Let’s consider a mass situated in a gravitational field and experiencing a gravitational force Fg. Suppose we apply an external force Fext that is in opposite direction to Fg. By pulling the mass some distance Δx in opposite direction to Fg, we can move the mass to a position of higher GPE.

And if we purposely choose Fext to be the same magnitude as Fg, 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, ΔU. In other words,

$\Delta U={{F}_{{ext}}}\Delta x$

Rearranging, we get

$\displaystyle {{F}_{{ext}}}=\frac{{\Delta U}}{{\Delta x}}$

If Δx is small enough, then

$\displaystyle {{F}_{{ext}}}=\frac{{dU}}{{dx}}$

So Fext is equal to the potential energy gradient (at position A). But ${{F}_{{ext}}}=-{{F}_{g}}$. So

$\displaystyle {{F}_{g}}=-\frac{{dU}}{{dx}}$

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 ${{F}_{g}}=mg$ and $U=m\phi$, we get

\displaystyle \begin{aligned}mg&=-\frac{{d(m\phi )}}{{dx}}\\g&=-\frac{{d\phi }}{{dx}}\end{aligned}

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 Φr 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).

Video Explanation

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

Gravitational Force and Gravitational Potential Energy Gradient

Concept Test

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