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Expression for gravitational potential with diagram.

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Answer 1

Gravitational potential

The gravitational potential at a point in the field of a body is defined as the work done in displacing a body of unit mass from infinity to that point in the field. It is denoted by V.

Let's consider that a body of mass placed at a distance $r$ from the centre of a mass $M$ [Kindly see the attachment for diagram]. Then gravitational force acting on the unit mass is given by,

$\implies \dfrac{GMm}{r^2}$

Where, $m$ refers unit mass. Then unit mass $m$ is equal to 1.

$\implies \dfrac{GM \times 1}{r^2} \\ \\ \implies F = \dfrac{GM}{r^2} \qquad .... .(1)$

The direction of this force is towards the centre of the body of mass $M.$

Let the unit mass be the distance from point $P$ to $Q$ through a distance $dr$ towards mass $M$, then work done is given by,

$\implies dW = \vec{F} \cdot d \vec{r} \\ \\ \implies dW = F \: dr \cos0$

Substituting the value of equation (1) in it, we get:

$\implies dW = \dfrac{GM}{r^2}dr \qquad .... .(2)$

Total work done in displacing the unit mass from infinity $(r = \infin)$ to the point $P$ whose distance from mass $M$ is $r$ can be calculated by integrating equation (2) between limits $r = \infin$ to $r = r.$

$\displaystyle\implies \int dW = \int\limits^r_\infin \dfrac{GM}{r^2}dr \\ \\ \displaystyle\implies W = GM\int\limits^r_\infin \: r^{ - 2} \: dr \\ \\ \displaystyle\implies W = - GM \bigg[ \dfrac{{r}^{ - 1} }{ - 1} \bigg]^r_\infin \\ \\ \displaystyle\implies W = - GM \bigg[ \dfrac{1}{r} \bigg]^r_\infin \\ \\ \displaystyle\implies W = - GM \bigg[ \dfrac{1}{r} - \frac{1}{ \infin} \bigg] \\ \\ \displaystyle\implies W = \frac{ -GM}{r}$

As we know that, the work done is equal to the gravitational potential. So, this work done is equal to the gravitational potential.

$\implies\boxed{V = \dfrac{ -GM}{r} }$

Hence, this is our required solution and expression for gravitational potential.