Subject: Physics
According to Newton’s law of gravitational, the force of attraction varies inversely as the square of the square of the distance of the body from the centre of the earth. This shows that in the space all around the earth, its gravitational pull can be experienced by other material bodies. The same is true for another material body.
Hence the space around a material body in which its gravitational pull can be experienced is called gravitational field.
The intensity of gravitational field measures the strength of the gravitational field. It is the force experienced by a body of unit mass at a point in the gravitational field.
To find gravitational field intensity at a point due to mass M at a distance r, let us keep an object of mass m at appoint i.e. at point P as shown in the figure. According to Newton’s law of gravitation, force between them is
\begin{align*} F &= \frac {GMm}{r^2} \\ \text {Intensity of gravitational field at that point will be} \\ E &= \frac Fm = \frac {GMm}{r^2m} = \frac {GM}{r^2} \\ \therefore E &= \frac {GM}{r^2} \\ \text {E is directed towards the centre of mass M.} \\ \end{align*}
Intensity of Gravitational Field on the Surface of the Earth
To find the intensity of gravitational field on the surface of the earth, place a body of mass m on its surface. Gravitational force in this body is
\begin{align*} F &= \frac {GMm}{R^2} \\ \end{align*} where G is the gravitational constant, M is the mass of the earth and R be its radius\begin{align*}\\ \therefore \text {Gravitational force intensity on the surface of the earth,} \\ E &= \frac fm = \frac {GMm}{mR^2} = \frac {GM}{R^2} = g \\ \end{align*}
Therefore the intensity of gravitational field at any point near the surface of the earth is equal to the acceleration due to gravity at that point. Its unit is Nkg^{-1} or ms^{-2}.
The work done in bringing a body of unit mass from infinity to the point considered without acceleration is defined as the gravitational potential at that point.
Gravitational potential Vg = \(\frac{w}{m}\)
Gravitational potential is a scalar quantity. Its SI-units is joule per kg.
Expression for Gravitational Potential at a point
Let us consider the earth as a perfect sphere of radius R and mass M. Let A be the point at a distance r from the centre of the earth, where the gravitational potential has to be calculated. Let us take a point C at a distance x from O as shown in the figure. Gravitational force of attraction on a body of unit mass at point C will be
F = \(\frac {Gm}{r^2}\)
If we displace the unit mass through a small distance dx from C to B, then small amount of work done has to be done which is given by
\begin{align*} dW &= F dx = \frac {GM}{x^2} dx \end{align*} Total work done in bringing the object of unit mass from \(\infty\) to the point A is obtained by integrating the above equation within the limits \begin{align*}x = \infty to x= r, \text {we get} \\ W &= \int _{infty}^r \frac {GM}{x^2} dx = -\left [ \frac {GM}{x} \right ]_{\infty }^r = -GM \left [\frac 1r - \frac {1}{\infty } \right ] \\ \text {or,} W &= - \frac {GM}{r} \\ \end{align*}This work done remains in the form of gravitational force when the object moves towards the earth. At the surface of the earth \begin{align*} r=R, then \\ V = - \frac {GM}{R} \end{align*}
Expression for Gravitational Potential Energy
Gravitational potential energy of an object at a point in the gravitational field is the work done to bring the object from infinity to that point without acceleration.
Let the earth be a perfect sphere of radius R and mass M. the mass of the earth can be supposed to be placed at a point A in the gravitational field, where OA = r as shown in the figure. Take two points B and C on the line of OP produced onwards such that OC = x and BC = dx.
Then the gravitational force of attraction on the body at point C will be
F = \(\frac{GMm}{x^2}\)
When the body is moved
ugh a very small distance CB=dx, then the small amount of work has to be done which is given by
dW= F dx =\(\frac {GMm}{x^2}\)
Total work done in bringing the body from infinity to point A is given by,
W = \(\underset{\infty}{\overset{r}{\LARGE\mathrm \int}}\) \(\frac {GMm}{x^2}\) = GM m \(\underset{\infty}{\overset{r}{\LARGE\mathrm \int}}\) x-2 dx
= -GM m [\(\frac{1}{x}\)]r\(\infty\) =- GM m [\(\frac{1}{r}\) - \(\frac{1}{\infty}\)]
W = - \(\frac{GM m}{r}\)
Since this work done is stored in the body as its gravitational potential energy U, the gravitational potential energy of the body at Point A in the gravitational field will be
U = - \(\frac{GM m}{r}\).... (ii)
\(\therefore\) From equation (i) and (ii), we get,
U = V_{A} \(\times\) m
i.e. gravitational potential energy =gravitational potential \(\times\) mass of the object
F = \(\frac{GMm}{x^2}\)
F = \(\frac {Gm}{r^2}\)
The space around a material body in which its gravitational pull can be experienced is called gravitational field.
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