Welcome, future physicists! Today, we're going to dive into one of the most fundamental concepts in gravitation: the acceleration due to gravity, often denoted by the symbol 'g'. This isn't just a number you memorise; it's a profound concept that explains why everything falls, why we stay on Earth, and how our planet behaves. Let's start right from the beginning!

### What Makes Things Fall? Introducing 'g'!

Have you ever wondered why, when you drop a pen, it always falls *down* towards the ground? Or why a ball thrown upwards eventually comes back down? The answer lies in a fundamental force of nature: gravity.

Sir Isaac Newton, by observing an apple fall from a tree, realised that there must be a force attracting objects towards the Earth. He formulated the famous Newton's Law of Universal Gravitation, which 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, this force (F) between two objects of masses M and m, separated by a distance 'r', is given by:
$$ F = frac{G M m}{r^2} $$
Here, G is the Universal Gravitational Constant, a fixed value.

Now, think about the pen falling. The Earth is pulling the pen, and the pen is pulling the Earth (though the Earth's huge mass means its acceleration is negligible!). This gravitational force causes the pen to accelerate downwards. This specific acceleration, caused solely by the Earth's gravitational pull, is what we call acceleration due to gravity (g).

So, in simple terms:
Acceleration due to gravity (g) is the acceleration experienced by an object due to the gravitational force exerted by a celestial body (like Earth), assuming only gravity is acting on it.

### Deriving the Value of 'g' at Earth's Surface

Let's find out what determines this magical 'g'. Imagine an object of mass 'm' resting on the Earth's surface.

1. Gravitational Force on the object: According to Newton's Law of Gravitation, the force exerted by the Earth (mass M_E, radius R_E) on the object (mass m) is:
$$ F_{grav} = frac{G M_E m}{R_E^2} $$
Here, R_E is the distance from the center of the Earth to the object on its surface.

2. Newton's Second Law: We also know from Newton's Second Law of Motion that force equals mass times acceleration (F = ma). In our case, the acceleration is 'g', so the force on the object is:
$$ F = m imes g $$

3. Equating the forces: Since both equations represent the same gravitational force acting on the object:
$$ m g = frac{G M_E m}{R_E^2} $$

4. Solving for 'g': Notice that the mass of the object 'm' appears on both sides, so it cancels out!
$$ oxed{g = frac{G M_E}{R_E^2}} $$

Key Insight: The acceleration due to gravity 'g' at a planet's surface is independent of the mass of the falling object! This is why a feather and a bowling ball, if dropped in a vacuum, fall at the same rate. (On Earth, air resistance makes the feather fall slower).

Let's plug in the approximate values for Earth:
* G ≈ 6.67 × 10⁻¹¹ N m²/kg²
* M_E (Mass of Earth) ≈ 5.97 × 10²⁴ kg
* R_E (Radius of Earth) ≈ 6.37 × 10⁶ m

If you calculate this, you'll get:
$$ g approx 9.8 ext{ m/s}^2 $$
This is the familiar value of 'g' we use in most calculations. It means that any object falling freely near the Earth's surface increases its speed by approximately 9.8 meters per second every second!

### Variations in 'g': It's Not Always 9.8 m/s²!

While 9.8 m/s² is a good average, the value of 'g' isn't constant everywhere on Earth. It varies due to several factors. Let's explore these fascinating variations.

#### 1. Variation with Height (Altitude)

Imagine you're climbing a very tall mountain, or even flying in an airplane. As you move away from the Earth's surface, you are increasing your distance from the Earth's center.

Recall our formula: $ g = frac{G M_E}{R^2} $.
Here, 'R' is the distance from the center of the Earth to the object.

* At the Earth's surface, R = R_E.
* At a height 'h' above the surface, R = R_E + h.

So, the acceleration due to gravity at height 'h' (let's call it g_h) will be:
$$ g_h = frac{G M_E}{(R_E + h)^2} $$

Comparing this with $ g = frac{G M_E}{R_E^2} $:
$$ g_h = g left( frac{R_E}{R_E + h}
ight)^2 $$
$$ g_h = g left( 1 + frac{h}{R_E}
ight)^{-2} $$

For small heights (h << R_E), we can use the binomial approximation (1+x)ⁿ ≈ 1 + nx.
Here, x = h/R_E and n = -2.
$$ g_h approx g left( 1 - frac{2h}{R_E}
ight) $$

What does this tell us?
* Since h is positive, the term (1 - 2h/R_E) is less than 1.
* Therefore, g_h < g.
* Conclusion: The acceleration due to gravity decreases as you go higher above the Earth's surface. This is intuitive – as you move away from the Earth, its gravitational pull weakens. This is why astronauts in orbit experience microgravity; they are far enough for 'g' to be significantly reduced, though not zero.

Example: If you are at the top of Mount Everest (approx. h = 8.8 km), what is the approximate 'g' value?
R_E ≈ 6400 km.
g_h = 9.8 (1 - 2 * 8.8 / 6400) = 9.8 (1 - 0.00275) = 9.8 * 0.99725 ≈ 9.773 m/s².
A slight but measurable decrease!

#### 2. Variation with Depth

What happens if you go *inside* the Earth, perhaps down a very deep mine shaft? This one is a bit more counter-intuitive.

Imagine a particle of mass 'm' at a depth 'd' below the Earth's surface. Its distance from the center of the Earth is now (R_E - d).
However, here's the trick: the gravitational force on this particle is *only* due to the mass of the Earth contained within a sphere of radius (R_E - d). The mass of the spherical shell *outside* this radius exerts no net gravitational force on the particle inside it (due to symmetry).

Let's assume for simplicity that the Earth has a uniform density, ρ (rho).
* Mass of Earth, M_E = Volume × Density = (4/3)πR_E³ρ
* Mass of Earth relevant at depth 'd', M'_E = (4/3)π(R_E - d)³ρ

Now, the acceleration due to gravity at depth 'd' (g_d) is:
$$ g_d = frac{G M'_E}{(R_E - d)^2} $$
Substitute M'_E:
$$ g_d = frac{G frac{4}{3}pi(R_E - d)^3
ho}{(R_E - d)^2} $$
$$ g_d = G frac{4}{3}pi (R_E - d)
ho $$

We know that $ g = frac{G M_E}{R_E^2} = frac{G frac{4}{3}pi R_E^3
ho}{R_E^2} = G frac{4}{3}pi R_E
ho $.
So, we can write:
$$ g_d = g frac{(R_E - d)}{R_E} $$
$$ oxed{g_d = g left( 1 - frac{d}{R_E}
ight)} $$

What does this tell us?
* Since 'd' is positive, the term (1 - d/R_E) is less than 1.
* Therefore, g_d < g.
* Conclusion: The acceleration due to gravity decreases as you go deeper inside the Earth.

Special Case: At the Center of the Earth (d = R_E)
If d = R_E, then g_d = g (1 - R_E/R_E) = g (1 - 1) = 0.
So, at the very center of the Earth, the acceleration due to gravity is zero! This makes sense, as you would be pulled equally in all directions, resulting in no net force.

#### 3. Variation Due to Earth's Shape (Oblateness)

Is the Earth a perfect sphere? Not quite! It's slightly flattened at the poles and bulges out at the equator, like a squashed ball. This shape is called an oblate spheroid.

