Variation of g — Explained

On this page

1. Conceptual Foundation: Newton's Law of Gravitation
2. Key Principles and Laws Governing Variation of 'g'
3. Real-World Applications:
4. Common Misconceptions:
5. NEET-Specific Angle:

Detailed Explanation

The acceleration due to gravity, 'g', is a cornerstone concept in classical mechanics, representing the acceleration imparted to objects solely by the gravitational attraction of a celestial body, typically Earth. While often approximated as a constant $9.8,\text{m/s}^2$ for introductory problems, a deeper understanding reveals its intricate variations, which are critical for NEET aspirants.

Conceptual Foundation: Newton's Law of Gravitation

At its heart, the concept of 'g' stems from Newton's Universal Law of Gravitation. This law states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Mathematically, the gravitational force $F_g$ between two masses $M$ (e.g., Earth) and $m$ (e.g., an object) separated by a distance $r$ is given by:

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

where $G$ is the universal gravitational constant ($6.

674 imes 10^{-11}, ext{N m}^2/ ext{kg}^2$).

According to Newton's second law of motion, force equals mass times acceleration ($F = ma$). For an object falling freely under gravity, the gravitational force is the net force, so $F_g = mg$. Equating these two expressions for $F_g$:

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

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

Here, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the object.

This fundamental equation shows that 'g' depends on the mass of the Earth and the distance from its center, but not on the mass of the object itself. This is why all objects, regardless of their mass, fall with the same acceleration in a vacuum.

Key Principles and Laws Governing Variation of 'g'

1
Variation with Altitude (Height above Earth's Surface):

As an object moves to a height $h$ above the Earth's surface, its distance from the center of the Earth becomes $r = R + h$, where $R$ is the radius of the Earth. The acceleration due to gravity at this height, $g_h$, can be expressed as:

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

We know that at the Earth's surface ($h=0$), $g = G \frac{M}{R^2}$.

Dividing $g_h$ by $g$:

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

So, g_h = g left(1 + \frac{h}{R}\right)^{-2}.

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

g_h approx g left(1 - \frac{2h}{R}\right)

This approximation is frequently used in NEET problems. It clearly shows that $g_h < g$, meaning 'g' decreases with increasing altitude. The rate of decrease is approximately $2gh/R$.

1
Variation with Depth (Below Earth's Surface):

Consider an object at a depth $d$ below the Earth's surface. Its distance from the center of the Earth is $r' = R - d$. When the object is inside the Earth, the gravitational force is exerted only by the mass of the Earth contained within a sphere of radius $r'$.

Assuming the Earth has a uniform density $ho$, the mass of the Earth $M = \frac{4}{3}pi R^3 \rho$. The mass of the inner sphere $M'$ at radius $r'$ is $M' = \frac{4}{3}pi (R-d)^3 \rho$. The acceleration due to gravity at depth $d$, $g_d$, is:

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

At the surface, $g = G \frac{M}{R^2} = G \frac{\frac{4}{3}pi R^3 \rho}{R^2} = G \frac{4}{3}pi R \rho$.

Dividing $g_d$ by $g$:

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

So, g_d = g left(1 - \frac{d}{R}\right).

This equation shows that $g_d < g$, meaning 'g' also decreases with increasing depth. At the center of the Earth ($d=R$), $g_d = g(1-R/R) = 0$. This is a crucial result: an object at the Earth's center experiences zero gravitational acceleration.

1
Variation with Latitude (Due to Earth's Rotation):

The Earth is not a perfect sphere; it's an oblate spheroid, flattened at the poles and bulging at the equator. This shape is a consequence of its rotation. When an object is on the surface of the rotating Earth, it experiences a centrifugal force (or more accurately, the gravitational force must provide the necessary centripetal force for circular motion).

This outward-acting pseudo-force effectively reduces the apparent weight of the object, and thus the effective 'g'. Consider an object of mass $m$ at latitude $lambda$. It moves in a circle of radius $r = R coslambda$ (where $R$ is the Earth's radius) with angular velocity $omega$.

The centripetal force required is $m r omega^2 = m (R coslambda) omega^2$. The component of this force acting radially outwards from the Earth's center is $m R omega^2 cos^2lambda$. The effective acceleration due to gravity, $g'$, at latitude $lambda$ is given by:

$$g' = g - Romega^2 cos^2lambda$$

where $g$ is the acceleration due to gravity if the Earth were not rotating (or at the poles, where $lambda = 90^circ$ and $coslambda = 0$).

* At the equator ($lambda = 0^circ$, $coslambda = 1$): $g'_{equator} = g - Romega^2$. This is the minimum value of 'g' due to rotation. * At the poles ($lambda = 90^circ$, $coslambda = 0$): $g'_{poles} = g$.

This is the maximum value of 'g' due to rotation. The Earth's rotation causes 'g' to be maximum at the poles and minimum at the equator. The difference $Romega^2$ is approximately $0.034,\text{m/s}^2$, which is small but significant for precise measurements.

Real-World Applications:

Satellite Orbits: — The precise calculation of 'g' at various altitudes is fundamental for determining the orbital mechanics of satellites, ensuring they stay in their intended paths.
Geodesy and Cartography: — Variations in 'g' are used to map the Earth's gravitational field, which helps in understanding its internal structure and creating accurate maps.
Oil and Mineral Exploration: — Local variations in 'g' (gravitational anomalies) can indicate the presence of denser or less dense materials underground, aiding in the discovery of mineral deposits and oil reserves.
Navigation Systems: — High-precision navigation systems, especially for submarines and aircraft, account for variations in 'g' to maintain accuracy.
Weight Measurement: — While mass is invariant, weight ($W = mg$) varies with 'g'. An object weighs slightly less at the equator than at the poles, and less on a mountain than at sea level.

