In our (presumably) $3$-dimensional universe, gravity is inversely proportional to distance squared. In a circular orbit, the force of gravity must be equal and opposite to the 'centripetal force'. $\frac {GM_1M_2}{r^2} = \frac {M_2v^2}r$. This allows the velocity of a planet's orbit to be calculated from its radius: $v = \frac {\sqrt {GM_1}}r$.
The kinetic energy of the planet is $\frac {M_2v^2}2$, while the gravitational potential energy is $\frac {-GM_1M_2}r$. This means that $KE =- \frac {GPE}2$. The total energy, $KE + GPE$, is equal to $-KE$.
This means that the planetary orbit is a stable equilibria: adding some energy to the system means that the planet moves further away from the sun, increasing its total energy; while taking energy away from the system means the planet approaches the sun, but either way it finds a new equilibrium.
Consider the case in a $4$-dimensional universe, where, presumably, gravity follows an inverse cube law. The equations now become:
$\frac {GM_1M_2}{r^3} = \frac {M_2v^2}r$ [force of gravity equal and opposite to centripetal force]
$v = \frac {\sqrt {GM_1}}r$ [velocity is inversely proportional to radius]
$GPE = -\frac {GM_1M_2}{2r^2}$
Total Energy = $KE + GPE = \frac {M_2v^2}2 + \frac {GM_1M_2}{2r^2} = 0$
The Kinetic and Gravitational Potential energies are equal and opposite to each other. In other words, the only possible orbits have no net energy, such that the input or extraction of even a small amount of energy would destabilise the orbit, with the planet spiralling towards or away from the sun inexorably. This would make life as we know it impossible, as planets couldn't maintain stable orbits.
The question: are there some other number of dimensions the universe could have, which would also lead to stable orbits?