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What's wrong in this derivation of potential at any point? Its showing GPE at any point is +ve but its supposed to be negative right?

Supoose a object is being moved from infinity to a point at distance $r$ from the center of earth. What's the GPE?

Direction of $F_\text{ext}$ and $\mathrm{d}x$ is opposite. So, $$ \begin{aligned} W &= \int F_\text{ext} \cdot \mathrm{d}x \cdot \cos180° \\ &= \int -\frac{GMm}{x^2} \mathrm{d}x \\ &= \frac{GMm}{x} |_{\infty}^r \\ &= GMm\left(\dfrac{1}{r} - \frac{1}{\infty}\right) \\ &= \frac{GMm}{r} \end{aligned} $$

As GPE is defined as the energy to take any object from inifnity to asked point, I have taken

  1. External force in opposite direction ($+x$ axis) of Gravitational force as kinetic energy should not be changed.
  2. direction of $\mathrm{d}x$ is in $-x$ axis
  3. So,angle is 180°
  4. Upper limit : $r$ lower limit: Infinity

Still the ans is +ve. Please explain what's the wrong assumption here?

Vincent Thacker
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Abrar Jamil
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1 Answers1

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When you did the integration (first line in the text box), you've used the limits $0$ to $\infty$. In the "standard" derivation, you integrate from $\infty$ to $0$, because the potential at a point is the work done when moving an object from infinitely far away (where it has GPE = 0 by definition) to that point.

Allure
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