Acceleration Due To Gravity MCQ - Practice Questions with Answers

The Gravitational Force exerted by the earth on a body is known as the gravitational pull of gravity. And this force will produce an acceleration in the motion of a body.

And this is known as the acceleration due to gravity.

This is denoted by g.

Let gravitational force exerted by the earth on the body of mass m resting on the surface of the earth is given by

$F=\frac{GMm}{R^2}$ …(1)

Where M = mass of the earth and R = radius of the earth

And If g is the acceleration due to gravity then this F can be written as

F=(mass)*(acceleration)=mg ...(2)

On Comparing 

We get

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

Now $\rho \to$ density of earth

$$\begin{array}{rl}& \text{ Then }M=\rho \ast \left(\frac{4}{3}\right)\pi R^3\\ & \text{ So }g=\frac{4}{3}\pi \rho GR\end{array}$$
 

• - Its average value is $9.8\mathrm{m}/{\mathrm{s}}^2$ or $981\mathrm{cm}/{\sec }^2$ or $32\mathrm{feet}/{\mathrm{s}}^2$ on the surface of the earth.
• - It is a vector quantity and its direction is always towards the center of the earth/Planet.
• - Dimension- $L{T}^{-2}$
• - Its value depends upon the mass, radius, and density of the Earth/Planet.

This means a Planet having more value of $\rho R$ than the earth will have a greater value of acceleration due to gravity
(g) than the earth

$$\begin{array}{rl}& If(\rho R{)}_P>(\rho R{)}_E\\ & \text{ ।.e }\Rightarrow g_P>g_E\end{array}$$
 

•  It is independent of the mass, shape, and density of the body situated on the surface of the Earth/planet.

The value of g will be the same for a light as well as heavy body if both are situated on the surface of the Earth/planet.

• The value of acceleration due to gravity (g) changes its value due to the following factors 

1. The shape of the earth

2. Height above the earth's surface

3. Depth below the earth's surface

4.  Axial rotation of the earth.

• Variation of 'g' due to the shape of the earth

Earth has an elliptical shape as shown in fig.

                       Where the Equatorial radius is about 21 km longer than the polar radius.

${}_{\text{Or }}{R}_e>R_p$
Where $R_e\to$ Radius of the equator
$R_p{\to }_{\text{Radius of pole }}$
${\mathrm{S}}_0{g}_p>g_e$

$$\mathrm{In}\mathrm{fact}{g}_p=g_e+0.018\mathrm{m}/{\mathrm{s}}^2$$
 

Or we can say Weight increases as the body is taken from the equator to the pole.

Let's study Variation in 'g' with height 

Value of g at the surface of the earth (at distance r=R from earth center)

$$g=\frac{GM}{R^2}$$
Value of $g$ at height $h$ from the surface of the earth (at a general distance $r=R+h$ from earth center)

$$g^{\prime }\alpha \frac{1}{r^2}$$

Where $r=R+h$
As we go above the surface of the earth, the value of $g$ decreases
So $g^{\prime }=\frac{GM}{r^2}$
Where $g^{\prime }\to$ gravity at a height h from the surface of the earth.
$R\to$ The radius of the earth
$h\to$ height above the surface
- Value of ' $g$ ' at $\mathrm{\infty }$
if $r=\mathrm{\infty }{g}^{\prime }=0$

No effect of Earth's gravitational pull at infinite distances.

• Value of g when h < < R

• Formula

1. Value of $g$

$$\begin{array}{rl}& g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2=g{(1+\frac{h}{R})}^{-2}\\ & g^{\prime }=g[1-\frac{2h}{R}]\end{array}$$

2. The absolute decrease in the value of $g$ with height

$$\mathrm{\Delta }g=g-g^{\prime }=\frac{2hg}{R}$$

3. The fractional decrease in the value of $g$ with height

$$\frac{\mathrm{\Delta }g}{g}=\frac{g-g^{\prime }}{g}=\frac{2h}{R}$$

4. Percentage decrease in the value of $g$ with height

$$\frac{\mathrm{\Delta }g}{g}\times 100\mathrm{\%}=\frac{2h}{R}\times 100\mathrm{\%}$$
 