* Equatorial Radius (R_e) > Polar Radius (R_p).
(Approx. R_e = 6378 km, R_p = 6357 km).

Since $ g = frac{G M_E}{R^2} $, and 'R' is in the denominator:
* At the poles, the radius (R_p) is smaller, so 'g' is slightly higher at the poles.
* At the equator, the radius (R_e) is larger, so 'g' is slightly lower at the equator.

This variation is independent of Earth's rotation, though rotation *causes* the oblateness. We're just looking at the geometric effect here.

#### 4. Variation Due to Earth's Rotation (Latitude)

This is perhaps the most complex variation at the fundamental level, involving concepts of frames of reference. Let's simplify it.

Imagine you're standing on the equator. As the Earth rotates, you are moving in a large circle. To stay in that circle, you need a centripetal force pulling you towards the center of the circle (which is also the center of the Earth). Where does this centripetal force come from? It comes from the gravitational pull of the Earth!

However, your *apparent* weight (and thus the apparent 'g' you feel) is the gravitational force *minus* the centripetal force required to keep you in circular motion.

* At the equator, your speed of rotation is maximum, so the centripetal force required is maximum. This effectively reduces the *apparent* 'g' you experience.
* As you move towards the poles, your circular path becomes smaller, and your speed of rotation decreases. At the poles themselves, you are essentially just rotating on the spot (the axis of rotation), so your speed is zero, and no centripetal force is "stolen" from gravity.
* Therefore, the apparent 'g' is highest at the poles and lowest at the equator due to Earth's rotation.

The exact formula for the effective acceleration due to gravity (g') at a latitude λ is:
$$ g' = g - omega^2 R_E cos^2lambda $$
Where:
* g is the acceleration due to gravity if the Earth wasn't rotating.
* ω (omega) is the angular speed of Earth's rotation.
* R_E is Earth's radius.
* λ (lambda) is the latitude (0° at equator, 90° at poles).

Key Takeaways:
* At the equator (λ = 0°, cos 0° = 1), g' = g - ω²R_E. This is the minimum value.
* At the poles (λ = 90°, cos 90° = 0), g' = g. This is the maximum value.

This effect, combined with the oblateness, makes 'g' noticeably different across various latitudes.

### Summary of Variations of 'g'

Let's quickly recap how 'g' changes:

Factor	Effect on 'g'	Reason
Height (Altitude)	Decreases	Increased distance from Earth's center, weaker gravitational pull.
Depth	Decreases	Only the mass of the inner sphere contributes to the pull; pull becomes zero at the center.
Earth's Shape	Higher at poles, Lower at equator	Earth is an oblate spheroid; R_polar < R_equatorial.
Earth's Rotation	Higher at poles, Lower at equator	Centripetal force requirement reduces effective 'g' at latitudes where rotation speed is significant.

### A Final Thought for JEE & CBSE Students

For both CBSE and JEE, a strong conceptual understanding of why 'g' varies is crucial. For JEE, you'll need to be comfortable with the derivations and applying the formulas to solve problems, especially those involving small heights or specific depths. Remember, these variations, though small in everyday life, are critical in fields like geodesy (study of Earth's shape) and satellite mechanics. Keep practicing, and you'll master this fundamental concept in no time!

Welcome, aspiring physicists, to a deep dive into one of the most fundamental concepts in gravitation: Acceleration due to gravity (g) and its variations. This topic is not just crucial for understanding how things fall around us, but it forms the bedrock for advanced concepts in orbital mechanics and astrophysics. For JEE Main and Advanced, a thorough understanding of the derivations and the nuances of each variation is absolutely essential. Let's begin our exploration!

1. Understanding Acceleration Due to Gravity (g) at the Earth's Surface

You've likely heard the term 'gravity' countless times, but let's precisely define what acceleration due to gravity (g) means. It is the acceleration experienced by an object due to the gravitational pull of a celestial body, like the Earth, when no other forces (like air resistance) are acting on it. It’s important to remember that 'g' is an acceleration, hence its unit is meters per second squared (m/s²).

Derivation of 'g' at the Earth's Surface

According to Newton's Universal Law of Gravitation, the force of attraction between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by:

$$ F = frac{G m_1 m_2}{r^2} $$

Where $G$ is the Universal Gravitational Constant ($6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$).

Now, consider an object of mass $m$ placed on the surface of the Earth. Let the Earth's mass be $M_E$ and its radius be $R_E$. The distance between the center of the Earth and the object (assuming it's a point mass at the surface) is $R_E$. The gravitational force acting on this object is:

$$ F_g = frac{G M_E m}{R_E^2} $$

We also know from Newton's Second Law of Motion that force is equal to mass times acceleration ($F = ma$). In this case, the acceleration is the acceleration due to gravity, 'g'. So, the gravitational force can also be written as:

$$ F_g = m g $$

Equating these two expressions for gravitational force:

$$ m g = frac{G M_E m}{R_E^2} $$

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, dropped in a vacuum, will fall at the same rate, a fact famously demonstrated by Galileo and later on the Moon by Apollo 15 astronauts.

So, the formula for acceleration due to gravity at the Earth's surface is:

$$ mathbf{g = frac{G M_E}{R_E^2}} $$

Using standard values ($G = 6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$, $M_E approx 5.97 imes 10^{24} ext{ kg}$, $R_E approx 6.37 imes 10^6 ext{ m}$), we get the familiar value:

$$ g approx 9.8 ext{ m/s}^2 $$

For many problems in JEE, you might be asked to use $g = 10 ext{ m/s}^2$ for simplification. Always check the problem statement!

2. Variation of 'g' with Altitude (Height above the Surface)

As we move away from the Earth's surface, the distance from the center of the Earth increases. How does this affect 'g'?

Derivation

Consider an object of mass $m$ at a height $h$ above the Earth's surface. The distance of this object from the center of the Earth is now $(R_E + h)$.

Using the general formula $g = GM/r^2$, where $r = (R_E + h)$, the acceleration due to gravity at height $h$, denoted as $g_h$, will be:

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

We can rewrite this expression by relating it to 'g' at the surface:

$$ g_h = frac{G M_E}{R_E^2 left(1 + frac{h}{R_E}
ight)^2} $$

Since $g = frac{G M_E}{R_E^2}$, we can substitute it into the equation:

$$ mathbf{g_h = g left(1 + frac{h}{R_E}
ight)^{-2}} $$

Approximation for Small Heights ($h ll R_E$)

When the height $h$ is very small compared to the Earth's radius ($h ll R_E$), we can use the binomial approximation: $(1+x)^n approx 1+nx$ for $|x| ll 1$.

Here, $x = h/R_E$ and $n = -2$. So, the approximation gives:

$$ left(1 + frac{h}{R_E}
ight)^{-2} approx 1 - 2frac{h}{R_E} $$

Substituting this back into the formula for $g_h$:

$$ mathbf{g_h approx g left(1 - frac{2h}{R_E}
ight)} $$

This approximate formula shows that 'g' decreases linearly with height for small altitudes. It's often used for problems involving heights like mountains or airplanes. For heights comparable to $R_E$ (e.g., satellites), the exact formula must be used.

Example 1: Calculate the percentage decrease in acceleration due to gravity when an object is raised to a height of 100 km above the Earth's surface. (Given $R_E = 6400 ext{ km}$).

