The gravitational potential (V) is the gravitational potential energy (U) per unit mass: where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.
Gravitational Potential due to a point mass
Gravitational Potential due to a point mass is given by:
Gravitational Potential due to a uniform ring
Example: Two rings having masses M and 2M, respectively, having the same radius are placed coaxially as shown in the figure. If the mass distribution on both the rings is non-uniform, then what is the gravitational potential at point P ?
Solution: Gravitational potential due to ring (M,R) at an axial point which is x unit away from the center, V=−x2+R2GMThus gravitational potential at P, VP=(−R2+R2GM)+(−(2R)2+R2G(2M)) VP=−GM[(2R21)+(5R22)] VP=−RGM[21+52]
Gravitational Potential inside and outside of a thin spherical shell
Gravitational field inside the shell: Eg=2R3−GM(3R2−r2)
Gravitational field outside the shell: Eg=r−GM
Gravitational Potential due to a solid sphere
Example The earth does not have a uniform density; it is most dense at its centre and least dense at its surface. An approximation of its density is ρ(r)=(A−Br), where A=12700kg/m3,B=1.50×10−3kg/m3 and r is the distance from the centre of earth. Use R=6.4×106 m for the radius of earth approximated as a sphere, Imagine dividing the earth into concentric, elementary spherical shells. Each shell has radius r, thickness dr, volume dV and mass dm=ρ(r)dV. By integrating dm from zero to R the mass of earth can be found. Knowing the fact that a uniform spherical shell gives no contribution to acceleration due to gravity inside it, we can also find g as a function of r. If B = 0, then find gravitational potential at the centre?
Solution: We have, ρ=A−Br dm=ρdv =(A−Br)dv Now, v=34πr3 ∴dv=4πr2dr Thus, dm=(A−Br)⋅4πr2dr We know, dU=−rGdm =−rG(A−Br)⋅4πr2dr =−4π⋅G(Ar−Br2)dr Integrating from 0 to R, ∴U=−4π⋅∫0R(Ar−Br2)dr =−4πG[2Ar2−3Br3]0R =−4π2GAR2(∵B=0) ∴=−2πGAR2