We have, gravitational field due to a point mass E=-\frac{{GM}}{{{{r}^{2}}}}

[The negative sign is used as gravitational force is attractive]

V=-\int_{{}}^{{}}{{Edr=-\frac{{GM}}{r}+C}}

when, r = ∞, V = 0; so C = 0

\displaystyle V= \displaystyle -\frac{{GM}}{r}

(b) Gravitational potential due to a homogeneous solid sphere

Case I:  When the point P lies outside the sphere

Consider a homogenous solid sphere of mass M and radius ‘R’. P is a point at a distance ‘r’ from the center of the sphere.

Let a particle of mass m₁ be kept fixed at point A and another particle of mass m₂ is taken from a point B to C.   Initially, the distance between the particle is r₁& finally it becomes AC = r₂. We have to calculate the change in gravitation in potential energy of the system of two particles.

Consider a small displacement when the distance between the particles changes from r to r + dr. In the figure this corresponds to the second particle going from D to E. 

Force acting on second particle is

\overset{\to }{\mathop{F}}\,\,=\,\frac{{G{{m}_{1}}{{m}_{2}}}}{{{{r}_{2}}}}along \overset{\to }{\mathop{{DA}}}\,

 dW   = \overset{\to }{\mathop{F}}\,.\,\overset{\to }{\mathop{{dr}}}\,         = – \frac{{G{{m}_{1}}{{m}_{2}}}}{{{{r}^{2}}}}\,dr

dU = – dW = \int{{\frac{{G{{m}_{1}}{{m}_{2}}\,\,dr}}{{{{r}^{2}}}}}}

\int\limits_{{{{r}_{1}}}}^{{{{r}_{2}}}}{{dU}}\,=\,G{{m}_{1}}{{m}_{2}}\,\int\limits_{{{{r}_{1}}}}^{{{{r}_{2}}}}{{\frac{{dr}}{{{{r}^{2}}}}}}= G{{m}_{1}}{{m}_{2}}\,\left( {-\frac{1}{r}} \right)_{{{{r}_{1}}}}^{{{{r}_{2}}}}

U(r₂) – U(r₁) = Gm₁m₂    \left( {\frac{1}{{{{r}_{1}}}}\,-\,\frac{1}{{{{r}_{2}}}}} \right)                 … (i) 

We choose the potential energy of the two particles system to be zero when the distance between them is infinity. This means U( \infty ) = 0

Hence U(r) = U(r) – U( \infty )

Now in equation (i)

Take r₁ = r  and r₂ = \infty

U( \infty ) – U(r) = Gm₁m₂ \left( {\frac{1}{r}\,-\,\frac{1}{\infty }} \right)          \      U(r) = -\,\frac{{G{{m}_{1}}{{m}_{2}}}}{{{{r}_{1}}}}  

Hence when two masses m₁ and m₂ separated by a distance their gravitational potential energy is U(r) = \frac{{-G{{m}_{1}}{{m}_{2}}}}{r}

Example-4:

Three particles each of mass m are placed at the corners of an equilateral triangle of side d. Calculate

(a) the potential energy of the system,

(b) work done on the system if the side of the triangle is changed from d to 2d.

Solution:

(a)     U=-\frac{{3Gmm}}{d} \displaystyle =-\frac{{3G{{m}^{2}}}}{d}

(b)        {{U}_{i}}=-\frac{{3G{{m}^{2}}}}{d},   {{U}_{f}}=-\frac{{3G{{m}^{2}}}}{{2d}}

Work done =w=\Delta U={{U}_{f}}-{{U}_{i}}

Example 9:

Two particles of masses M and m are initially at rest and infinitely separated. When they move towards each other due to gravitational attraction, their relative velocity at any instant in terms of ‘d’ distance between them at that instant is

(1) {{\left( {\frac{{2Gd}}{{M\ +\ m}}} \right)}^{{1/2}}}                                   

(2) {{\left[ {\frac{{2G(M\ +\ m)}}{d}} \right]}^{{1/2}}}

(3) \frac{{2G(M\ +\ m)}}{d}

(4) \frac{{2Gd}}{{M\ +\ m}}

Solution:

The problem may be treated in the centre of mass system coordinates. In this system the displacement and velocities are measured relative to the centre of  mass. The reduced mass \mu \ \,=\ \,\frac{{m\ M}}{{m\ +\,M}}

The kinetic energy  = \frac{1}{2}\ \mu \ {{v}^{2}} = \frac{1}{2}\ \left( {\frac{{m\ M}}{{m\ +\ M}}} \right)\ {{v}^{2}}

Work done in moving the masses from infinite distance to a separation distance d is

G\ M\ m\ \int\limits_{\infty }^{d}{{\frac{{dr}}{{{{r}^{2}}}}}}\ \,=\ \,\,-\ \frac{{G\ M\ m}}{d}

  \frac{1}{2}\ \frac{{m\ M}}{{(M\ +\ m)}}\ {{v}^{2}}\ -\ \frac{{G\ M\ m}}{d}\ =\ 0

∴v  = \sqrt{{\frac{{2\ G\ (M\ +\,m)}}{d}}}.

∴ (2)

Illustration

Find the potential energy of a system of four particles each of mass m placed at the vertices of a square of side  a.  Also find gravitational potential at the center of square.