Solution:

Given $h = 100 ext{ km}$ and $R_E = 6400 ext{ km}$. Since $h ll R_E$, we can use the approximate formula for the change in 'g'.

Percentage decrease = $frac{g - g_h}{g} imes 100\%$

Using $g_h approx g left(1 - frac{2h}{R_E}
ight)$:

$frac{g - g left(1 - frac{2h}{R_E}
ight)}{g} = frac{g left[1 - left(1 - frac{2h}{R_E}
ight)
ight]}{g} = frac{2h}{R_E}$

Percentage decrease = $frac{2h}{R_E} imes 100\%$

Substitute values: $frac{2 imes 100 ext{ km}}{6400 ext{ km}} imes 100\% = frac{200}{6400} imes 100\% = frac{1}{32} imes 100\% approx mathbf{3.125\%}$.

3. Variation of 'g' with Depth (Below the Surface)

What happens to 'g' as we go inside the Earth, like in a deep mine or tunnel?

Derivation

Let's consider an object of mass $m$ at a depth $d$ below the Earth's surface. Its distance from the center of the Earth will be $(R_E - d)$.

To calculate 'g' at this depth, we use a very important concept from gravitation: only the mass of the sphere *inside* the radius containing the object contributes to the gravitational force. The gravitational effect of the spherical shell outside this radius cancels out.

So, at depth $d$, only the mass of the sphere of radius $(R_E - d)$ contributes to the gravitational acceleration. Let this effective mass be $M'$.

Assume the Earth has a uniform density $
ho$. Then, the total mass of the Earth $M_E = frac{4}{3} pi R_E^3
ho$.

The mass of the inner sphere $M' = frac{4}{3} pi (R_E - d)^3
ho$.

We can express $M'$ in terms of $M_E$:

$$ frac{M'}{M_E} = frac{frac{4}{3} pi (R_E - d)^3
ho}{frac{4}{3} pi R_E^3
ho} = frac{(R_E - d)^3}{R_E^3} $$

So, $M' = M_E left(frac{R_E - d}{R_E}
ight)^3$.

Now, the acceleration due to gravity at depth $d$, denoted $g_d$, is:

$$ g_d = frac{G M'}{(R_E - d)^2} $$

Substitute the expression for $M'$:

$$ g_d = frac{G left[ M_E left(frac{R_E - d}{R_E}
ight)^3
ight]}{(R_E - d)^2} $$

$$ g_d = frac{G M_E}{R_E^3} (R_E - d) $$

We know that $g = frac{G M_E}{R_E^2}$. So, we can write $frac{G M_E}{R_E^3} = frac{g}{R_E}$.

Substituting this back, we get:

$$ mathbf{g_d = frac{g}{R_E} (R_E - d)} $$

Or, by factoring out $R_E$ from the parenthesis:

$$ mathbf{g_d = g left(1 - frac{d}{R_E}
ight)} $$

This formula shows that 'g' decreases linearly with depth. At the center of the Earth ($d = R_E$), $g_d = g(1 - R_E/R_E) = g(0) = 0$. This makes sense: at the center, the gravitational pull from all directions cancels out.

Comparison: Altitude vs. Depth

Notice that for small $h$ and $d$, the decrease in 'g' is $Delta g_{height} = g frac{2h}{R_E}$ and $Delta g_{depth} = g frac{d}{R_E}$. This implies that for the same distance from the surface, 'g' decreases twice as fast when moving upwards as it does when moving downwards (for small distances).

Example 2: At what depth below the Earth's surface does the acceleration due to gravity become 75% of its value at the surface? (Given $R_E = 6400 ext{ km}$).

Solution:

We are given $g_d = 0.75g$. We need to find $d$.

Using the formula for variation with depth: $g_d = g left(1 - frac{d}{R_E}
ight)$

Substitute $g_d = 0.75g$:

$0.75g = g left(1 - frac{d}{R_E}
ight)$

$0.75 = 1 - frac{d}{R_E}$

$frac{d}{R_E} = 1 - 0.75 = 0.25$

$d = 0.25 R_E$

$d = 0.25 imes 6400 ext{ km} = mathbf{1600 ext{ km}}$

4. Variation of 'g' due to the Shape of the Earth

The Earth is not a perfect sphere. It's an oblate spheroid – it bulges at the equator and is flattened at the poles. This non-spherical shape has a direct impact on the value of 'g'.

The equatorial radius ($R_E$) is greater than the polar radius ($R_P$).

($R_E approx 6378 ext{ km}$, $R_P approx 6357 ext{ km}$)

Since $g = frac{G M_E}{R^2}$, and $g$ is inversely proportional to the square of the radius ($g propto 1/R^2$), a larger radius means a smaller 'g'.

• At the poles, $R$ is minimum ($R_P$), so $g_{pole}$ is maximum.


• At the equator, $R$ is maximum ($R_E$), so $g_{equator}$ is minimum.

The difference is approximately $g_{pole} - g_{equator} approx 0.018 ext{ m/s}^2$. This is a small but measurable difference. For instance, an object would weigh slightly more at the poles than at the equator due to this variation.

5. Variation of 'g' due to the Rotation of the Earth

This is arguably the most complex and important variation for JEE Advanced. The Earth is constantly rotating about its own axis. This rotation introduces a centrifugal pseudo-force in the non-inertial frame of reference attached to the Earth, which affects the apparent weight of objects and thus the effective acceleration due to gravity.

Derivation

Consider an object of mass $m$ at a latitude $lambda$ on the Earth's surface. Latitude ($lambda$) is the angle made by the line joining the object to the center of the Earth with the equatorial plane.

The Earth rotates with an angular velocity $omega$. For an object at latitude $lambda$, its circular path around the Earth's axis has a radius $r = R_E cos lambda$.

As the Earth rotates, the object experiences a centrifugal force (an outward pseudo-force) given by $F_c = m omega^2 r = m omega^2 R_E cos lambda$. This force acts radially outwards from the axis of rotation.

The true gravitational force $F_g = mg$ acts towards the center of the Earth. The centrifugal force acts away from the axis of rotation. The resultant force (and hence the effective 'g', denoted $g'$) is the vector sum of $mg$ and the centrifugal force. However, it's simpler to think about the effective acceleration $g'$ as the vector sum of $g$ (towards the center) and the acceleration due to centrifugal force ($omega^2 R_E cos lambda$ outwards from the axis).

Let's consider the component of the centrifugal force along the line joining the object to the center of the Earth. The angle between the centrifugal force vector (perpendicular to the axis of rotation) and the line from the object to the Earth's center is $lambda$.

The effective gravitational force $F'_{g}$ acting on the mass $m$ is the resultant of the true gravitational force $F_g = mg$ (directed towards the center) and the centrifugal force $F_c = momega^2 R_E coslambda$ (directed outwards, perpendicular to the axis of rotation).
The component of $F_c$ directed radially outwards from the Earth's center is $F_c coslambda = momega^2 R_E cos^2lambda$.

Therefore, the effective acceleration due to gravity $g'$ at latitude $lambda$ is approximately given by:

$$ mathbf{g' = g - omega^2 R_E cos^2 lambda} $$

Here:

• $g$ is the acceleration due to gravity if the Earth were not rotating.