Common Misconceptions:

'g' is a universal constant: — While $G$ (universal gravitational constant) is constant, 'g' is specific to a celestial body and varies even on its surface.
'g' increases with depth: — Students often mistakenly assume that getting closer to the Earth's center means stronger gravity. However, the 'shell theorem' (which underlies the depth variation derivation) dictates that only the mass inside the sphere of the object's radius contributes to the net gravitational force, leading to a decrease.
Ignoring Earth's rotation: — For many problems, the rotational effect is small and ignored, but it's a real physical phenomenon that causes measurable differences in 'g' at different latitudes. NEET questions might specifically test this.

NEET-Specific Angle:

For NEET, understanding the formulas for variation of 'g' with altitude, depth, and latitude is paramount. Students should be able to: * Apply the exact formula g_h = g left(1 + \frac{h}{R}\right)^{-2} and its approximation g_h approx g left(1 - \frac{2h}{R}\right) for altitude.

Know when to use the approximation (typically for $h ll R$, e.g., $h < 5$). * Apply the formula g_d = g left(1 - \frac{d}{R}\right) for depth. Understand that 'g' is maximum at the surface and zero at the center.

* Apply the formula $g' = g - Romega^2 cos^2lambda$ for latitude. Understand the implications for 'g' at poles and equator. * Compare the rates of decrease of 'g' with altitude and depth. For small changes, 'g' decreases twice as fast with height as it does with depth (i.

e., $2h/R$ vs $d/R$). Solve numerical problems involving these formulas. Answer conceptual questions about the direction of change (increase/decrease) and the reasons behind it. * Recognize that the Earth's non-uniform density and irregular shape also cause minor local variations, though these are usually beyond the scope of basic NEET problems.

Often confused with

Side-by-side differences the NEET paper likes to test.

Variation of g vs Variation of g with Altitude vs. Depth

Aspect	Variation of g	Variation of g with Altitude vs. Depth
Formula (for small changes)	$g_h approx g(1 - 2h/R)$	$g_d = g(1 - d/R)$
Rate of decrease (for small changes)	Decreases approximately as $2gh/R$	Decreases approximately as $gd/R$
Maximum value	At Earth's surface ($h=0$)	At Earth's surface ($d=0$)
Minimum value	Approaches zero as $h o infty$	Zero at Earth's center ($d=R$)
Reason for decrease	Increased distance from Earth's center, weakening gravitational pull.	Decreased effective mass of Earth pulling the object (mass above cancels out).

The acceleration due to gravity 'g' decreases both with increasing altitude and increasing depth from the Earth's surface. However, the rate and underlying physical reasons differ. For small changes, 'g' decreases twice as fast with height as it does with depth.

At infinite height, 'g' approaches zero, while at the Earth's center, 'g' becomes exactly zero. The altitude variation is due to increased distance from the entire Earth's mass, whereas depth variation is due to the reduction of the effective gravitating mass.

Why it is tested: For NEET, understanding these distinct variations is crucial for solving numerical problems and conceptual questions. Students must be able to apply the correct formula based on whether the object is above or below the surface and compare the magnitudes of 'g' at different points.

Questions students ask

5 answered on this topic.

Why does 'g' decrease with altitude?

The acceleration due to gravity, 'g', is inversely proportional to the square of the distance from the center of the Earth ($g = GM/r^2$). As you move to a higher altitude, your distance 'r' from the Earth's center increases. Since 'r' is in the denominator and squared, even a small increase in 'r' leads to a noticeable decrease in 'g'. This is a direct consequence of Newton's Law of Universal Gravitation, where gravitational force weakens with increasing separation.

Why does 'g' decrease with depth, and become zero at the Earth's center?

When an object is at a certain depth below the Earth's surface, the gravitational force it experiences is only due to the mass of the Earth contained within a sphere whose radius extends from the center to the object's position.

The mass of the Earth outside this sphere (the 'shell' above the object) exerts no net gravitational force on the object. As you go deeper, the effective mass pulling you downwards decreases, leading to a reduction in 'g'.

At the Earth's center, you are surrounded by Earth's mass in all directions, and the net gravitational pull is zero, hence 'g' is zero.

How does Earth's rotation affect 'g'?

The Earth's rotation creates a centrifugal effect that acts outwards, opposing the inward pull of gravity. This effect is maximum at the equator, where the rotational speed is highest, and zero at the poles. Consequently, the effective acceleration due to gravity ('g'') is reduced at the equator ($g' = g - Romega^2$) and is maximum at the poles ($g' = g$). This means objects weigh slightly less at the equator than at the poles due to this rotational effect.

When should I use the approximation $g_h = g(1 - 2h/R)$ for altitude?

The approximation $g_h = g(1 - 2h/R)$ is derived using the binomial expansion and is valid when the height 'h' is much smaller than the Earth's radius 'R' ($h ll R$). Typically, this means when 'h' is less than about 5% of 'R'. For larger heights, the exact formula $g_h = g(1 + h/R)^{-2}$ must be used to maintain accuracy. NEET problems often specify conditions or provide options that guide which formula to use.

Is 'g' truly constant at a given location?

No, 'g' is not truly constant even at a given location over time, though its variations are usually very small and often ignored in basic physics. Factors like tidal forces from the Moon and Sun, changes in local atmospheric pressure, and even very slight seismic activity can cause minuscule fluctuations in 'g'. However, for NEET purposes, 'g' at a specific latitude and altitude is generally considered constant unless otherwise specified.