Let’s consider a mass situated in a gravitational field and experiencing a gravitational force *F*_{g}. Suppose we apply an external force *F*_{ext} that is in opposite direction to *F*_{g}. By pulling the mass some distance Δ*x* in opposite direction to *F*_{g}, we can move the mass to a position of higher GPE.

And if we purposely choose *F*_{ext} to be the same magnitude as *F*_{g}, we can keep the mass’s KE constant during the move. This will allow us to equate the (positive) work done by the external force to the increase in GPE, Δ*U*. In other words,

Rearranging, we get

If Δ*x *is small enough, then

So *F*_{ext} is equal to the potential energy gradient (at position A). But . So

This means that the gravitational force (at a point) is equal to the negative of the potential energy gradient (at that point)!

If we substitute and , we get

This means that gravitational field strength (at a point) is equal to the negative of the potential gradient (at that point).

Graphically, this means that the gradient of the *Φ*–*r* graph is related to *g*…

while the area under the *g-r* graph is related to *Φ*.

Note that we included the negative sign for *g* to indicate that the direction of *g* is in the negative *r* direction (i.e. towards the centre of the Earth).

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**Video Explanation**

Mr Chua, can you explain the F=-dU/dx formula?

Gravitational Force and Gravitational Potential Energy Gradient

**Concept Test**