7.5 GPE Gradient and Potential Gradient

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 Δx 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, 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 F_ext 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 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

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