91Ó°ÊÓ

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

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

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.

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:

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

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,

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

So,change in potential energy is,

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

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

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.