• $omega$ is the angular speed of Earth's rotation ($2pi$ radians in 24 hours).


• $R_E$ is the radius of the Earth.


• $lambda$ is the latitude of the location.

Special Cases:

1. At the Poles ($lambda = 90^circ$):
   
   $cos 90^circ = 0$. So, $g' = g - omega^2 R_E (0)^2 = g$.
   
   There is no effect of Earth's rotation on 'g' at the poles. This makes sense as objects at the poles are on the axis of rotation and have no circular motion.
   


2. At the Equator ($lambda = 0^circ$):
   
   $cos 0^circ = 1$. So, $g' = g - omega^2 R_E (1)^2 = g - omega^2 R_E$.
   
   The effect of rotation is maximum at the equator, leading to the greatest reduction in 'g'.

The angular velocity of Earth is $omega = frac{2pi}{T} = frac{2pi}{24 imes 3600 ext{ s}} approx 7.29 imes 10^{-5} ext{ rad/s}$.

Calculating $omega^2 R_E$ at the equator: $(7.29 imes 10^{-5})^2 imes 6.4 imes 10^6 approx 0.0338 ext{ m/s}^2$.

This means $g_{equator}$ is about $9.8 - 0.0338 = 9.7662 ext{ m/s}^2$. This value is close to the observed value of 'g' at the equator.

Condition for a body to fly off at the Equator:

If the Earth rotates fast enough, the centrifugal force at the equator could balance or exceed the gravitational force. In such a scenario, objects at the equator would become weightless or even fly off the surface. This happens if $g' = 0$, i.e., $g - omega^2 R_E = 0$.

So, $omega^2 R_E = g implies omega = sqrt{frac{g}{R_E}}$.

This 'critical' angular velocity would be much higher than Earth's current rotational speed.

Example 3: If the Earth suddenly stops rotating, how would the acceleration due to gravity change at a place with latitude $30^circ$?

Solution:

Initial acceleration due to gravity at latitude $lambda$ is $g' = g - omega^2 R_E cos^2 lambda$.

If the Earth stops rotating, $omega = 0$. The new acceleration due to gravity, let's call it $g''$, would be $g'' = g$.

The change in acceleration due to gravity is $Delta g = g - g' = g - (g - omega^2 R_E cos^2 lambda) = omega^2 R_E cos^2 lambda$.

Given $lambda = 30^circ$. So, $cos 30^circ = sqrt{3}/2$, and $cos^2 30^circ = 3/4$.

The change would be $Delta g = omega^2 R_E left(frac{3}{4}
ight)$.

We know $omega^2 R_E approx 0.0338 ext{ m/s}^2$ (calculated previously for the equator).

So, $Delta g = 0.0338 imes frac{3}{4} approx mathbf{0.02535 ext{ m/s}^2}$.

The acceleration due to gravity would increase by approximately $0.02535 ext{ m/s}^2$ at latitude $30^circ$ if the Earth stopped rotating.

6. Combined Effect and Conclusion

In reality, all these factors (altitude, depth, shape, and rotation) combine to give the actual value of 'g' at any specific location. For most practical purposes and JEE problems, these effects are considered independently or in specific combinations. The value of 'g' that we often use, $9.8 ext{ m/s}^2$, is an average value on the surface, already accounting for the Earth's shape and rotation effects to some extent.

Understanding these variations is not just theoretical; it has practical implications in areas like precise weighing, missile trajectories, satellite launches, and even pendulum clock mechanisms. This deep dive should equip you with the fundamental derivations and conceptual clarity needed to tackle a wide range of problems in Newtonian gravitation for your JEE preparations.

The topic "Acceleration due to gravity and its variation" is fundamental for both board exams and JEE. Mastering its formulas and conditions is crucial. Here are some mnemonics and short-cuts to help you remember the key concepts and avoid common pitfalls.

Mnemonics & Short-Cuts for Acceleration Due to Gravity (g)

Quick recall of formulas and their specific conditions can save significant time in exams. Pay close attention to the small details, like the factor of '2' in certain formulas.

1. 
   Base Value of 'g' on Earth's Surface:
   
   
   
   • Formula: $g = frac{GM}{R^2}$
   
   
   • Mnemonic: "Gravity is Generally Measured on Radius Squared."
     
     
     
     • (G: Universal Gravitational Constant, M: Mass of Earth, R: Radius of Earth)
     
     
     
     
   
   
   
   


2. 
   Variation with Altitude (Height 'h'):
   
   
   
   • Case 1: For small heights ($h ll R$)
     
     
     
     • Formula: $g_h = g(1 - frac{2h}{R})$
     
     
     • Mnemonic: "Going High (small h), 'g' is reduced by '2H' over R."
       
       
       
       • The '2' factor is critical here.
       
       
       
       
     
     
     
     
   
   
   • Case 2: For large heights (general formula)
     
     
     
     • Formula: $g_h = gleft(frac{R}{R+h}
       ight)^2 = frac{GM}{(R+h)^2}$
     
     
     • Mnemonic: "For Really high altitudes, remember the Ratio Squared: (R/(R+h))^2."
     
     
     
     
   
   
   
   


3. 
   Variation with Depth ('d'):
   
   
   
   • Formula: $g_d = g(1 - frac{d}{R})$
   
   
   • Mnemonic: "Going Deep, 'g' is reduced by 'D' over R (NO '2')."
     
     
     
     • This highlights the key difference from the small height formula.
     
     
     
     
   
   
   
   


4. 
   JEE Specific: Height vs. Depth Distinction
   
   
   
   • This is a common point of confusion and frequently tested.
   
   
   • Short-cut/Mnemonic: "For small changes, Height Has '2', Depth Doesn't."
     
     
     
     • If $g_h = g_d$ for small h and d, then $1 - frac{2h}{R} = 1 - frac{d}{R} implies 2h = d$. So, 'g' decreases by the same amount at height 'h' as it does at depth '2h'.
     
     
     
     
   
   
   
   


5. 
   Variation with Shape (Non-sphericity of Earth):
   
   
   
   • Earth is an oblate spheroid (flattened at poles, bulging at equator). Thus $R_{equator} > R_{pole}$.
   
   
   • Since $g propto frac{1}{R^2}$, 'g' is maximum at poles and minimum at the equator due to shape.
   
   
   • Mnemonic: "Poles are Pointed (smaller R), so 'g' is Peak at Poles."
     
     
     
     • Conversely, the Equator is Extended (larger R), so 'g' is Empty (lesser) there.
     
     
     
     
   
   
   
   


6. 
   Variation with Rotation (Latitude $lambda$):
   
   
   
   • Formula: $g' = g - Romega^2 cos^2lambda$
   
   
   • Mnemonic: "Rotation Reduces 'g' by R-omega-squared-Cosine-squared-Lambda." (R: Earth's radius, $omega$: Angular speed of Earth, $lambda$: Latitude)
   
   
   • Extremes:
     
     
     
     • At Equator ($lambda=0^circ$, $cos^2 0^circ = 1$): Maximum reduction ($g' = g - Romega^2$). "Equator's Effect is Enormous."
     
     
     • At Poles ($lambda=90^circ$, $cos^2 90^circ = 0$): No reduction ($g' = g$). "Poles are Paralyzed (unaffected) by rotation."
     
