The gravitational potential $G_\text{pot}$ has units of energy per unit mass:

$$ \bigg[\rm\frac{J}{kg}\bigg] = \bigg[\rm\frac{kg\cdot m^2}{s^2\cdot kg}\bigg] = \bigg[\rm\frac{m^2 }{s^2}\bigg]. $$

The gravitational force is $F = - \nabla G_\text{pot}$ so this would lead me to believe that unit-wise, due to the gradient, we have a similar expression to the above, apart from an additional $\rm m$ in the denominator:

$$ \bigg[\rm\frac{J}{kg\cdot m}\bigg] = \bigg[\rm\frac{kg\cdot m}{s^2\cdot kg}\bigg] = \bigg[\rm\frac{m }{s^2}\bigg]. $$

But force has units of Newtons:

$$ \bigg[\rm N\bigg] = \bigg[\rm\frac{kg\cdot m}{s^2}\bigg] \neq \bigg[\rm\frac{m}{s^2}\bigg] $$

So why am I missing a $\rm kg$ in my units when I take the gradient of the gravitational potential?

  • $\begingroup$ Force doesn't have units of Joules. By some strange historical coincidence, the SI unit of force is the Newton. $\endgroup$ – Emilio Pisanty Jun 24 '18 at 11:03

You are using a wrong relation. The relation is not

  • "force equals the negative gradient of gravitational potential" but

  • "force equals the negative gradient of gravitational potential energy":

$$F=-\nabla U= -\frac{dU}{dx}$$

The $U$ here is potential energy, not potential. A potential is rather a potential energy per mass.

Had you used potential energy to derive the force unit, you would indeed have gotten the correct force unit of $[\mathrm{\frac{kg \; m}{s^2}}]=[\mathrm{N}]$. But using potential to derive the unit, you get not the unit of force but that of force per mass, $[\mathrm{\frac{kg \; m}{s^2}/kg}]=[\mathrm{\frac{m}{s^2}}]=[\mathrm{\frac{N}{kg}}]$.

This is why (due to the "per-mass" feature) you are lacking one $\mathrm{kg}$ in the derived unit.

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I find the easiest way to remember the dimensions of the units is to start from the second law:

$$ F = ma $$

then work (which is a form of energy) is force times distance.

The units of acceleration are m/sec$^2$ so that's $LT^{-2}$. And multiplying by mass gives us the dimensions of force $MLT^{-2}$, then multiplying by distance gives us the dimensions of energy $ML^2T^{-2}$.

When you take the gradient of the potential energy, $dU/dx$, you are in effect dividing by $L$, so you get back $MLT^{-2}$. And those are of course the dimensions of force.

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  • 3
    $\begingroup$ It's hard to beat $E = m * c²$ as mnemonic. $\endgroup$ – Eric Duminil Jun 24 '18 at 18:21
  • $\begingroup$ @EricDuminil true, but I first learned dimensions long before I'd even heard of relativity :-) $\endgroup$ – John Rennie Jun 25 '18 at 4:22
  • $\begingroup$ we usually hear the formula in popular culture long before we learn anything about dimensional analysis, though. $\endgroup$ – Eric Duminil Jun 25 '18 at 5:10

You started with the gravitational potential, which is the potential energy per unit mass. As a result you obtain the force per unit mass, which in the acceleration, with unit $m/s^2$.

By the way, the mksi unit of force is the Newton ($kgm/s^2$).

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The gravitational force is given by

$$\mathbf{F} = - m\cdot\nabla G_{\text{pot}}$$

The negative gradient of a potential is equal a Field Strength, and the force acting on a mass is equal $- m \cdot \nabla G_{\text{pot}} $.

Another example the Electric Force, where $V$ is the electric potential: $\mathbf{F} = q \cdot \mathbf{E} = - q \cdot \nabla V$

So your $2^{nd}$ equation should be corrected to

$$\Bigg[ \cfrac{kg\cdot m}{s^2}\Bigg] = \Bigg[ kg \cdot \cfrac{J}{kg\cdot m}\Bigg] = \Bigg[ kg \cdot \cfrac{kg \cdot \frac{m^2}{s^2}}{kg\cdot m}\Bigg]$$

$$\Bigg[ \cfrac{kg\cdot m}{s^2}\Bigg] = \Bigg[ \cfrac{kg \cdot m}{s^2}\Bigg]$$

Knowing the potential $\mathbf U$, given in Joule, let's you compute the force by

$$\mathbf F = - \nabla U\quad .$$

btw: it is wrong to to see Gravitation as a classical force -> see Einstein's Relativity Theory

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  • $\begingroup$ It is definitely not wrong, just not accurate. There's a better theory that describes more phenomena, but a classical approximation is enough in many normal life situations. If you can't see relativistic/quantum effects, it is correct within the scope of classical physics. You're standing on a building made with classical physics, so it better be right "enough" haha. $\endgroup$ – FGSUZ Jun 24 '18 at 21:56

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