Question

Figure 13-44 shows four particles,each of mass 20.0 g, that form a square with an edge length of d =0.600 m. Ifdis reduced to0.200 m, what is the change in the gravitational potential energy of the four-particle system?

Short answer

The change in gravitational potential energyis$-4.8\times {10}^{-13}\text{J.}$

Step-by-step solution

Step 1: Given

Mass of Zero plant is$M=20g\left(\frac{{10}^{-3}kg}{1g}\right)=\text{0.020kg}$

The side of the square is $\text{d}=0.6\text{m}$

Determining the concept

Using the formula for gravitational potential energy, find the change in potential energy.

The formula is as follows:

$\text{U}=-\frac{\text{GMm}}{\text{R}}$

where, M, and m are masses, R is the radius, G is the gravitational constant and U is potential energy.

Determining thechange in gravitational potential energy

First, the initial potential energy on the surface ofZero planet can be found as,${\text{U}}_{\text{1}}=\frac{\text{Gmm}}{\text{R}}$

Now, the distance between the corner points,

$\begin{array}{rcl}AB& =& \sqrt{d^2+d^2}\\ & =& d\sqrt{2}\\ & & \end{array}$

So,thegravitational potential energy is as follows:

$\begin{array}{rcl}{U}_1& =& -\left(\frac{4Gmm}{d}\right)-\frac{2Gmm}{d\sqrt{2}}\\ & =& -\frac{4\times 6.67\times {10}^{-11}N·k{g}^{-2}·m^2\times {\left(0.020kg\right)}^2}{0.6m}\\ & & -\frac{2\times 6.67\times {10}^{-11}N·k{g}^{-2}·m^2\times {\left(0.020m\right)}^2}{\sqrt{2}\times 0.6m}\\ & =& -2.4\times {10}^{-13}\text{J}\\ & & \end{array}$

Now,thedistance is reduced to 0.2 m.

As,thedistance reduces to$\frac{1}{3}$of the original distance$\left(\frac{0.6m}{3}=0.2m\right)$, so, thegravitational potential energy will increase by 3 because it is inversely proportional to distance.

So,thenew gravitational potential energy is,

$\begin{array}{rcl}{U}_2& =& -3\times 2.4\times {10}^{-13}J\\ & =& -7.2\times {10}^{-13}J\end{array}$

So,change in potential energy is,

$\begin{array}{rcl}\Delta U& =& U_2-U_1\\ & =& -7.2\times {10}^{-13}J-\left(-2.4\times {10}^{-13}J\right)\\ & =& -4.8\times {10}^{-13}J\\ & & \end{array}$

Hence,the change in gravitational potential energyis$-4.8\times {10}^{-13}\text{J.}$

Therefore, gravitational potential energy is inversely proportional to the distance between the objects. If the distance is reduced, the potential energy will increase. In this problem, this concept can be used to calculate the change in gravitational potential energy.