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Gravitational Field Intensity - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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.

  • 54 Questions around this concept.

Solve by difficulty

A body of mass 60g experiences a gravitational force 3.0\mathrm{~N} when placed at a particular point. The magnitude of the gravitational field intensity at that point is:
 

Two bodies of masses m\: and \: 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

What is the equation for the gravitational field due to a uniform circular ring of radius 'a' at a point on its axis?

Dependence of intensity of gravitational field (E) of earth with distance (r) from centre of earth is correctly represented by

Starting from the centre of the earth having radius R, the variation of (acceleration due to gravity) is shown by

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The gravitational field intensity due to a uniform solid sphere at a distance r (for outside the solid sphere) from its centre is proportional to: 

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

\vec{I}=\frac{\vec{F}}{m}

\vec{I}\rightarrow G.field\: Intensity

m\rightarrow mass\: of\: object

\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-

        

Field lines are radially Inward 

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 .

a= radius of the ring

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<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<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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