Question
Explain gravitational potential at a point in gravitational field. Give relation between gravitational field intensity and gravitational potential.
✓
Answer
Gravitational potential at a point in a gravitational field of a body is defined as the amount of work done in bringing a body of unit mass from infinity to that point without acceleration.
Consider two points A and B, distance dr apart in a uniform gravitational field of intensity $\vec{\text{I}}$ Let the direction of $\vec{\text{I}}$ be along AB.
Gravitational force on the particle of mass m placed at B will be, $\vec{\text{F}}=\text{m}\vec{\text{I}}.$
Work done by gravitational force for the displacement of the particle from B to A (i.e. a displacement $\overrightarrow{\text{dr}})$ will be
$\text{dW}=\vec{\text{F}}.\overrightarrow{\text{dr}}=\text{m}\vec{\text{I}}.\ \overrightarrow{\text{dr}}=\text{mI}\text{ dr }\cos180^\circ$
Change in gravitational potential,
$\text{dV}=\frac{\text{dW}}{\text{m}}=-\text{Idr}$
$\Rightarrow\text{I}=-\frac{\text{dV}}{\text{dr}}$
$\text{Here,}\frac{\text{dV}}{\text{dr}}$ is called gravitational potential gradient.
Consider two points A and B, distance dr apart in a uniform gravitational field of intensity $\vec{\text{I}}$ Let the direction of $\vec{\text{I}}$ be along AB.
Gravitational force on the particle of mass m placed at B will be, $\vec{\text{F}}=\text{m}\vec{\text{I}}.$
Work done by gravitational force for the displacement of the particle from B to A (i.e. a displacement $\overrightarrow{\text{dr}})$ will be
$\text{dW}=\vec{\text{F}}.\overrightarrow{\text{dr}}=\text{m}\vec{\text{I}}.\ \overrightarrow{\text{dr}}=\text{mI}\text{ dr }\cos180^\circ$
Change in gravitational potential,
$\text{dV}=\frac{\text{dW}}{\text{m}}=-\text{Idr}$
$\Rightarrow\text{I}=-\frac{\text{dV}}{\text{dr}}$
$\text{Here,}\frac{\text{dV}}{\text{dr}}$ is called gravitational potential gradient.
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