Gravitational Field Intensity MCQ - Practice Questions with Answers

• Gravitational field-

It is space or surroundings 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

$$\overrightarrow{I}=\frac{\overrightarrow{F}}{m}$$

$\overrightarrow{I}\to$ G. Field Intensity
$m\to$ mass of object
$\overrightarrow{f}\to$ 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 : $\frac{\text{ Newton }}{\mathrm{kg}}$ or $\frac{\mathrm{m}}{{\mathrm{s}}^2}$

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

• 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 $F=\frac{GmM}{r^2}$

The corresponding I is given by 

$$\begin{array}{rl}& I=\frac{F}{m}=\frac{GMm}{r^{2}m}\\ & I=\frac{GM}{r^2}\end{array}$$

Where $G\to$ Gravitational const

$$\begin{array}{rl}M& \to \text{ mass of earth }\\ \text{ 1. }I& \propto \frac{1}{r^2}\end{array}$$
This means As the distance (r) of test mass from point (M) Increases I decreases.

$$\text{ 2. }I=0\text{ at }(r=\mathrm{\infty })$$
 

• 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 

${\overrightarrow{I}}_{net}={\overrightarrow{I}}_1+{\overrightarrow{I}}_2+{\overrightarrow{I}}_3+\dots \dots$.

• Point of zero intensity-

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

And P is the point where net Intensity $={\overrightarrow{I}}_{net}={\overrightarrow{I}}_1+{\overrightarrow{I}}_2=0$
Then P is the point of zero intensity
Let point $P$ is at distance $\times$ from m 1
Then For point $\mathrm{P}{\overrightarrow{I}}_{\text{net }}={\overrightarrow{I}}_1+{\overrightarrow{I}}_{2=0}$

$$-\frac{G{m}_1}{x^2}+\frac{G{m}_2}{(d-x{)}^2}=0$$

Then

$$x=\frac{\sqrt{m_1}d}{\sqrt{m_1}+\sqrt{m_2}}$$

And

$$(d-x)=\frac{\sqrt{m_2}d}{\sqrt{m_1}+\sqrt{m_2}}$$
 

• Gravitational field line-

          Field line of Isolated mass-

Field lines are radially Inward 

$\begin{array}{rl}& I=\frac{GM}{r^2}\\ & g=\frac{GM}{R^2}\Rightarrow I=g\end{array}$

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.

The value of g is larger when lines are close together

While the Value of g is smaller when they are far apart

3. The lines never cross

4. Lines do not form closed loops

For Uniform circular ring

At the center of the ring

$$I=0$$

At a point on its Axis

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

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

a= radius of the ring

For Uniform disc

$\mathrm{\Theta }\to$ Angle with axis
$a\to$ Radius of disc
At the center of the disc

$$I=0$$

At a point on its axis

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

For Spherical shell

$R\to$ Radius of shell
$r\to$ Position of Pt.
$M\to$ Mass of a spherical shell
- Inside the surface

$$\begin{array}{c}r<R\\ I=0\end{array}$$

- on the surface

$$\begin{array}{rl}& r=R\\ & I=\frac{GM}{R^2}\end{array}$$

Outside the surface $r>R$

$$I=\frac{GM}{r^2}$$
 

For  Uniform solid sphere-

- Inside surface $r<R$

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

- on the surface

$$\begin{array}{rl}& r=R\\ & I=\frac{GM}{R^2}\end{array}$$

- Outside surface $(r>R)$

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