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III. Long Answer Questions · Q4

Q.Derive the expression for gravitational potential energy.

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Step 1. Hold m1m_1 fixed and move m2m_2 from separation r′r' to rr. The infinitesimal work done for a step drdr is dW=F dr=Gm1m2r2 drdW=F\,dr=\dfrac{Gm_1m_2}{r^2}\,dr, since work is done against the attractive gravitational force.

Step 2. Integrating from r′r' to rr:

W=∫r′rGm1m2r2 dr=[−Gm1m2r]r′r=−Gm1m2r+Gm1m2r′.W=\int_{r'}^{r}\frac{Gm_1m_2}{r^2}\,dr=\left[-\frac{Gm_1m_2}{r}\right]_{r'}^{r}=-\frac{Gm_1m_2}{r}+\frac{Gm_1m_2}{r'}.

Step 3. This work equals the change in potential energy, W=U(r)−U(r′)W=U(r)-U(r'), where U(r)=−Gm1m2/rU(r)=-Gm_1m_2/r.

Step 4. Choosing the reference point at r′→∞r'\to\infty (so U(∞)=0U(\infty)=0, the standard convention) makes the second term vanish, giving the final, standard result

U(r)=−Gm1m2r.U(r)=-\frac{Gm_1m_2}{r}. …

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