If the gravitational field in the space is given as $\left(-\frac{K}{r^{2}}\right)$. Taking the reference point to be at $\mathrm{r}=2 \mathrm{~cm}$ with gravitational potential $\mathrm{V}=10 \mathrm{~J} / \mathrm{kg}$. Find the gravitational potential at $\mathrm{r}=3 \mathrm{~cm}$ in SI unit (Given, that $\mathrm{K}=6 \mathrm{~Jcm} / \mathrm{kg}$)
Solution
<p>$E = - {K \over {{r^2}}}$</p>
<p>$\Delta V = - \int\limits_{r = 2\,cm}^{3\,cm} {E.\,dr}$</p>
<p>$= \int\limits_2^3 {{k \over {{r^2}}}dr}$</p>
<p>$$ = \left[ { - {K \over r}} \right]_2^3 = \left( {{K \over 6}} \right) = {6 \over 6} = 1$$ J/kg</p>
<p>${V_f} - {V_i} = 1$</p>
<p>$\Rightarrow {V_f} - 10 = 1$</p>
<p>${V_f} = 11$ J/kg</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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