3$, then the orbit is not stable. One can, with a bit more work, show that $d = -3$ is also unstable.
So for dimensions $n \geq 4$, the orbit is unstable. It appears, however, that for $d = -1$ or $-2$, the orbit is stable, so this gives us the result that orbits in 3-dimensions (our world) and also that of 2-dimensions are stable, in disagreement with the video's statement. I might be wrong, though.
cheers.
A: One brief point to add to the answers posted above, even though I can't pretend to understand all the math:
As far as I know, orbits in 2D are stable in the sense the orbiting body does not escape or collapse to the primary. See e.g.
https://www.reddit.com/r/askscience/comments/q8fmo/what_would_orbits_look_like_in_a_2d_universe/
In fact, as the 2D force decays to 1/r, its potential is logarithmic -- meaning that escape velocity is infinite. This is rather easy to show; even I can do it.
However, if I have understood correctly, it seems that the orbits are usually not closed, but shaped like flower petals. For near-circular orbits that wouldn't necessarily be too much of an issue.
A: I assume this is talking about Newtonian gravity (i.e., not relativity).  Let's consider the effective potential:
$$V_\text{eff}(r) = \frac{L^2}{2mr^2} + V(r)$$
where $V$ is the ordinary potential energy, and $L$ is the angular momentum.  First, you may ask why the effective potential has this form.  Remember that for a single particle, $L = mr^2 \omega$, so this is equivalently,
$$V_\text{eff}(r) = \frac{\omega^2 r^2}{2m} + V(r)$$
This first term appears from the equations of motion for a free particle.  Phrasing it in terms of angular momentum is convenient because under central forces, angular momentum is a conserved quantity.
Why do we use the effective potential?  Because it helps us talk solely about the radial motions of a particle, lumping the angular motions in with the real potential.  A local extremum of the effective potential tells us about an equilibrium distance.
Now, in 3d, the potential $V(r)$ for gravity is $-GMm/r$.  What this means is that, as $r \to 0$, the effective potential will eventually blow up, thanks to the angular momentum part, overcoming the gravitational part and forcing the particle outward again unless it lies on a direct infall trajectory.
In 2d, the potential is different.  Why is this?  Newtonian gravity deals with differential equations of the form $\nabla^2 V \propto \rho$.  The point-source solution to this equation (the Green's function) is proportional to $\ln r$--compare, for example, the electric potential of an infinite line charge.  This is exactly the same geometry and differential equation, at least in structure.
Let's check for a second that this is the case.  Let $V = C \ln r$ in 2d for some constant $C$.  Then the gravitational force is
$$F = - \frac{\partial V}{\partial r} = -C/r$$
which is inward for all positive $C$.  This is important.  In 2d, then, our effective potential looks like,
$$V_\text{eff} = Kr^{-2} + C \ln r$$
for two constants $K, C$.  The force is
$$F_{\text{eff}} = 2 K r^{-3} - C r^{-1} = -r^{-1} (-2K r^{-2} + C)$$
So $r_\text{eq} = \sqrt{2K/C}$. But is this equilibrium stable?
$$\frac{\partial F_\text{eff}}{\partial r} = -6 Kr^{-4} + C r^{-2}$$
At $r_\text{eq}$, this evaluates to $-6C^2/4K + C^2/2K = -C^2/K$.
Hm.  That would suggest the equilibrium point is stable.  So, perhaps someone has a reference to suggest this.  I'm stuck.
