Derivation : Acceleration Due to Gravity with Depth ( Below ) | Derivation, Formula, Numericals, MCQs & FAQs | Class 11 Physics

Variation of Acceleration Due to Gravity with Depth (Below the Earth's Surface)

When a body is taken to a depth d below the Earth's surface, the acceleration due to gravity decreases. This variation of gravity with depth can be derived by assuming that the Earth is a homogeneous sphere of uniform density.

Derivation of Variation of g with Depth

The value of acceleration due to gravity on the Earth's surface is:

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

where:

• G = Universal Gravitational Constant
• M = Mass of Earth
• R = Radius of Earth

For a homogeneous Earth of density ρ,

$$ M=\frac{4}{3}\pi R^3\rho $$

Substituting the value of M:

$$ g=\frac{G\left(\frac{4}{3}\pi R^3\rho\right)}{R^2} $$ $$ g=\frac{4}{3}\pi GR\rho $$

Let a body be at a depth d below the Earth's surface. The effective radius becomes:

$$ r=R-d $$

The mass enclosed within radius (R − d) is:

$$ M'=\frac{4}{3}\pi (R-d)^3\rho $$

Therefore, acceleration due to gravity at depth d is:

$$ g_d=\frac{GM'}{(R-d)^2} $$ $$ g_d=\frac{G\left(\frac{4}{3}\pi (R-d)^3\rho\right)}{(R-d)^2} $$ $$ g_d=\frac{4}{3}\pi G(R-d)\rho $$

Dividing by the expression of g:

$$ \frac{g_d}{g} = \frac{\frac{4}{3}\pi G(R-d)\rho} {\frac{4}{3}\pi GR\rho} $$ $$ \frac{g_d}{g} = \frac{R-d}{R} $$ $$ \frac{g_d}{g} = 1-\frac{d}{R} $$

Hence,

$$ \boxed{g_d=g\left(1-\frac{d}{R}\right)} $$

Conclusion

Thus, the value of acceleration due to gravity decreases linearly with depth below the Earth's surface.

$$ \boxed{g_d=g\left(1-\frac{d}{R}\right)} $$

Important Facts

1. At the Surface of the Earth

For d = 0,

$$ g_d=g $$

2. At the Centre of the Earth

For d = R,

$$ g_{centre} = g\left(1-\frac{R}{R}\right) = 0 $$

Hence, acceleration due to gravity at the centre of the Earth is zero.

3. Weight at the Centre of the Earth

$$ W=mg=0 $$

Therefore, the weight of a body becomes zero at the centre.

4. Decrease in Gravity with Depth

$$ g-g_d=\frac{gd}{R} $$

5. Percentage Decrease in Gravity

$$ \%\text{ decrease} = \frac{g-g_d}{g}\times100 $$ $$ = \frac{d}{R}\times100 $$

Key Formulae

$$ g_d=g\left(1-\frac{d}{R}\right) $$ $$ g-g_d=\frac{gd}{R} $$ $$ \%\text{ decrease} = \frac{d}{R}\times100 $$ $$ g_{centre}=0 $$ $$ W_{centre}=0 $$

Frequently Asked Questions (FAQ)

Q1. What is the formula for acceleration due to gravity at depth d?

Answer:

$$ g_d=g\left(1-\frac{d}{R}\right) $$

Q2. How does gravity vary with depth?

Gravity decreases linearly with depth below the Earth's surface.

Q3. What is the value of gravity at the centre of the Earth?

Answer:

$$ g_{centre}=0 $$

Q4. Why is gravity zero at the centre of the Earth?

Because gravitational pulls from all directions cancel each other.

Q5. What is the weight of a body at the Earth's centre?

Answer:

$$ W=mg=0 $$

Q6. What is the percentage decrease in gravity at depth d?

$$ \frac{d}{R}\times100 $$

Quiz Questions (MCQs)

1. The value of acceleration due to gravity at the centre of the Earth is:
   (a) g
   (b) g/2
   (c) 0
   (d) Infinite
   Answer: (c) 0


2. Gravity at depth d is:
   (a) g(1 + d/R)
   (b) g(1 − d/R)
   (c) gR/d
   (d) g
   Answer: (b)


3. The weight of a body at the Earth's centre is:
   (a) Maximum
   (b) Minimum
   (c) Zero
   (d) Infinite
   Answer: (c)


4. Gravity inside the Earth:
   (a) Increases linearly
   (b) Decreases linearly
   (c) Remains constant
   (d) Becomes infinite
   Answer: (b)


5. If a body is taken to a depth R/2, the value of gravity becomes:
   (a) g
   (b) g/2
   (c) 2g
   (d) 0
   Answer: (b)

Solved Numericals

Numerical 1

A mine is located 1600 km below the Earth's surface. Calculate the value of g at that depth.

Given:

$$ R=6400\ km $$ $$ d=1600\ km $$

Solution:

$$ g_d=g\left(1-\frac{d}{R}\right) $$ $$ g_d=9.8\left(1-\frac{1600}{6400}\right) $$ $$ g_d=9.8(0.75) $$ $$ \boxed{g_d=7.35\ m/s^2} $$

Numerical 2

Find the acceleration due to gravity at a depth equal to one-fourth of Earth's radius.

$$ d=\frac{R}{4} $$ $$ g_d=g\left(1-\frac{1}{4}\right) $$ $$ g_d=\frac{3g}{4} $$ $$ g_d=7.35\ m/s^2 $$

Numerical 3

At what depth will gravity become half of its surface value?

$$ \frac{g}{2}=g\left(1-\frac{d}{R}\right) $$ $$ \frac{1}{2}=1-\frac{d}{R} $$ $$ \frac{d}{R}=\frac{1}{2} $$ $$ \boxed{d=\frac{R}{2}} $$

Numerical 4

Calculate the percentage decrease in gravity at a depth of 800 km.

$$ R=6400\ km $$ $$ d=800\ km $$ $$ \%\text{ decrease} = \frac{d}{R}\times100 $$ $$ = \frac{800}{6400}\times100 $$ $$ =12.5\% $$ $$ \boxed{12.5\%} $$

Notes : Class 11 Physics – Chapter 7: Gravitation (NCERT Topics)

1. Introduction
2. Kepler's Laws
3. Universal Law of Gravitation
4. Newton's law of Gravitation in Vector form
5. Principle of Superposition : Gravitation force
6. Shell Theorem and Gravitation Shielding 
7. The Gravitational Constant
8. Acceleration Due to Gravity of the Earth
9. Mass and Mean Density of Earth
10. Acceleration Due to Gravity Above the Surface of Earth
11. Acceleration Due to Gravity Below the Surface of Earth
12. Gravitational Potential Energy
13. Escape Speed
14. Earth Satellites
15. Orbiting velocity of Satellite
16. Total Energy of Satellite