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

Expert verified

The change in gravitational potential energyis-4.8×10-13J.

Step by step solution

01

Step 1: Given

Mass of Zero plant isM=20g10-3kg1g=0.020kg

The side of the square is d=0.6m

02

Determining the concept

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

The formula is as follows:

U=-GMmR

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

03

Determining thechange in gravitational potential energy

First, the initial potential energy on the surface ofZero planet can be found as,U1=GmmR

Now, the distance between the corner points,

AB=d2+d2=d2

So,thegravitational potential energy is as follows:

U1=-4Gmmd-2Gmmd2=-4×6.67×10-11N·kg-2·m2×0.020kg20.6m-2×6.67×10-11N·kg-2·m2×0.020m22×0.6m=-2.4×10-13J

Now,thedistance is reduced to 0.2 m.

As,thedistance reduces to13of the original distance0.6m3=0.2m, so, thegravitational potential energy will increase by 3 because it is inversely proportional to distance.

So,thenew gravitational potential energy is,

U2=-3×2.4×10-13J=-7.2×10-13J

So,change in potential energy is,

ΔU=U2-U1=-7.2×10-13J--2.4×10-13J=-4.8×10-13J

Hence,the change in gravitational potential energyis-4.8×10-13J.

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.

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