JEE Main & Advanced Physics Gravitation Gravitational Field

Gravitational Field

Category : JEE Main & Advanced

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.

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