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

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

$\vec{I} \rightarrow$ 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 : $\frac{\text { Newton }}{\mathrm{kg}}$ 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 $F=\frac{G m M}{r^2}$

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

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

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

• 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 

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

• 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 $=\vec{I}_{n e t}=\vec{I}_1+\vec{I}_2=0$
Then P is the point of zero intensity
Let point $P$ is at distance $\times$ from m 1
Then For point $\mathrm{P} \vec{I}_{\text {net }}=\vec{I}_1+\vec{I}_{2=0}$

$$
-\frac{G m_1}{x^2}+\frac{G m_2}{(d-x)^2}=0
$$

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 

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

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

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

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

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 a spherical shell
- Inside the surface

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

- on the surface

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

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 $(r>R)$

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