The potential energy of a particle in a centr | vedclass

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(A) The relation between central force $F$ and potential energy $U(r)$ is given by $F = -\frac{dU}{dr}$.Given $U(r) = r^{-2} - r^{-1}$.Calculating the

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$\frac{1}{27}$

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(A) The relation between central force $F$ and potential energy $U(r)$ is given by $F = -\frac{dU}{dr}$.Given $U(r) = r^{-2} - r^{-1}$.Calculating the force: $F = -\frac{d}{dr}(r^{-2} - r^{-1}) = -(-2r^{-3} + r^{-2}) = \frac{2}{r^3} - \frac{1}{r^2}$.For the maximum attractive force, we find the extremum of $F$ by setting $\frac{dF}{dr} = 0$.$\frac{dF}{dr} = \frac{d}{dr}(2r^{-3} - r^{-2}) = -6r^{-4} + 2r^{-3} = 0$.$2r^{-3} = 6r^{-4} \Rightarrow \frac{2}{r^3} = \frac{6}{r^4} \Rightarrow r = 3$.Substituting $r = 3$ into the force equation:$F = \frac{2}{(3)^3} - \frac{1}{(3)^2} = \frac{2}{27} - \frac{1}{9} = \frac{2-3}{27} = -\frac{1}{27}$.The magnitude of the force is $|F| = \frac{1}{27}$.

The potential energy of a particle in a central field has the form $U(r) = \frac{1}{r^2} - \frac{1}{r}$, where '$r$' is the distance from the centre of the field. The magnitude of the maximum attractive force in Newton is

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