Let's study Variation in 'g' with depth

Value of g at the surface of the earth (at d=0)

$$g=\frac{GM}{R^2}=\frac{4}{3}\pi \rho gR$$

Value of $g$ at depth $d$ from the surface of the earth (at a general distance $r=(R-d)$ from earth centre) $=g^{\prime }$
And $g^{\prime }\alpha (R-d)$
This means Value of g ' decreases on going below the surface of the earth.
So $g^{\prime }=g[1-\frac{d}{R}]$
- Value of ' $g$ ' at the centre of the earth

At the centre

$$[d=R]$$

So $g^{\prime }=0$

I.e Acceleration due to gravity at the centre of the earth becomes zero.

• - The absolute decrease in the value of $g$ with depth

  $$\mathrm{\Delta }g=g-g^{\prime }=\frac{dg}{R}$$

  - The fractional decrease in the value of $g$ with depth

  $$\frac{\mathrm{\Delta }g}{g}=\frac{g-g^{\prime }}{g}=\frac{d}{R}$$

  
  The value of $g$ decreases with depth.
  - Percentage decrease in the value of $g$ with depth

  $$\frac{\mathrm{\Delta }g}{g}\times 100\mathrm{\%}=\frac{d}{R}\times 100\mathrm{\%}$$

  
  Note- The rate of decrease of gravity outside the earth ( $h\ll \mathrm{R}$ ) is double that of inside the

Let's study Variation in 'g' due to the Rotation of the earth

As the earth rotates about its axis

Let its angular velocity be $\omega$ about an axis as shown in the figure.
So if a body is placed on its surface then it will move along the circular path.
So the apparent weight of the body will decrease as it will experience a centrifugal force due to rotation.
We can calculate this apparent weight of the body using force balance.
Let calculate its value for a body of mass $m$ situated at a point $P$ (having latitude $=\lambda$ )
As shown in the figure.
- $\lambda =$ The angle between the equatorial plane at that point and the line joining that point to the earth's center.
- For the poles $\lambda =90$ and for equator $\lambda =0$

$$\begin{array}{rl}& r=R\mathrm{Cos}\lambda \\ & F_c=m{\omega }^{2}r=m{\omega }^{2}R\cos \lambda \end{array}$$

By applying Newton's 2nd law along with the line joining point $P$ and center.

$$mg-F_c\mathrm{Cos}\lambda =m{g}^{\prime }$$

Where $g$ ' is the acceleration value due to gravity at point $P$.

- The apparent weight of the body decreases with an increase in angular velocity ( $\omega$ )
- The apparent weight of the body varies from point to point because each point has different latitudes and the magnitude of centrifugal force varies with the latitude of the place.
- For Pole, $\lambda =90$

So $g_{\text{pole }}=g$
I.e value of $g$ at the poles is independent of the angular velocity of the earth.
- For equator $\lambda =0$

$$g_{\text{equator }}=g-{\omega }^{2}R$$

I.e Decrease in the value of $g$ is maximum at the equator
- Weightlessness due to rotation of the earth-

Weightlessness means $g^{\prime }=0$
So $g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$
As $\lambda =0$ (For equator)
$O=g-{\omega }^{2}R{\cos }^{2}0$

$$\begin{array}{rl}& g-{\omega }^{2}R=0\\ & \omega =\sqrt{\frac{g}{R}}\end{array}$$

Where $\omega \to$ Angular velocity for which a body at the equator will become weightless
- The time period of Rotation of the earth for which the body at the equator will become weightless

$$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R}{g}}$$

Where $R\to$ Radius of the earth
And using $R=6400\times {10}^3\mathrm{m}$

$$g=10\mathrm{m}/{\mathrm{s}}^2$$

We get $\omega =\frac{1}{800}\frac{\mathrm{rad}}{\sec }$
And $T=1.40\mathrm{hr}$
- Relation of gravity at the poles and equator

After considering the effect of rotation, and the elliptical shape of the earth

$$g_p=g_e+0.052\mathrm{m}/{\mathrm{s}}^2$$

Where $g_p\to$ gravity at the pole
$g_e\to$ gravity at equator