Acceleration due to Gravity — Explained

Detailed Explanation

The concept of acceleration due to gravity, denoted by $g$, is a cornerstone of classical mechanics and a direct consequence of Newton's Universal Law of Gravitation. It describes the acceleration experienced by an object solely under the influence of gravitational force. For NEET aspirants, a deep understanding of its derivation, standard value, and various factors influencing its magnitude is indispensable.

1. Conceptual Foundation: Linking Gravity and Acceleration

At its heart, acceleration due to gravity is the specific acceleration an object undergoes when the only force acting upon it is gravity. Consider an object of mass $m$ near the surface of a planet of mass $M$ and radius $R$.

According to Newton's Universal Law of Gravitation, the gravitational force $F_g$ exerted by the planet on the object is given by:

$$F_g = \frac{G M m}{r^2}$$

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

67 imes 10^{-11}, ext{N m}^2/ ext{kg}^2$) and$r$is the distance between the center of the planet and the center of the object. If the object is on or very near the surface,$r$can be approximated as the radius of the planet,$R$.

Simultaneously, according to Newton's Second Law of Motion, the net force acting on an object is equal to the product of its mass and its acceleration ($F = ma$). In this case, the acceleration is $g$, so the gravitational force can also be written as:

$$F_g = m g$$

2. Derivation of Acceleration due to Gravity ($g$)

By equating these two expressions for the gravitational force, we can derive a formula for $g$:

$$m g = \frac{G M m}{R^2}$$

Notice that the mass of the object, $m$, cancels out from both sides. This is a profound result: the acceleration due to gravity is independent of the mass of the falling object. This means a feather and a hammer, if dropped in a vacuum, would fall with the same acceleration.

Thus, the acceleration due to gravity at the surface of a planet is:

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

For Earth, taking $M_E = 5.97 \times 10^{24},\text{kg}$ (mass of Earth) and $R_E = 6.37 \times 10^6,\text{m}$ (mean radius of Earth), and $G = 6.

67 imes 10^{-11}, ext{N m}^2/ ext{kg}^2$, we get:$$g approx \frac{(6.67 \times 10^{-11},\text{N m}^2/\text{kg}^2) \times (5.97 \times 10^{24},\text{kg})}{(6.37 \times 10^6,\text{m})^2} approx 9.8,\text{m/s}^2$$ This is the standard value often used in calculations near the Earth's surface.

3. Factors Affecting Acceleration due to Gravity

While $g approx 9.8,\text{m/s}^2$ is a useful approximation, its actual value varies. Understanding these variations is crucial for NEET.

**a) Variation with Altitude (Height $h$):**

As an object moves above the Earth's surface to a height $h$, its distance from the Earth's center becomes $R_E + h$. The acceleration due to gravity at this altitude, $g_h$, is:

$$g_h = \frac{G M_E}{(R_E + h)^2}$$

We can rewrite this in terms of $g$ at the surface:

$$g_h = \frac{G M_E}{R_E^2 (1 + h/R_E)^2} = \frac{g}{(1 + h/R_E)^2}$$

For small heights ($h ll R_E$), we can use the binomial approximation $(1+x)^{-n} approx 1-nx$ for $x ll 1$.

Here, $x = h/R_E$ and $n=2$.

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

This shows that $g$ decreases with increasing altitude. The decrease is approximately linear for small heights.

**b) Variation with Depth (Depth $d$):**

When an object is taken to a depth $d$ below the Earth's surface (e.g., inside a mine), its distance from the center is $R_E - d$. However, for calculating gravity inside the Earth, we only consider the mass of the Earth contained within a sphere of radius $(R_E - d)$.

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

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

We know that $g = \frac{G M_E}{R_E^2} = \frac{G \frac{4}{3}pi R_E^3 \rho}{R_E^2} = G \frac{4}{3}pi R_E \rho$.

Dividing $g_d$ by $g$:

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

So,

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

This indicates that $g$ also decreases with increasing depth.

At the center of the Earth ($d = R_E$), $g_d = 0$. This is because at the center, the gravitational forces from all parts of the Earth cancel out.

**c) Variation with Shape of Earth (Latitude $lambda$):**

The Earth is not a perfect sphere; it's an oblate spheroid, meaning it's flattened at the poles and bulges at the equator. The equatorial radius is slightly larger than the polar radius ($R_E \text{ (equator)} > R_E \text{ (pole)}$). Since $g = GM/R^2$, a larger radius implies a smaller $g$. Therefore, $g$ is slightly less at the equator than at the poles due to the Earth's shape.

**d) Variation with Rotation of Earth (Latitude $lambda$):**

The Earth rotates about its axis. An object on the surface at latitude $lambda$ experiences a centrifugal force (or rather, the gravitational force provides the centripetal force required for circular motion).

This effectively reduces the apparent weight of the object and thus the effective acceleration due to gravity. The effective acceleration due to gravity $g'$ at latitude $lambda$ is given by:

$$g' = g - R_E omega^2 cos^2lambda$$

where $omega$ is the angular velocity of the Earth's rotation.

At the equator ($lambda = 0^circ$, $coslambda = 1$), $g' = g - R_E omega^2$. At the poles ($lambda = 90^circ$, $coslambda = 0$), $g' = g$. This shows that $g$ is minimum at the equator and maximum at the poles due to rotation.

The effect of rotation is more significant than the effect of shape, but both contribute to $g_{\text{equator}} < g_{\text{pole}}$.

4. Real-World Applications and Significance

Projectile Motion: — The constant (or near-constant for short ranges) acceleration due to gravity is fundamental to understanding the trajectory of projectiles, from a thrown ball to a ballistic missile.
Satellite Orbits: — While orbital motion involves centripetal force, the gravitational force providing this centripetal force is directly related to $g$ at that altitude.
Weight: — An object's weight is defined as $W = mg$. Since $g$ varies, an object's weight also varies slightly depending on its location on Earth.
Geodesy and Geophysics: — Precise measurements of $g$ are used to study the Earth's interior, detect mineral deposits, and understand tectonic plate movements.

5. Common Misconceptions

$g$ vs $G$: — Students often confuse the universal gravitational constant ($G$) with acceleration due to gravity ($g$). $G$ is a fundamental constant of nature, always the same everywhere. $g$ is a local acceleration, specific to a celestial body and varying with location.
$g$ is constant: — While $g approx 9.8,\text{m/s}^2$ is a good approximation for many problems, it's crucial to remember that $g$ is not truly constant and varies with altitude, depth, and latitude.
Mass vs. Weight: — Mass is an intrinsic property of an object and remains constant. Weight ($mg$) depends on $g$, and thus changes with location.

6. NEET-Specific Angle

For NEET, questions frequently test the understanding of:

The basic formula $g = GM/R^2$.
The approximate value of $g$ ($9.8,\text{m/s}^2$ or $10,\text{m/s}^2$).
The variations of $g$ with altitude and depth, especially the approximate formulas for small $h$ and $d$.
The combined effect of Earth's shape and rotation on $g$ at poles vs. equator.
Conceptual questions distinguishing $g$ from $G$, and mass from weight.
Problems involving percentage change in $g$ due to small changes in $h$ or $d$.
Graphical representation of $g$ variation (e.g., from center to outside the Earth). The value of $g$ increases linearly from the center ($g=0$) to the surface ($g=GM/R^2$) and then decreases as $1/r^2$ outside the surface. Mastery of these nuances is key to scoring well on gravitation questions.