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

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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

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.

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

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