Physics > Gravitation > 4.0 Gravitational potential

  Gravitation
    1.0 Newton's law of gravitation
    2.0 Variation of acceleration due to gravity
    3.0 Gravitational field
    4.0 Gravitational potential
    5.0 Gravitational potential energy
    6.0 Satellites
    7.0 Kepler's law of planetary motion
    8.0 Problem solving technique

4.4 Gravitational potential due to a uniform ring at a point on its centre


The gravitational potential due to a uniform ring of mass $M$ and radius $R$ at a distance $r$ from the centre is given as,
$$V = - \frac{{GM}}{{\sqrt {{R^2} + {r^2}} }}\quad \quad \left( {0 \leqslant r \leqslant \infty } \right)$$

Proof:



Gravitational field at point $P$ due to circular ring of mass $M$ and radius $R$ is,
$$E = \frac{{GMr}}{{{{\left( {{R^2} + {r^2}} \right)}^{\frac{3}{2}}}}}$$
Similarly, force on mass $m$ at point $P$ is, $$F = Em$$$$F = \frac{{GMmr}}{{{{\left( {{R^2} + {r^2}} \right)}^{\frac{3}{2}}}}}$$
Work done for displacement $dr$ is, $$dW = \overrightarrow F .\,d\overrightarrow r $$$$dW = - \frac{{GMmr}}{{{{\left( {{R^2} + {r^2}} \right)}^{\frac{3}{2}}}}}dr$$
Potential is given by, $$dV = - \frac{{dW}}{m}$$ or $$dV = \frac{{GMrdr}}{{{{\left( {{R^2} + {r^2}} \right)}^{\frac{3}{2}}}}}$$
Integrating with proper limits we get,
$$\int\limits_{{V_\infty }}^{{V_r}} {dV} = \int\limits_\infty ^r {\frac{{GMrdr}}{{{{\left( {{R^2} + {r^2}} \right)}^{\frac{3}{2}}}}}} $$$$\left[ V \right]_{{V_\infty }}^{{V_r}} = GM\left[ { - \frac{1}{{\sqrt {{R^2} + {r^2}} }}} \right]_\infty ^r$$$${V_r} - {V_\infty } = - GM\left( {\frac{1}{{\sqrt {{R^2} + {r^2}} }} - \frac{1}{\infty }} \right)$$ As $\left( {{V_\infty } = 0} \right)$, $${V_r} = - \frac{{GM}}{{\sqrt {{R^2} + {r^2}} }}$$
The $V$ vs $r$ graph can be drawn as,


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