<< Chapter < Page | Chapter >> Page > |
A change in gravitational potential (ΔV) is equal to the negative of work by gravity on a unit mass,
$$\Delta V=-E\Delta r$$
For infinitesimal change, we can write the equation,
$$\Rightarrow dV=-Edr$$
$$\Rightarrow E=-\frac{dV}{dr}$$
Thus, if we know potential function, we can find corresponding field strength. In words, gravitational field strength is equal to the negative potential gradient of the gravitational field. We should be slightly careful here. This is a relationship between a vector and scalar quantity. We have taken the advantage by considering field in one direction only and expressed the relation in scalar form, where sign indicates the direction with respect to assumed positive reference direction. In three dimensional region, the relation is written in terms of a special vector operator called “grad”.
Further, we can see here that gravitational field – a vector – is related to gravitational potential (scalar) and position in scalar form. We need to resolve this so that evaluation of the differentiation on the right yields the desired vector force. As a matter of fact, we handle this situation in a very unique way. Here, the differentiation in itself yields a vector. In three dimensions, we define an operator called “grad” as :
$$\mathrm{grad}=\left(\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k\right)$$
where " $\frac{\partial}{\partial x}$ ” is partial differentiation operator with respect to "x". This is same like normal differentiation except that it considers other dimensions (y,z) constant. In terms of “grad”,
$$E=-\mathrm{grad}\phantom{\rule{2pt}{0ex}}V$$
Gravitational potential energy of a particle of mass “m” is related to gravitational potential of the field by the equation,
$$U=mV$$
This relation is quite handy in calculating potential energy and hence “self energy” of a system of particles or a rigid body. If we recall, then we calculated “self energy” of a system of particles by a summation process of work in which particles are brought from infinity one by one. The important point was that the gravitational force working on the particle kept increasing as more and more particles were assembled. This necessitated to calculate work by gravitational forces due to each particle present in the region, where they are assembled.
Now, we can use the “known” expressions of gravitational potential to determine gravitational potential energy of a system, including rigid body. We shall derive expressions of potential energy for few regular geometric bodies in the next module. One of the important rigid body is spherical shell, whose gravitational potential is given as :
For a point inside or on the shell of radius “a”,
$$V=-\frac{GM}{a}$$
This means that potential inside the shell is constant and is equal to potential at the surface.
For a point outside shell of radius “a” (at a linear distance, “r” from the center of shell) :
$$V=-\frac{GM}{r}$$
This means that shell behaves as a point mass for potential at a point outside the shell. These known expressions allow us to calculate gravitational potential energy of the spherical shell as explained in the section below.
The self potential energy is equal to work done by external force in assembling the shell bit by bit. Since zero gravitational potential energy is referred to infinity, the work needs to be calculated for a small mass at a time in bringing the same from infinity.
In order to calculate work, we draw a strategy in which we consider that some mass has already been placed symmetrically on the shell. As such, it has certain gravitational potential. When a small mass “dm” is brought, the change in potential energy is given by :
$$dU=Vdm=-\frac{Gm}{R}dm$$
We can determine total potential energy of the shell by integrating the expressions on either side of the equation,
$$\Rightarrow \int dU=-\frac{G}{R}\int mdm$$
Taking constants out from the integral on the right side and taking into account the fact that initial potential energy of the shell is zero, we have :
$$\Rightarrow U=-\frac{G}{R}[\frac{{m}^{2}}{2}\underset{0}{\overset{M}{]}}$$
$$\Rightarrow U=-\frac{G{M}^{2}}{2R}$$
This is total potential energy of the shell, which is equal to work done in bringing mass from infinity to form the shell. This expression, therefore, represents the self potential energy of the shell.
In the same manner, we can also find “self energy” of a solid sphere, if we know the expression for the gravitational potential due to a solid sphere.
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?