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The concept of potential energy is linked to a system – not to a single particle or body. So is the case with gravitational potential energy. True nature of this form of energy is often concealed in practical consideration and reference to Earth. Gravitational energy is not limited to Earth, but is applicable to any two masses of any size and at any location. Clearly, we need to expand our understanding of various physical concepts related with gravitational potential energy.
Here, we shall recapitulate earlier discussions on potential energy and apply the same in the context of gravitational force.
The change in the gravitational potential energy of a system is related to work done by the gravitational force. In this section, we shall derive an expression to determine change in potential energy for a system of two particles. For this, we consider an elementary set up, which consists of a stationary particle of mass, " ${m}_{1}$ " and another particle of mass, " ${m}_{2}$ ", which moves from one position to another.
Now, we know that change in potential energy of the system is equal to negative of the work by gravitational force for the displacement of second particle :
$$\Delta U=-{W}_{G}$$
On the other hand, work by gravitational force is given as :
$${W}_{G}=\int {F}_{G}dr$$
Combining two equations, the mathematical expression for determining change in potential energy of the system is obtained as :
$$\Rightarrow \Delta U=-\underset{{r}_{1}}{\overset{{r}_{2}}{\int}}{F}_{G}dr$$
In order to evaluate this integral, we need to set up the differential equation first. For this, we assume that stationary particle is situated at the origin of reference. Further, we consider an intermediate position of the particle of mass " ${m}_{2}$ " between two positions through which it is moved along a straight line. The change in potential energy of the system as the particle moves from position “r” to “r+dr” is :
$$dU=-{F}_{G}dr$$
We get the expression for the change in gravitational potential energy by integrating between initial and final positions of the second particle as :
$$\Delta U=-\underset{{r}_{1}}{\overset{{r}_{2}}{\int}}{F}_{G}dr$$
We substitute gravitational force with its expression as given by Newton's law of gravitation,
$$F=-\frac{G{m}_{1}{m}_{2}}{{r}^{2}}$$
Note that the expression for gravitational force is preceded by a negative sign as force is directed opposite to displacement. Now, putting this value in the integral expression, we have :
$$\Rightarrow \Delta U=\underset{{r}_{1}}{\overset{{r}_{2}}{\int}}\frac{G{m}_{1}{m}_{2}dr}{{r}^{2}}$$
Taking out constants from the integral and integrating between the limits, we have :
$$\Rightarrow \Delta U=G{m}_{1}{m}_{2}[-\frac{1}{r}\underset{{r}_{1}}{\overset{{r}_{2}}{]}}$$
$$\Rightarrow \Delta U={U}_{2}-{U}_{1}=G{m}_{1}{m}_{2}[\frac{1}{{r}_{1}}-\frac{1}{{r}_{2}}]$$
This is the expression of gravitational potential energy change, when a particle of mass “ ${m}_{2}$ ” moves from its position from “ ${r}_{1}$ ” to “ ${r}_{2}$ ” in the presence of particle of mass “ ${m}_{1}$ ”. It is important to realize here that “ $1\u2215{r}_{1}$ ” is greater than “ $1\u2215{r}_{2}$ ”. It means that the change in gravitational potential energy is positive in this case. In other words, it means that final value is greater than initial value. Hence, gravitational potential energy of the two particles system is greater for greater linear distance between particles.
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