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Question: Two objects have masses of \( 5mg \) and \( 7mg \) . How much does the gravitational potential energ...

Two objects have masses of 5mg5mg and 7mg7mg . How much does the gravitational potential energy between the objects change if the distance between them changes from 90m90m to 2000m2000m ?

Explanation

Solution

It is known that the gravitational potential energy of two masses is proportional to the product of the masses and inversely proportional to the inverse of the distance between them.
The expression for gravitational potential for two masses is given by, U=−GMmrU = \dfrac{{ - GMm}}{r} where, MM is the mass of the one body and mm is the mass of another body separated by a distance rr . Where GG is the gravitational constant and its value is   G=6.674×10−11  m3⋅kg−1⋅s−2\;G = 6.674 \times {10^ - }^{11}\;{m^3} \cdot k{g^ - }^1 \cdot {s^ - }^2 .

Complete step by step answer:
We have given here, the mass of first body is m=5mg=5×10−3kgm = 5mg = 5 \times {10^{ - 3}}kg and mass of another body is, M=7mg=7×10−3kgM = 7mg = 7 \times {10^{ - 3}}kg .
The distance between them at first is r1=90m{r_1} = 90m and after changes it is r2=2000m{r_2} = 2000m
Now we know that the gravitational potential energy of two masses is proportional to the product of the masses and inversely proportional to the inverse of the distance between them. The expression for gravitational potential for two masses is given by, U=−GMmrU = \dfrac{{ - GMm}}{r} where, MM is the mass of the one body and mm is the mass of another body separated by a distance rr . Where, GG is the gravitational constant and its value is   G=6.674×10−11  m3⋅kg−1⋅s−2\;G = 6.674 \times {10^ - }^{11}\;{m^3} \cdot k{g^ - }^1 \cdot {s^ - }^2
So, for r1=90m{r_1} = 90m gravitational potential will be,
U1=−6.674×10−11×7×10−3×5×10−390J{U_1} = \dfrac{{ - 6.674 \times {{10}^ - }^{11} \times 7 \times {{10}^{ - 3}} \times 5 \times {{10}^{ - 3}}}}{{90}}J
Calculating the value we have,
U1=−2.595×10−17J{U_1} = - 2.595 \times {10^ - }^{17}J
Now, for r2=2000m{r_2} = 2000m gravitational potential of the system will be,
U2=−6.674×10−11×7×10−3×5×10−32000J{U_2} = \dfrac{{ - 6.674 \times {{10}^ - }^{11} \times 7 \times {{10}^{ - 3}} \times 5 \times {{10}^{ - 3}}}}{{2000}}J
Calculating the value we have,
U2=−0.117×10−17J{U_2} = - 0.117 \times {10^ - }^{17}J
So, change in gravitational potential will be,
ΔU=U2−U1\Delta U = {U_2} - {U_1}
SO, change will be,
ΔU=[−0.117×10−17−(−2.595×10−17)]J\Delta U = [ - 0.117 \times {10^ - }^{17} - ( - 2.595 \times {10^ - }^{17})]J
Calculating the value we have,
ΔU=2.478×10−17J\Delta U = 2.478 \times {10^ - }^{17}J
So, the change in potential energy will be, 2.478×10−17J2.478 \times {10^ - }^{17}J.

Note:
We can see that the value of the potential energy is very small and is negligible in the real world. To realize gravitational potential physically energy masses must be in order of the gravitational constant. Also, the distance between the masses must be low.