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Standard XII

Physics

Gravitational Potential

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Question

Find the potential energy of the gravitational interaction of two mass points of masses $m_{1}$ and $m_{2}$ located at a distance $r$ from each other.

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Solution

The gravitational potential due to $m_{1}$ at the point of location of $m_{2}$ ,
$V_{2} = \int_{r}^{\infty} \to G \cdot d \to r = \int_{r}^{\infty} - \frac{γ m_{1}}{x^{2}} d x = - \frac{γ m_{1}}{r}$
So, $U_{21} = m_{2} V_{2} = - \frac{γ m_{1} m_{2}}{r}$
Similarly $U_{12} = - \frac{γ m_{1} m_{2}}{r}$
Hence, $U_{12} = U_{21} = U = - \frac{γ m_{1} m_{2}}{r}$

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Find the potential energy of the gravitational interaction of two mass points of masses m1 and m2 located at a distance r from each other.

The gravitational potential due to m1 at the point of location of m2,V2=∫∞r→G⋅d→r=∫∞r−γm1x2dx=−γm1rSo, U21=m2V2=−γm1m2rSimilarly U12=−γm1m2rHence, U12=U21=U=−γm1m2r

Find the potential energy of the gravitational interaction of two mass points of masses $m_{1}$ and $m_{2}$ located at a distance $r$ from each other.