     
     
     
   
   
   
   

Intuitive Understanding: Acceleration due to Gravity and its Variation

Understanding the concept of acceleration due to gravity (g) goes beyond memorizing formulas; it involves grasping the fundamental reasons why objects fall and why this acceleration isn't constant everywhere.

What is 'g'? The Earth's Pull

Imagine holding a ball and letting it go. It falls. Why? Because the Earth pulls it downwards. This pull is what we call gravity. As the ball falls, it speeds up, meaning its velocity increases. This increase in velocity over time is called acceleration. The specific acceleration an object experiences solely due to Earth's gravitational pull is termed acceleration due to gravity (g).

* At the Earth's surface, on average, 'g' is approximately 9.8 m/s². This means for every second an object falls, its downward speed increases by 9.8 meters per second (ignoring air resistance).
* JEE & CBSE Note: While the value 9.8 m/s² is common, for certain problems, 10 m/s² is often used for simplification unless specified otherwise.

Intuition Behind Variations in 'g'

The acceleration due to gravity isn't constant across the Earth or even as you move away from its surface. Think of Earth's gravity like a giant magnet's pull: its strength depends on distance and the distribution of its "magnetic" material (mass).

1. Variation with Height (Altitude)

* Analogy: Imagine a strong magnet. The closer you are to it, the stronger the pull. As you move away, the pull weakens.
* Intuition: When you climb a mountain, fly in a plane, or go into space, you are moving further away from the Earth's center. Since gravity's strength decreases with the square of the distance from the center of the mass, the gravitational pull, and thus 'g', becomes weaker.
* Practical Implication: Astronauts on the International Space Station (ISS) experience 'microgravity' not because there's no gravity, but because they are constantly falling around the Earth, and 'g' is significantly reduced compared to the surface.

2. Variation with Depth

* Analogy: This is less intuitive. Imagine the Earth as a giant onion with layers. When you are on the surface, all the onion layers below you pull you down. Now, if you dig a tunnel and go inside, some of the onion layers are now *above* you.
* Intuition: As you go deeper into the Earth, the mass *above* you starts pulling you in the opposite direction (upwards or sideways), partially cancelling out the downward pull from the remaining mass *below* you. Effectively, only the mass in the sphere *below* your current position contributes to the net gravitational force. As you go deeper, the amount of this contributing mass decreases, leading to a reduction in 'g'. At the very center of the Earth, the net gravitational pull is zero from all directions, so 'g' becomes zero.

3. Variation due to Earth's Shape (Equatorial Bulge)

* Intuition: Earth isn't a perfect sphere; it's an oblate spheroid, meaning it bulges at the equator and is flattened at the poles. This bulge is due to its rotation.
* At the equator, you are further away from the Earth's center compared to the poles.
* Since gravity weakens with distance, 'g' is slightly less at the equator simply because of this greater distance.

4. Variation due to Earth's Rotation (Centrifugal Effect)

* Analogy: Imagine spinning on a merry-go-round. You feel an outward push. This is a "fictitious" centrifugal force.
* Intuition: The Earth rotates around its axis. When you stand on the Earth, you are also rotating with it. This rotation causes a small outward "centrifugal" effect, especially noticeable at the equator where the rotational speed is highest.
* This outward effect slightly counteracts the inward gravitational pull, effectively reducing the *apparent* acceleration due to gravity. At the poles, there's no circular motion, so this effect is zero.
* JEE & CBSE Note: Both the greater distance from the center and the centrifugal effect contribute to 'g' being minimum at the equator and maximum at the poles.

By visualizing these effects, you can build a strong intuitive foundation for understanding the complex variations of 'g', which is crucial for tackling related problems in competitive exams.

Real-World Applications of Acceleration Due to Gravity and its Variation

Understanding the acceleration due to gravity (g) and how it varies with altitude, depth, and latitude is not just theoretical knowledge but forms the backbone of numerous real-world applications across science, engineering, and technology.

1. 
   Space Exploration and Satellite Technology:
   
   
   
   • Rocket Launches and Trajectory: The launch of rockets and positioning of satellites heavily rely on precise calculations involving the variation of 'g' with altitude. As a rocket ascends, 'g' decreases, impacting the required thrust and fuel consumption.
   
   
   • Orbital Mechanics: The stability and period of satellite orbits (e.g., geostationary, LEO - Low Earth Orbit) are directly determined by the effective gravitational force at their respective altitudes. Engineers must account for the diminishing 'g' to ensure satellites maintain their intended paths.
   
   
   • Escape Velocity: The concept of escape velocity, crucial for sending probes out of Earth's gravitational influence, is a direct application of Earth's gravitational potential, which is derived from 'g'.
   
   
   
   


2. 
   Geophysical Surveys and Resource Exploration:
   
   
   
   • Oil, Gas, and Mineral Prospecting: Gravimeters are highly sensitive devices used to detect minute variations in 'g' on the Earth's surface. Denser rock formations (often associated with mineral deposits) cause a slightly higher 'g', while less dense formations (like sedimentary basins containing oil and gas) result in a slightly lower 'g'. This helps geologists locate potential reserves.
   
   
   • Mapping Subsurface Structures: Variations in 'g' can reveal hidden geological features such as fault lines, salt domes, and the boundaries of different rock types, which is vital for understanding Earth's crust.
   
   
   
   


3. 
   Geodesy and Global Positioning Systems (GPS):
   
   
   
   • Earth's Shape and Gravitational Field: Geodesists study the Earth's precise shape (geoid) by measuring 'g' variations across its surface. The Earth is an oblate spheroid, causing 'g' to be slightly greater at the poles than at the equator due to centrifugal force and differences in radius.
   
   
   • Accurate Mapping and Navigation: High-precision GPS and navigation systems require an extremely accurate model of Earth's gravitational field. Variations in 'g' affect the relativistic corrections needed for precise timing signals from satellites, impacting the accuracy of your smartphone's location services.
   
   
   
   


4. 
   Civil Engineering and Structural Design:
   
   
   
   • Load Calculations: For large structures like bridges, skyscrapers, and dams, the force of gravity is a primary factor in design. While not necessarily focused on *variation* for a single structure, understanding the fundamental value of 'g' is critical for calculating loads and ensuring structural integrity.
   
   
   • Seismic Monitoring: Changes in local gravity fields can sometimes precede or accompany seismic activity, providing data for earthquake prediction research.
   
   
   
   

JEE Tip: Questions often link these applications to theoretical concepts, such as how 'g' varies with depth (e.g., why 'g' is maximum at the surface) or height (e.g., calculating satellite orbital periods). Understand the underlying physics thoroughly.

Understanding abstract physics concepts often becomes easier with relatable analogies. Here, we use simple everyday scenarios to grasp the nuances of acceleration due to gravity (g) and its variations.

1. Acceleration due to Gravity (g) at Earth's Surface

• 
  Analogy: Imagine a giant, incredibly powerful magnet hidden deep inside the Earth. Everything near its surface (like us) feels a constant "pull" towards its center. This pull is analogous to 'g', the acceleration due to gravity, which causes objects to fall towards the Earth.
  


• 
  Concept: Near the Earth's surface, 'g' is approximately constant (9.8 m/s²) and acts vertically downwards, pulling objects towards the Earth's center.
  

