Understanding the Relation Between Gravitational Field and Potential

The relationship between gravitational field and potential is a fundamental concept in physics, relevant for solving various types of problems in gravitation. Understanding this relation allows for the calculation of gravitational effects on masses in fields around objects like Earth, planets, and stars. This concept is essential for success in exams such as JEE, where both definitions and derivations are frequently tested.

Newton’s Law of Universal Gravitation and Gravitational Field

Newton’s law of universal gravitation states that every pair of masses $M_1$ and $M_2$, separated by a distance $R$, attract each other with a force given by $F = G \dfrac{M_1 M_2}{R^2}$, where $G$ is the universal gravitational constant. This force acts along the line joining the centres of the two bodies.

The gravitational field at a point is defined as the force experienced by a unit mass placed at that point. For a mass $M$ at a distance $r$, the magnitude of gravitational field $E$ at that point is $E = \dfrac{GM}{r^2}$. The direction is always towards the mass producing the field.

The unit of gravitational field is $\mathrm{N\,kg^{-1}}$, which is equivalent to an acceleration ($\mathrm{m\,s^{-2}}$). Detailed study of these concepts can be found in Understanding Gravitation.

Gravitational Potential and Its Definition

Gravitational potential at a point is defined as the work done to bring a unit mass from infinity to that point in the gravitational field. For a mass $M$, the gravitational potential $V$ at a distance $r$ from its centre is $V = -\dfrac{GM}{r}$. The negative sign indicates the attractive nature of gravity.

The SI unit of gravitational potential is $\mathrm{J\,kg^{-1}}$. Gravitational potential measures the potential energy per unit mass at a certain point in the field. Full details on gravitational potential energy are available in Gravitational Potential Energy.

Derivation: Relation Between Gravitational Field and Potential

Consider a mass $m$ moved from point $A$ to point $B$ in a gravitational field. The work done by the field is $dW = F \cdot dr = mE \cdot dr$, where $dr$ is a small displacement in the field direction.

For conservative forces such as gravity, $dU = -dW$. The change in gravitational potential energy $dU$ equals $-mE \cdot dr$, and the corresponding change in potential is $dV = \dfrac{dU}{m} = -E \cdot dr$.

Therefore, the relation is expressed as:

$\displaystyle E = -\dfrac{dV}{dr}$

This means the gravitational field at any point is equal to the negative gradient of gravitational potential at that point. In vector form, $\vec{E} = -\nabla V$.

Relation Between Gravitational Field and Potential Energy

The gravitational potential energy $U$ of a mass $m$ at a distance $r$ from mass $M$ is $U = mV$, where $V$ is the gravitational potential at that point. Hence, $U = -\dfrac{GMm}{r}$. This is discussed further in Gravitational Potential Energy.

Gravitational Field and Potential Due to Spherical Mass

For a uniform solid sphere of mass $M$ and radius $a$, the gravitational field and potential vary depending on the location (outside, on the surface, or inside the sphere).

For a point outside the sphere ($r \geq a$):

Gravitational field: $E = \dfrac{GM}{r^2}$

Gravitational potential: $V = -\dfrac{GM}{r}$

For a point on the surface ($r = a$):

$E = \dfrac{GM}{a^2}$, $V = -\dfrac{GM}{a}$

For a point inside the uniform solid sphere ($r < a$):

$E = \dfrac{GM r}{a^3}$

$V = -\dfrac{GM}{2a^3}(3a^2 - r^2)$

The potential inside a sphere varies quadratically with $r$, while the field increases linearly with $r$ up to the surface. This variation is important in planetary and stellar physics, and for exam preparation such as with Gravitation Revision Notes.

Solved Example: Calculating Field from Potential

Given the gravitational potential at a point is $V = -2 \times 10^7\,\mathrm{J\,kg^{-1}}$, and the potential at a nearby point $0.5\,\mathrm{m}$ away is $V' = -1.8 \times 10^7\,\mathrm{J\,kg^{-1}}$, calculate the field.

• Potential difference $\Delta V = V' - V = 0.2 \times 10^7\,\mathrm{J\,kg^{-1}}$
• Distance $\Delta r = 0.5\,\mathrm{m}$
• Field $E = -\dfrac{\Delta V}{\Delta r} = -\dfrac{0.2 \times 10^7}{0.5} = -4 \times 10^6\,\mathrm{N\,kg^{-1}}$

Thus, the gravitational field between the two points is $4 \times 10^6\,\mathrm{N\,kg^{-1}}$, directed towards the mass producing the field.

Key Points on Gravitational Field and Potential Relations

• Gravitational field is the negative derivative of potential with respect to distance
• Both field and potential are central for solving gravitation problems
• Potential is scalar; field is vector
• The field points in the direction of decreasing potential
• Integration of field gives potential difference

Further practice on these concepts can be found in the Gravitation Practice Paper.

Summary: Importance of the Relationship

The derivation $E = -\dfrac{dV}{dr}$ links the vector property of gravitational field to the scalar characteristic of gravitational potential. Mastery of this relationship is essential for advanced gravitation concepts and numerical problem solving. For acceleration-related details, refer to Acceleration Due to Gravity.