Acceleration due to Gravity and its Variation

The acceleration due to gravity, denoted by 'g', is the acceleration experienced by an object due to the gravitational force exerted by a celestial body, typically the Earth. Near the Earth's surface, this acceleration is approximately 9.8 m/s². It's crucial to understand that this value isn't constant everywhere and can vary due to several factors.

Derivation of Acceleration due to Gravity

Let's consider an object of mass 'm' at a distance 'r' from the center of the Earth, which has a mass 'M' and radius 'R'. The gravitational force exerted by the Earth on the object is given by Newton's law of universal gravitation:

$F = G \frac{Mm}{r^2}$

According to Newton's second law of motion, this force is also equal to the product of the object's mass and its acceleration (which is 'g' in this case):

$F = mg$

Equating these two expressions for force:

$mg = G \frac{Mm}{r^2}$

The mass of the object 'm' cancels out, giving us the expression for acceleration due to gravity:

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

Here, G is the universal gravitational constant ($6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2$), M is the mass of the Earth, and r is the distance from the center of the Earth.

If we consider an object at the Earth's surface, then $r = R$ (the radius of the Earth). So, the acceleration due to gravity at the surface is:

$g_{surface} = G \frac{M}{R^2}$

We can also express 'g' in terms of the Earth's density ($\rho$). Assuming the Earth is a uniform sphere, its mass M can be written as:

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

Substituting this into the expression for $g_{surface}$:

$g_{surface} = G \frac{(\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3}\pi GR\rho$

This shows that 'g' is directly proportional to the radius and density of the planet, assuming uniform density.

Variation in Acceleration due to Gravity

The value of 'g' is not uniform across the Earth's surface and varies due to the following primary reasons:

1. Altitude (Height above the surface)

As an object moves to a higher altitude (away from the Earth's surface), the distance 'r' from the center of the Earth increases. Since 'g' is inversely proportional to the square of the distance ($g \propto 1/r^2$), the acceleration due to gravity decreases with increasing altitude.

Let $g_h$ be the acceleration due to gravity at a height 'h' above the Earth's surface. The distance from the center of the Earth is $r = R+h$.

$g_h = G \frac{M}{(R+h)^2}$

We can rewrite this in terms of $g_{surface}$:

$g_h = G \frac{M}{R^2(1 + h/R)^2} = g_{surface} \left( \frac{1}{(1 + h/R)^2} \right) = g_{surface} (1 + h/R)^{-2}$

For small heights (h << R), we can use the binomial approximation $(1+x)^n \approx 1+nx$ for $|x| \ll 1$. Here, $x = h/R$ and $n = -2$.

$g_h \approx g_{surface} (1 - 2h/R)$

This approximation clearly shows that 'g' decreases linearly with height 'h' for small altitudes.

Example: At the top of Mount Everest (approx. 8.8 km above sea level), the acceleration due to gravity is slightly less than at sea level.

2. Depth below the surface

When an object is at a depth 'd' below the Earth's surface, its distance from the center of the Earth is $r = R-d$. However, the gravitational force experienced by the object is only due to the mass of the Earth within the sphere of radius $r = R-d$. The mass of the shell between radius 'r' and 'R' exerts no net force on the object at radius 'r' (Shell Theorem).

Let $M_r$ be the mass of the Earth within radius $r = R-d$. Assuming uniform density $\rho$:

$M_r = \frac{4}{3}\pi r^3 \rho$

The acceleration due to gravity at depth 'd' (at radius 'r') is:

$g_d = G \frac{M_r}{r^2} = G \frac{(\frac{4}{3}\pi r^3 \rho)}{r^2} = \frac{4}{3}\pi G r \rho$

Substituting $r = R-d$:

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

We can relate this to $g_{surface}$:

$g_{surface} = \frac{4}{3}\pi GR\rho$

So, $g_d = \frac{R-d}{R} g_{surface} = \left(1 - \frac{d}{R}\right) g_{surface}$

This formula shows that acceleration due to gravity decreases linearly with depth 'd' from the surface. At the center of the Earth ($d=R$), $g_d$ becomes zero.

Example: In a mine shaft, the acceleration due to gravity is slightly less than at the surface.

3. Shape of the Earth (Ellipsoidal Shape)

The Earth is not a perfect sphere but is slightly flattened at the poles and bulges at the equator. This is due to its rotation. The equatorial radius is larger than the polar radius.

Since $g \propto 1/R^2$, and the equatorial radius ($R_e$) is greater than the polar radius ($R_p$), the acceleration due to gravity is less at the equator than at the poles.

$g_{equator} < g_{pole}$

The average value of 'g' at the equator is about 9.78 m/s², while at the poles, it is about 9.83 m/s².

4. Rotation of the Earth (Centrifugal Effect)

The Earth rotates on its axis. This rotation causes a centrifugal acceleration, which acts outwards, opposing the gravitational force. This effect is maximum at the equator and zero at the poles.

The centrifugal acceleration ($a_c$) at a point on the Earth's surface at latitude $\lambda$ is given by $a_c = \omega^2 R \cos\lambda$, where $\omega$ is the angular velocity of the Earth's rotation and R is the radius of the Earth.

The effective acceleration due to gravity ($g_{eff}$) is the vector difference between the true gravitational acceleration ($g_{true}$) and the centrifugal acceleration ($a_c$):

$g_{eff} = g_{true} - a_c$ (approximately, as $a_c$ is small compared to $g_{true}$)

At the equator ($\lambda = 0^\circ$), $\cos\lambda = 1$, so $a_c$ is maximum, and $g_{eff}$ is minimum. $g_{equator} = g_{true, equator} - \omega^2 R_e$

At the poles ($\lambda = 90^\circ$), $\cos\lambda = 0$, so $a_c = 0$, and $g_{eff}$ is maximum (equal to $g_{true}$). $g_{pole} = g_{true, pole}$

The Earth's rotation accounts for about 0.02 m/s² of the difference between the poles and the equator. The remaining difference is due to the Earth's ellipsoidal shape.

5. Local Variations (Irregularities in Earth's Mass Distribution)

The Earth's crust is not uniform in density. The presence of dense materials like mineral deposits or less dense regions like underground caves can cause local variations in the gravitational field and hence in the acceleration due to gravity. These effects are usually very small and localized.

Summary of Variations

The value of 'g' is highest at the poles and lowest at the equator. It decreases as we go up in altitude or down in depth (though the behavior with depth is different from altitude).

Factor	Effect on 'g'	Formula/Relation
Altitude (h)	Decreases	$g_h \approx g_{surface} (1 - 2h/R)$ for $h \ll R$
Depth (d)	Decreases (linearly)	$g_d = g_{surface} (1 - d/R)$
Latitude (Rotation)	Decreases from Pole to Equator	$g_{eff} = g_{true} - \omega^2 R \cos\lambda$
Shape (Ellipsoidal)	Decreases from Pole to Equator	$R_{equator} > R_{pole}$ implies $g_{equator} < g_{pole}$