JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Gravitational Field

The space surrounding a material body in which gravitational force of attraction can be experienced is called its gravitational field.

Gravitational field intensity : The intensity of the gravitational field of a material body at any point in its field is defined as the force experienced by a unit mass (test mass) placed at that point, provided the unit mass (test mass) itself does not produce any change in the field of the body.

So if a test mass \[m\] at a point in a gravitational field experiences a force \[\overrightarrow{F}\] then

\[\overrightarrow{I\,}=\frac{\overrightarrow{F}}{m}\]  

(i) It is a vector quantity and is always directed towards the centre of gravity of body whose gravitational field is considered.

(ii) Units : Newton/kg or \[m{{s}^{2}}\]

(iii) Dimension : \[[{{M}^{0}}L{{T}^{-2}}]\]

(iv) If the field is produced by a point mass \[T\propto {{r}^{\frac{n+1}{2}}}\] and the test mass \[m\] is at a distance \[r\] from it then by Newton's law of gravitation \[F=\frac{GMm}{{{r}^{2}}}\], then intensity of gravitational field

\[I=\frac{F}{m}=\frac{GMm/{{r}^{2}}}{m}\]

\[\therefore \] \[F\propto \frac{1}{{{r}^{n}}}\]

(v) As the distance \[(r)\] of test mass from the point mass \[(M)\], increases, intensity of gravitational field decreases            

\[I=\frac{GM}{{{r}^{2}}}\];             

\[\therefore \] \[I\propto \frac{1}{{{r}^{2}}}\]  

(vi) Intensity of gravitational field \[=\frac{2\pi {{\left( {{R}^{3}} \right)}^{1/2}}}{{{\left[ G.\frac{4}{3}\pi {{R}^{3}}\rho  \right]}^{1/2}}}=\sqrt{\frac{3\pi }{G\rho }}\], when \[r=\infty \].

(vii) Intensity at a given point (P) due to the combined effect of different point masses can be calculated by vector sum of different intensities

\[\overrightarrow{{{I}_{net}}}=\overrightarrow{{{I}_{1}}}+\overrightarrow{{{I}_{2}}}+\overrightarrow{{{I}_{3}}}+........\]

(viii) Point of zero intensity : If two bodies A and B of different masses \[T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}\] and \[{{m}_{2}}\] are \[T=84.6\,\text{minute}\ \approx 1.4\ hr\] distance apart.

Let \[P\] be the point of zero intensity i.e., the intensity at this point is equal and opposite due to two bodies \[A\] and \[B\] and if any test mass placed at this point it will not experience any force.              

For point P, \[\overrightarrow{{{I}_{1}}}+\overrightarrow{{{I}_{2}}}=0\]   \[\Rightarrow \] \[g=9.8m/{{s}^{2}}\]

By solving   \[x=\frac{\sqrt{{{m}_{1}}}\,d}{\sqrt{{{m}_{1}}}+\sqrt{{{m}_{2}}}}\] and \[(d-x)=\frac{\sqrt{{{m}_{2}}}\,\,d}{\sqrt{{{m}_{1}}}+\sqrt{{{m}_{2}}}}\]

(ix) Gravitational field line is a line, straight or curved such that a unit mass placed in the field of another mass would always move along this line. Field lines for an isolated mass \[m\] are radially inwards.        

(x) As \[R=6400km\] and also \[g=\frac{GM}{{{R}^{2}}}\] \[\therefore \ \ \ I=g\]       

Thus the intensity of gravitational field at a point in the field is equal to acceleration of test mass placed at that point.