The gravitational field is a region of space where a mass feels a force due to the presence of another mass. The gravitational potential is a measure of how much energy a mass has at a point in the field. The intensity of the gravitational field, denoted by E, is a measure of the strength of the gravitational force per unit mass at a point in the field.
We usually think that body A can pull another body B with gravity even if they are far apart. This is known as action at a distance. But this idea has some difficulties when we deal with very large distances. We now believe that a body cannot influence another body directly if they are not touching. The interaction between two bodies happens in two steps. First, we suppose that the body A makes a gravitational field around itself. This field is real and has its own energy and momentum. The field points in a certain direction at every point in space and its strength changes from place to place. Second, we suppose that when a body B is in the gravitational field, the field pushes or pulls it. The force that the field applies on the body depends on the direction and the strength of the field. We measure the strength of the gravitational field E at a point by the equation.
\(\displaystyle \overrightarrow{E}=\frac{{\overrightarrow{F}}}{m}\)
Where F is the force exerted by the field on a body of mass m placed in the field. Quite often the intensity of the gravitational field is abbreviated as gravitational field. Its SI unit is N kg-1
If a mass m is placed close to the surface of the earth, the force on it is mg. We say that the earth has set up a gravitational field and this field exerts a force on the mass. The intensity of the field is
\(\displaystyle \overrightarrow{E}=\frac{{\overrightarrow{F}}}{m}=\frac{{m\overrightarrow{g}}}{m}=\overrightarrow{g}\)
Thus, the intensity of the gravitational field near the surface of the earth is equal to the accelerati011 due to gravity. It should be clearly understood that the intensity of the gravitational field and the accelerati011 due to gravity are two separate physical quantities having equal magnitudes and directions.
Relation between Gravitational field and Potential
To derive the relation between gravitational field and potential, we can use the definition of work done by a force. The work done by a force F on a particle of mass m when it is displaced from r to r + dr is given by:
\(\displaystyle dW = F \cdot dr\)
The work done is equal to the change in kinetic energy of the particle, which is zero if the particle moves without acceleration. Therefore, we have:
\(\displaystyle dW = 0\)
The force on the particle is due to the gravitational field E, which is equal to the negative of the gradient of the gravitational potential V. That is:
\(\displaystyle F = -m \nabla V\)
where \(\displaystyle \nabla\) is the del operator that gives the vector of partial derivatives of a scalar function. Substituting this into the work done equation, we get:
\(\displaystyle dW = -m \nabla V \cdot dr = 0\)
Expanding the dot product, we get:
\(\displaystyle -m \left(\frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy + \frac{\partial V}{\partial z} dz \right) = 0\)
Dividing by -m and rearranging the terms, we get:
\(\displaystyle \frac{\partial V}{\partial x} dx = -\frac{\partial V}{\partial y} dy – \frac{\partial V}{\partial z} dz\)
Integrating both sides with respect to x, we get:
\(\displaystyle V = -\int \frac{\partial V}{\partial y} dy – \int \frac{\partial V}{\partial z} dz + C\)
where C is an arbitrary constant of integration. This equation shows that the gravitational potential is a function of x, y, and z. To eliminate the constant C, we can use the boundary condition that the potential is zero at infinity. That is:
\(\displaystyle V(\infty, \infty, \infty) = 0\)
Substituting this into the equation, we get:
\(\displaystyle 0 = -\int_{\infty}^{\infty} \frac{\partial V}{\partial y} dy – \int_{\infty}^{\infty} \frac{\partial V}{\partial z} dz + C\)
Solving for C, we get:
\(\displaystyle C = \int_{\infty}^{\infty} \frac{\partial V}{\partial y} dy + \int_{\infty}^{\infty} \frac{\partial V}{\partial z} dz\)
Substituting this back into the equation for V, we get:
\(\displaystyle V = -\int \frac{\partial V}{\partial y} dy – \int \frac{\partial V}{\partial z} dz + \int_{\infty}^{\infty} \frac{\partial V}{\partial y} dy + \int_{\infty}^{\infty} \frac{\partial V}{\partial z} dz\)
Simplifying, we get:
\(\displaystyle V = -\int_{\infty}^{x} \frac{\partial V}{\partial x} dx\)
This equation shows that the gravitational potential at a point x is equal to the negative of the work done by the gravitational field per unit mass to move a particle from infinity to x. Comparing this with the definition of gravitational field, we get:
\(\displaystyle E = -\frac{dV}{dx}\)
This is the relation between the gravitational field and the potential for a one-dimensional case. For a general case, we can write this relation in vector form as:
\(\displaystyle \vec{E} = -\nabla V\)
where \(\displaystyle \vec{E}\) is the gravitational field vector and \(\displaystyle \nabla V\) is the gradient vector of the gravitational potential. This relation shows that the gravitational field is equal to the negative of the rate of change of the gravitational potential in the direction of the field.