Gravitational Field Intensity - Practice Questions & MCQ

• 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

$$
\begin{aligned}
& \vec{I}=\frac{\vec{F}}{m} \\
& \vec{I} \rightarrow \text { G.field Intensity } \\
& m \rightarrow \text { mass of object } \\
& \vec{f} \rightarrow \text { Gravitational Force }
\end{aligned}
$$

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 : $\frac{N e w t o n}{k g}$ or $\frac{m}{s^2}$
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 $F=\frac{G m M}{r^2}$
And the corresponding I is given by

$
\begin{aligned}
& I=\frac{F}{m}=\frac{G M m}{r^2 m} \\
& I=\frac{G M}{r^2}
\end{aligned}
$

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.
2. $I=0$ at $(r=\infty)$

• 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 

$\vec{I}_{n e t}=\vec{I}_1+\vec{I}_2+\vec{I}_3+\ldots \ldots \ldots$

• 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 $=\vec{I}_{\text {net }}=\vec{I}_1+\vec{I}_{2=0}$
Then P is the point of zero intensity
Let point $P$ is at distance x from m 1
Then For point $\mathrm{P} \vec{I}_{\text {net }}=\vec{I}_1+\vec{I}_{2=0}$
$-\frac{G m_1}{x^2}+\frac{G m_2}{(d-x)^2}=0$
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
As $I=\frac{G M}{r^2}$

$
g=\frac{G M}{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

3. The lines never cross

4. Lines do not form closed loops

For Uniform circular ring

At the center of ring

$
I=0
$

At a point on its Axis

$
I=\frac{G M r}{\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 .
$\mathrm{a}=$ radius of the ring

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{2 G M}{a^2}(1-\cos \Theta)
$
 

For Spherical shell

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

$
\begin{aligned}
& r<R \\
& I=0
\end{aligned}
$

- on the surface

$
\begin{aligned}
& r=R \\
& I=\frac{G M}{R^2}
\end{aligned}
$

- $\quad$ Outside the surface $r>R$

$
I=\frac{G M}{r^2}
$
 

For  Uniform solid sphere-

• - Inside surface $r<R$

  $
  I=\frac{G M r}{R^3}
  $

  - on the surface

  $
  \begin{aligned}
  & r=R \\
  & I=\frac{G M}{R^2}
  \end{aligned}
  $

  - Outside surface $(r>R)$

  $
  I=\frac{G M}{r^2}
  $