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Physics Question on Gravitational Potential Energy

Three uniform spheres, with masses mA=350 kg,mB=2000 kgm_{A}=350\, kg , m_{B}=2000\, kg, and mC=500 kgm_{C}=500 \,kg have the (x,y)( x , y ) coordinates (0,0)cm,(−80,0)cm(0,0) cm ,(-80,0) cm and (40,0)cm(40,0) cm respectively. The gravitational potential energy, UU, of the system and change in its value in terms of increase or decrease, if the sphere of mass is removed, may be given as

A

U=−1.92×10−4JU=-1.92\times 10^{-4}J and its value shall decrease if the sphere B is removed

B

U=−1.92×10−4JU=-1.92\times 10^{-4}J and its value shall increase if the sphere B is removed

C

U=−1.43×10−4JU=-1.43\times 10^{-4}J and its value shall decrease if mg is removed

D

U=−1.43×10−4JU=-1.43\times 10^{-4}J and its value shall increase if mBm_{B} is removed

Answer

U=−1.43×10−4JU=-1.43\times 10^{-4}J and its value shall increase if mBm_{B} is removed

Explanation

Solution

U=−G[mAmBrAB+mBmCrBC+mCmArCA]U=-G\left[\frac{m_{A} m_{B}}{r_{A B}}+\frac{m_{B} m_{C}}{r_{B C}}+\frac{m_{C} m_{A}}{r_{C A}}\right] =6.7×10−11[350×200080×10−2+2000×500120×10−2+350×50040×10−2]=6.7 \times 10^{-11}\left[\frac{350 \times 2000}{80 \times 10^{-2}}+\frac{2000 \times 500}{120 \times 10^{-2}}+\frac{350 \times 500}{40 \times 10^{-2}}\right] =−6.7×10−11[8750+8333.33+4375]×10+2=-6.7 \times 10^{-11}[8750+8333.33+4375] \times 10^{+2} =−6.7×10−9×21458.33=-6.7 \times 10^{-9} \times 21458.33 =−143770.81×10−9=-143770.81 \times 10^{-9} =−1.43×10−4J=-1.43 \times 10^{-4} J Hence gravitational potential energy U=−1.43×10−4JU=-1.43 \times 10^{-4} J and its value shall increase if mBm_{B} removed.