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In the following figure potential differenceV_{B}-V_{A} is equals to 

Option: 1

 

-GM\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]

 

 

 


Option: 2

-GM\left [ \frac{1}{r_{2}}-\frac{1}{r_{1}} \right ]


Option: 3

GM\left [ \frac{1}{r_{2}}+\frac{1}{r_{1}} \right ]


Option: 4

-GM\left [ \frac{1}{r_{1}}+\frac{1}{r_{2}} \right ]


Answers (1)

best_answer

As we learn

Gravitational Potential difference -

\Delta V=V_{B}-V_{A}=\frac{W_{A\rightarrow B}}{m}

\Delta V=-GM\left [ \frac{1}{r_{B}}-\frac{1}{r_{A}} \right ]

- wherein

r_{B}\rightarrow the distance of mass at B

r_{A}\rightarrow distance of mass at A

 

 \bigtriangleup V=V_{B}-V_{A}=\frac{-GM}{r_{2}}-\left ( \frac{-GM}{r_{1}} \right )

=\frac{-GM}{r_{2}}- \frac{-GM}{r_{1}}

V_{B}-V_{A}=-GM\left [ \frac{1}{r_{2}}-\frac{1}{r_{1}} \right ]

 

 

Posted by

Divya Prakash Singh

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