2. Variation of 'g' with Altitude (Height)

• 
  Analogy: Consider the strength of a radio signal or Wi-Fi. The closer you are to the transmitter/router, the stronger the signal. As you move further away, the signal weakens.
  


• 
  Concept: Similarly, as you move higher above the Earth's surface (increase altitude), your distance from the Earth's center increases. Since gravitational force (and thus 'g') depends inversely on the square of the distance from the center (g ∝ 1/r²), 'g' decreases with increasing altitude.
  


• 
  JEE/CBSE Tip: For small heights (h << R), the approximation g' = g(1 - 2h/R) is often used in problems. For larger heights, the exact formula g' = g(R/(R+h))² is required.
  

3. Variation of 'g' with Depth

• 
  Analogy: Imagine you're inside a large, hollow sphere made of a dense material (like a lead ball). If you are at the surface, you feel the gravitational pull of the entire sphere towards its center. Now, imagine tunneling inwards. According to the Shell Theorem (a key concept in gravitation), the part of the sphere *above* you (forming a spherical shell) exerts *no net gravitational force* on you. Only the mass of the sphere *below* you (closer to the center than your current position) effectively pulls you.
  


• 
  Concept: As you go deeper into the Earth, the effective mass pulling you towards the center decreases. Consequently, 'g' decreases linearly with depth, becoming zero at the Earth's center (where the net gravitational pull from all directions cancels out).
  


• 
  Insight: This is a common point of confusion. While moving away from the center (altitude) also decreases 'g', the *reason* for decrease with depth is different – it's due to reduction in effective mass pulling you.
  

4. Variation of 'g' with Earth's Rotation and Latitude

• 
  Analogy: Think about riding a fast-spinning merry-go-round. When you're near the center (like the Earth's poles), you feel less outward push. But as you move towards the edge (like the Earth's equator), you feel a stronger outward push (centrifugal force) trying to throw you off. This outward push effectively makes you feel lighter, reducing your apparent weight.
  


• 
  Concept: The Earth's rotation creates a centrifugal force that acts outwards, opposing gravity. This effect is maximum at the equator (largest radius of rotation) and zero at the poles. Therefore, the effective acceleration due to gravity is minimum at the equator and maximum at the poles.
  


• 
  JEE/CBSE Tip: The Earth's rotation also causes it to bulge at the equator and flatten at the poles. This means the surface points at the equator are further from the center of mass (larger R) than at the poles (smaller R), further contributing to a lower 'g' at the equator. The formula is g' = g - Rω²cos²λ, where λ is the latitude.
  

These analogies help build an intuitive understanding of how and why 'g' varies, which is crucial for solving problems in gravitation.

Prerequisites for Acceleration Due to Gravity and its Variation

To effectively grasp the concepts of acceleration due to gravity (g) and its variations, a solid understanding of certain fundamental principles from earlier chapters and basic mathematics is essential. Revisiting these concepts will build a strong foundation, making the current topic much clearer and easier to master for both board exams and competitive tests like JEE Main.

Here are the key prerequisites:

• 
  Newton's Law of Universal Gravitation:
  
  
  
  • You must be thoroughly familiar with the statement and mathematical form of Newton's Law of Universal Gravitation: F = G(m₁m₂)/r².
  
  
  • Understand the meaning of each term: G (universal gravitational constant), m₁ and m₂ (masses of interacting bodies), and r (distance between their centers).
  
  
  • This law is the fundamental basis for defining gravitational force and subsequently, acceleration due to gravity.
  
  
  
  


• 
  Newton's Second Law of Motion:
  
  
  
  • Recall the fundamental relationship between force, mass, and acceleration: F = ma.
  
  
  • This law is crucial for understanding how gravitational force acting on an object leads to its acceleration. By equating gravitational force (from Newton's Law of Gravitation) to ma, we derive the expression for 'g'.
  
  
  
  


• 
  Concepts of Mass and Weight:
  
  
  
  • Understand the distinction between mass (an intrinsic property of matter, constant everywhere) and weight (the force of gravity acting on an object, which can vary).
  
  
  • Recall that weight is essentially the gravitational force an object experiences: W = mg. This equation directly connects 'g' to the concept of weight.
  
  
  
  


• 
  Basic Algebra and Equation Manipulation:
  
  
  
  • Proficiency in rearranging equations, solving for unknowns, and substituting values is critical for deriving formulas for 'g' and solving related problems.
  
  
  • This includes handling powers, square roots, and basic proportionality.
  
  
  
  


• 
  Basic Vector Concepts:
  
  
  
  • Understand that force and acceleration are vector quantities, possessing both magnitude and direction.
  
  
  • Gravitational force and acceleration always act towards the center of the gravitating body (e.g., Earth).
  
  
  
  


• 
  Density and Volume of a Sphere:
  
  
  
  • For understanding variations of 'g' inside Earth or other celestial bodies, you should know the formula for the volume of a sphere (V = 4/3 πr³) and the concept of density (ρ = Mass/Volume).
  
  
  • This knowledge allows you to calculate the mass of a portion of a spherical body, which is essential for deriving 'g' at a certain depth.
  
  
  
  


• 
  Basic Calculus (JEE Specific - Intuitive Understanding):
  
  
  
  • While direct calculus might not be extensively used for basic derivations of 'g', an intuitive understanding of how quantities vary with position (e.g., rate of change) can be beneficial for understanding the more complex variations of 'g' (e.g., for non-uniform density distributions or advanced problems involving integration).
  
  
  • JEE Tip: While not strictly needed for foundational concepts, familiarity with basic differentiation and integration can aid in solving more complex JEE problems related to gravitational potential and fields, which are closely linked to 'g'.
  
  
  
  

By ensuring you have a firm grip on these prerequisite topics, you will find "Acceleration due to gravity and its variation" to be a logical and straightforward extension of your existing knowledge, rather than a completely new and challenging concept.

Understanding the acceleration due to gravity (g) and its variations is fundamental to the study of gravitation. Several other topics in physics are either prerequisites, direct consequences, or applications of this concept. Mastering these related areas will provide a holistic understanding and improve your problem-solving skills for both board exams and competitive tests like JEE.

Here are the key related topics:

• 
  Newton's Law of Universal Gravitation:
  
  
  
  • This is the bedrock upon which the concept of 'g' is built. The acceleration due to gravity (g = GM/R²) is directly derived from Newton's law of universal gravitation (F = GMm/R²) by considering the force per unit mass.
  
  
  • A strong grasp of the gravitational constant (G), masses of celestial bodies, and the distance between them is crucial for understanding why 'g' exists and how it varies.
  
  
  
  


• 
  Gravitational Field and Potential:
  
  
  
  • The acceleration due to gravity 'g' is essentially the gravitational field strength at a point. It represents the gravitational force experienced by a unit mass placed at that point.
  
  
  • Variations in 'g' directly lead to variations in gravitational potential energy (U = -GMm/R) and gravitational potential (V = -GM/R). Problems often involve finding potential given 'g' or vice-versa, especially for extended bodies.
  
  
  • JEE Focus: Questions frequently link 'g' with the gradient of gravitational potential (g = -dV/dR).
  
  
  
  


• 
  Escape Velocity:
  
  
  
  • Escape velocity is the minimum speed required for an object to escape the gravitational field of a celestial body and never return.
  
