Is the index of time in the unit equivalent to force dependent on the number of dimensions When the unit of force is represented as $\text{distance per time}^n$ in 3 dimensions $n$ is equal to 2. Is n also equal to 2 for higher or lower dimensions (2D, 4D, 5D, ...) or is it dependent on the number of dimensions?
This question was prompted by my discovery that in 3D the relationship force due to gravity is proportional to the $\text{distance}^{-2}$, but in 2D it is proportional to $\text{distance}^{-1}$.
Many Thanks!
 A: The dimension of force is $\text{mass} \cdot \text{distance} / \text{time}^2$ and this is independent of the dimension of the space (as $\vec F = m \vec a$ holds in any dimension).
The relations $\text{distance}^{-d+1}$ for gravitation in $d$ dimension has a different reason. The fundamental relation holding in all dimensions is
$$ \nabla \cdot \vec F_G = \gamma \rho. $$
Where $\gamma$ is the gravitation constant.
Integrating over a ball containing a spherically symmetric mass centred around the coordinate origin we get:
$$ \gamma m = \int_{B(R)} d^dr\, \nabla \cdot \vec F_G = \int_{\partial B(R)} d\sigma(r) \vec n \cdot \vec F_G $$
Where $\sigma$ is the surface measure on the boundary of the integration sphere and $\vec n$ is the outer surface normal. As the gravitational field of a spherically symmetric mass distribution will be spherically symmetric, we can easily compute the integral on the right hand side (the integrand is constant), and get:
$$ \gamma m = \sigma(\partial B(R)) F_G(R).$$
By $\sigma(\partial B(R)) = R^{d - 1} \sigma(\partial B(1))$ we get:
$$ F_G(R) = \frac{\gamma m}{\sigma(S^{d-1}) R^{d-1}} $$
Where $\sigma(S^{d-1})$ is the surface of the $(d-1)$ dimensional sphere.
The dimension of the force remains the same, the quantity that changes in dimension is the mass density ($\text{mass}/\text{length}^d$) and therefore the dimension of the gravitation constant must be adapted (as $\nabla \cdot \vec F$ will always have the dimension $\text{mass}/\text{time}^2$)!
