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# Gravitational Field Intensity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Gravitational field Intensity is considered one the most difficult concept.

• Gravitational field due to Point mass is considered one of the most asked concept.

• 36 Questions around this concept.

## Solve by difficulty

What is the unit of gravitational-field Intensity?

What is the dimension of Gravitational field Intensity :

What is the effect of increasing the mass of an object on the strength of the gravitational field at a point in space?

Two bodies of masses are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

Which of the following is true regarding the Properties of the Gravitational field line-

What is the formula for the gravitational field due to a point mass at a distance r from the mass?

What is the direction of the gravitational field due to the point mass?

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What is the difference between gravitational force and gravitational field due to point mass?

What is the gravitational field at the centre of a uniform circular ring?

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What is the direction of the gravitational field at a point on the axis of a uniform circular ring?

## Concepts Covered - 6

Gravitational field Intensity
• Gravitational field-

It is space or surrounding in which a material body feels the gravitational force of attraction.

• Gravitational field Intensity-

It is the force experienced by a unit mass at a point in the field.

It is denoted by I

If the mass of a body is m then I is given by

$\dpi{100} \vec{I}=\frac{\vec{F}}{m}$

$\dpi{100} \vec{I}\rightarrow G.field\: Intensity$

$\dpi{100} m\rightarrow mass\: of\: object$

$\dpi{100} \vec{f}\rightarrow Gravitational\: Force$

1.  It is a vector quantity

2.   If the field is produced by a body M the direction of its Gravitational field Intensity is always towards the center of gravity of M.

3. $Unit : \frac{Newton}{kg}\: or\: \frac{m}{s^{2}}$

4. $Dimension : \left [ M^{0}LT^{-2} \right ]$

Gravitational field due to Point mass
• Gravitational field due to Point mass

If the point mass M is producing the field and test mass is at distance r as shown in fig

So Force is given as $F=\frac{GmM}{r^{2}}$

And the corresponding I is given by

$I= \frac{F}{m}= \frac{GMm}{r^{2}m}$

$I= \frac{GM}{r^{2}}$

Where $G \rightarrow Gravitational\: const$

$M\rightarrow mass\: of\: earth$

1.  $I\propto \frac{1}{r^{2}}$

Means As the distance (r) of test mass from point (M) Increases I decreases.

1. $I= 0\: at\: \left ( r= \infty \right )$

• Superposition of Gravitational field

The net Intensity at a given point due to different point masses (M1,M2,M3…) can be calculated  by doing the vector sum of their individual intensities

$\vec{I}_{net}=\vec{I_{1}}+\vec{I_{2}}+\vec{I_{3}}+........$

• Point of zero intensity-

Let m1 and m2 are separated at a distance d from each other

And P is the point where net Intensity=$\vec{I}_{net}=\vec{I_{1}}+\vec{I_{2}}$=0

Then P is the point of zero intensity

Let point P is at distance x from m1

Then For point P $\vec{I}_{net}=\vec{I_{1}}+\vec{I_{2}}$=0

$- \, \frac{Gm_{1}}{x^{2}}\, +\, \frac{Gm_{2}}{\left ( d-x \right )^{2}}=0$

Then $x= \frac{\sqrt{m_{1}}\: d}{\sqrt{m_{1}}+\sqrt{m_{2}}}$

And $\left ( d-x \right )= \frac{\sqrt{m_{2}}\: d}{\sqrt{m_{1}}+\sqrt{m_{2}}}$

• Gravitational field line-

Field line of Isolated mass-

As  $I= \frac{GM}{r^{2}}$

$g= \frac{GM}{R^{2}}\Rightarrow I=g$

So we can say that the intensity of the gravitational field at a point P in the field of Isolated mass is equal to the acceleration of test mass placed at that point P.

Properties of Gravitational field line-

1. The line includes arrows which represent the direction of the gravitational field.

2. The magnitude of the gravitational field is proportional to the number of field lines crossing a unit area perpendicular to them.

Value of g is larger when lines are close together

While Value of g smaller when they are far apart

1. The lines never cross

2. Lines do not form closed loops

Gravitational field due to uniform circular ring

For Uniform circular ring

At the center of ring

$I=0$

At a point on its Axis

$I= \frac{GMr}{\left ( a^{2}+r^{2} \right )^{\frac{3}{2}}}$

Where,

$r\rightarrow$ The distance of the point P along the Axis of the ring, from its center .

Gravitational field Intensity due to uniform disc

For Uniform disc

$\Theta \rightarrow$ Angle with axis

$a \rightarrow$ Radius of disc

At the center of the disc

$I=0$

At a point on its axis

$I=\frac{2GM}{a^{2}}\left ( 1-\cos \Theta \right )$

Gravitational field Intensity due to spherical shell/hollow sphere

For Spherical shell

$R\rightarrow$ Radius of shell

$r\rightarrow$ Position of Pt.

$M\rightarrow$ Mass of spherical shell

• Inside the surface

$r

$I=0$

• on the surface

$r=R$

$I= \frac{GM}{R^{2}}$

•                Outside the surface $r>R$

$I= \frac{GM}{r^{2}}$

Gravitational field Intensity due to uniform solid sphere

For  Uniform solid sphere-

•  Inside surface  $r

$I= \frac{GMr}{R^{3}}$

• on the surface

$r=R$

$I= \frac{GM}{R^{2}}$

• Outside surface $\left ( r>R \right )$

$I= \frac{GM}{r^{2}}$

## Study it with Videos

Gravitational field Intensity
Gravitational field due to Point mass
Gravitational field due to uniform circular ring

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