  
  • Its derivation involves equating the kinetic energy of the object with the negative of its gravitational potential energy. Since gravitational potential is dependent on 'g' and the radius, understanding 'g' is crucial for calculating escape velocity (v_e = √(2GM/R) = √(2gR)).
  
  
  
  


• 
  Orbital Velocity and Satellite Motion:
  
  
  
  • For a satellite in orbit, the gravitational force (F = mg_orbit) provides the necessary centripetal force (F = mv²/r). Therefore, the orbital velocity (v_o = √(GM/r)) is directly derived from the balance involving the acceleration due to gravity at the orbital radius.
  
  
  • Concepts like geostationary satellites and polar satellites rely heavily on a precise understanding of 'g' at specific altitudes.
  
  
  • JEE Focus: Problems often combine variations of 'g' with orbital mechanics to determine orbital parameters or energy changes.
  
  
  
  


• 
  Weight and Weightlessness:
  
  
  
  • Weight is defined as the force exerted by gravity on an object (W = mg). Variations in 'g' (e.g., with altitude, depth, or due to Earth's rotation) directly explain why an object's weight can change.
  
  
  • The phenomenon of weightlessness experienced in orbiting spacecraft or during free fall is a direct consequence of the apparent absence of the normal force, even though 'g' might still be present.
  
  
  
  


• 
  Simple Harmonic Motion (Pendulum):
  
  
  
  • While a separate chapter, the period of a simple pendulum (T = 2π√(L/g)) is inversely proportional to the square root of 'g'.
  
  
  • Experiments to measure 'g' often utilize pendulums, and understanding how 'g' varies is essential for predicting changes in a pendulum's period at different locations (e.g., at the equator vs. poles).
  
  
  • CBSE Relevance: This is a common practical application of 'g' often discussed in lab experiments.
  
  
  
  

Common Exam Traps in Acceleration due to Gravity and its Variation

Understanding the variation of acceleration due to gravity (g) is crucial, but exams often set traps to test your conceptual clarity and attention to detail. Be vigilant for the following common pitfalls:

• 
  Trap 1: Ignoring Earth's Rotation for Apparent 'g'
  
  Many students forget that the measured or 'apparent' acceleration due to gravity (g') at a latitude λ on Earth is affected by the Earth's rotation.
  
  Mistake: Using only g (at poles) for all locations without considering the centrifugal effect.
  
  Correction: The effective 'g' at latitude λ is given by g' = g - ω²R cos²λ, where ω is the angular speed of Earth and R is its radius. This means 'g' is maximum at poles (λ=90°) and minimum at the equator (λ=0°). This is a frequent JEE Main trap.
  


• 
  Trap 2: Incorrectly Applying Height vs. Depth Formulas
  
  Students often mix up the formulas for variation with height and depth, or use approximations when exact formulas are needed.
  
  Mistake (Height): Using the approximation g_h ≈ g(1 - 2h/R) when 'h' is comparable to 'R' (e.g., h = R/2). This approximation is valid only for h << R.
  
  Correction (Height): For any height 'h', the exact formula is g_h = g (R / (R+h))².
  
  Mistake (Depth): Using the height formula for depth or vice versa.
  
  Correction (Depth): For depth 'd', the formula is g_d = g (1 - d/R). This formula is exact assuming uniform density or a spherically symmetric mass distribution, and is valid right up to the center of the Earth.
  


• 
  Trap 3: Confusing 'g' and 'G'
  
  A basic but persistent error is mixing up the gravitational constant (G) with the acceleration due to gravity (g).
  
  Mistake: Using G where g is required, or vice versa, especially in problems involving units or calculation of forces.
  
  Correction: G is a universal constant (6.67 × 10^-11 Nm²/kg²), while g is the acceleration experienced by a body due to gravity, which varies with location and mass of the celestial body (approx 9.8 m/s² on Earth's surface). Remember g = GM/R².
  


• 
  Trap 4: Assuming 'g' is always zero at the center of Earth
  
  While 'g' is indeed zero at the center of a uniformly dense sphere, questions involving tunnels or non-uniform density can be tricky.
  
  Mistake: Directly assuming g=0 at the center without considering the path or density variation.
  
  Correction: For a uniform spherical body, 'g' varies linearly with distance 'r' from the center for points inside, i.e., g_r = g (r/R). So, 'g' is zero only at r=0.
  


• 
  Trap 5: Ignoring the Mass and Radius of Other Celestial Bodies
  
  When a problem involves calculating 'g' on planets other than Earth, students sometimes inadvertently use Earth's parameters.
  
  Mistake: Using Earth's Mass (M_E) or Radius (R_E) instead of the specific planet's M_P and R_P.
  
  Correction: Always use the parameters relevant to the celestial body in question: g_planet = GM_P/R_P².
  

JEE Main Tip: Be very careful with the wording of problems related to height (above the surface), depth (below the surface), and effective 'g' due to rotation. Exact formulas are preferred for JEE unless 'h << R' is explicitly stated or clearly implied by the numerical values.

Understanding the acceleration due to gravity (g) and its variations is fundamental for both JEE Main and board exams. This section encapsulates the essential points you must remember.

Key Takeaways: Acceleration due to Gravity and its Variation

• Definition of 'g':
  
  
  
  • Acceleration experienced by a freely falling body near the Earth's surface due to the Earth's gravitational pull.
  
  
  • On Earth's surface, its value is approximately $9.8 , ext{m/s}^2$ (or $980 , ext{cm/s}^2$ or $32 , ext{ft/s}^2$). For calculations, $10 , ext{m/s}^2$ is often used unless specified otherwise.
  
  
  • Formula on Earth's surface: $mathbf{g = frac{GM}{R^2}}$, where G is the universal gravitational constant, M is the mass of Earth, and R is the radius of Earth.
  
  
  
  



• Variation with Altitude (Height 'h'):
  
  
  
  • As you move above the Earth's surface, the acceleration due to gravity decreases.
  
  
  • Exact Formula: $mathbf{g_h = frac{GM}{(R+h)^2} = g left( frac{R}{R+h}
    ight)^2 = g left(1 + frac{h}{R}
    ight)^{-2}}$.
  
  
  • Approximation (for h << R): $mathbf{g_h approx g left(1 - frac{2h}{R}
    ight)}$. This approximation is valid only when 'h' is very small compared to 'R'.
    
    JEE Tip: For numerical problems, always use the exact formula unless the options clearly suggest an approximation, or the question specifies 'h << R'.
  
  
  
  



• Variation with Depth ('d'):
  
  
  
  • As you move below the Earth's surface towards the center, the acceleration due to gravity decreases linearly (assuming uniform density).
  
  
  • Formula: $mathbf{g_d = g left(1 - frac{d}{R}
    ight)}$.
  
  
  • At the center of the Earth (d=R), $mathbf{g_d = 0}$.
  
  
  • At the surface (d=0), $mathbf{g_d = g}$.
  
  
  • Important Note: The maximum value of 'g' is at the Earth's surface.
  
  
  
  



• Variation due to Earth's Rotation (Latitude '$lambda$'):
  
  
  
  • The Earth's rotation causes a centrifugal force, reducing the effective 'g'. This effect depends on latitude.
  
  
  • Formula for effective 'g': $mathbf{g' = g - Romega^2 cos^2lambda}$, where $omega$ is the angular velocity of Earth, and $lambda$ is the latitude.
  
  
  • At the poles ($lambda = 90^circ$): $mathbf{g' = g}$. Effect of rotation is zero.
  
  
  • At the equator ($lambda = 0^circ$): $mathbf{g' = g - Romega^2}$. This is the minimum value due to rotation.
  
  
  • Consequently, 'g' is maximum at the poles and minimum at the equator due to rotation.
  
  
  
  



• Variation due to Shape of Earth:
  
  
  
  • The Earth is not a perfect sphere; it's an oblate spheroid, bulging at the equator and flattened at the poles.
  
  
  • The equatorial radius (R$_e$) is greater than the polar radius (R$_p$).
  
  
  • Since $g propto 1/R^2$, a smaller radius implies a larger 'g'. Therefore, $g_{poles} > g_{equator}$ due to shape alone.
  
  
  • This effect, combined with the rotation, makes the acceleration due to gravity measurably higher at the poles than at the equator.
  
  
  
  

Mastering these variations and their associated formulas is crucial for solving problems in Gravitation. Always pay attention to the specific conditions given in the problem (e.g., small height vs. large height) to choose the correct formula.

This section provides a structured approach to solving problems involving the acceleration due to gravity (g) and its variations. Mastering these steps is crucial for both board exams and JEE, as it covers fundamental concepts.

1. 
   

   Identify the Base Value of 'g':

   
   
   
   
   • Recall the fundamental formula for acceleration due to gravity on the surface of a celestial body (like Earth): g = GM/R², where G is the universal gravitational constant, M is the mass of the body, and R is its radius.
   
   
   • For Earth, a standard value of g ≈ 9.8 m/s² (or 10 m/s² if specified) is used.
   
   
   
   


2. 
   

   Determine the Type of Variation:

   
   

   Problems typically fall into one of three categories:

   
   
   
   
   • Variation with Altitude (Height 'h'): Object is above the surface.
   
   
   • Variation with Depth (Depth 'd'): Object is below the surface.
   
   
   • Variation with Latitude (Due to Earth's Rotation): Object is on the surface, but the effect of Earth's spin is considered.
   
   
   
   


3. 
   

   Apply the Correct Formula based on Variation Type:

   
   
   

   a) Variation with Altitude 'h' (above surface):

   
   
   
   
   • Exact Formula (JEE & CBSE):
     

     g_h = GM/(R+h)² = g * (R/(R+h))²

     
     

     This formula is universally applicable, regardless of how large 'h' is compared to 'R'.

     
     
   
   
   • Approximation (JEE & sometimes CBSE for small 'h'):
     

     g_h ≈ g * (1 - 2h/R)

     
     

     This approximation is valid only when h << R (typically h < 5% of R). Caution: Using this for large 'h' will lead to significant errors. Always check the magnitude of 'h' relative to 'R'.

     
     
   
   
   
   
   

   b) Variation with Depth 'd' (below surface):

   
   
   
   
   • Formula (JEE & CBSE):
     

     g_d = g * (1 - d/R)

     
     

     This formula assumes a uniform density Earth for simplicity. It shows that 'g' decreases linearly with depth and becomes zero at the Earth's center (d=R).

     
     
   
   
   
   
   

   c) Variation with Latitude 'λ' (due to Earth's rotation - Primarily JEE):

   
   
   
   
   • Formula:
     

     g' = g - Rω²cos²λ

     
     

     Where 'ω' is the angular velocity of Earth, 'R' is Earth's radius, and 'λ' is the latitude of the location.

     
     
     
     
     • At Poles (λ=90°), cosλ=0, so g' = g (no effect of rotation).
     
     
     • At Equator (λ=0°), cosλ=1, so g' = g - Rω² (minimum 'g' due to rotation).
     
     
     
     
   
   
   
   


4. 
   

   Handle Percentage Change Questions:

   
   
   
   
   • If a problem asks for the percentage change (increase or decrease) in 'g', use the general formula:
     

     % Change = [(g_final - g_initial) / g_initial] * 100%

     
     
   
   
   • For small height 'h', the percentage decrease in 'g' is approximately (2h/R) * 100%.
   
   
   • For depth 'd', the percentage decrease in 'g' is (d/R) * 100%.
   
   
   
   


5. 
   

   Pay Attention to Units and Standard Values:

   
   
   
   
   • Always ensure all quantities are in consistent units (e.g., SI units).
   
   
   • Recall standard values:
     
     
     
     • Earth's Radius (R) ≈ 6.4 x 10⁶ m (or 6400 km)
     
     
     • Universal Gravitational Constant (G) ≈ 6.67 x 10⁻¹¹ Nm²/kg²
     
     
     • Angular velocity of Earth (ω) ≈ 7.27 x 10⁻⁵ rad/s (for rotation problems)
     
     
     
     
   
   
   
   


6. 
   

   JEE Specific Considerations:

   
   
   
   
   • Problems might involve comparing 'g' values on different planets or celestial bodies, requiring the use of g = GM/R² for each.
   
   
   • Occasionally, problems might involve non-uniform density distributions for Earth, requiring integration or a more advanced approach (less common for basic problems).
   
   
   
   

By systematically applying these steps and understanding the underlying physics, you can confidently approach a wide range of problems concerning acceleration due to gravity and its variations.

Welcome to the CBSE Focus Areas for 'Acceleration due to gravity and its variation'. This section highlights the key concepts and derivations frequently tested in the CBSE Board Examinations.

For CBSE, a strong emphasis is placed on understanding the fundamental definition of acceleration due to gravity and its variations with altitude (height), depth, and to a lesser extent, latitude and the Earth's shape. Derivations for height and depth variations are particularly important.

1. Acceleration due to Gravity (g) on Earth's Surface

• Definition: It is the acceleration experienced by an object falling freely under the influence of Earth's gravitational pull.


• Formula: $g = frac{GM}{R^2}$, where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.


• Value: Approximately $9.8 m/s^2$. You should be familiar with this value and its units.

2. Variation of 'g' with Altitude (Height)

This is a crucial derivation for CBSE boards. You must be able to derive and apply the formula.

• Consider an object at height 'h' above the Earth's surface. The distance from the center of Earth becomes (R+h).


• Derivation Steps:
  
  
  
  1. Start with $g = frac{GM}{R^2}$ for surface.
  
  
  2. For height h, $g_h = frac{GM}{(R+h)^2}$.
  
  
  3. Divide $g_h$ by g: $frac{g_h}{g} = frac{R^2}{(R+h)^2} = left(1 + frac{h}{R}
     ight)^{-2}$.
  
  
  4. For small heights (h << R): Use binomial expansion (1+x)^n approx 1+nx$.
  
  
  5. Resulting formula: $g_h approx gleft(1 - frac{2h}{R}
     ight)$.
  
  
  
  


• Key Point: Acceleration due to gravity decreases with increasing altitude.

3. Variation of 'g' with Depth

Another important derivation for CBSE. Understand the concept of the effective mass contributing to gravity inside the Earth.

• Consider an object at depth 'd' below the Earth's surface. The distance from the center of Earth becomes (R-d).


• Derivation Steps:
  
  
  
  1. Assume Earth has uniform density $
     ho = frac{M}{frac{4}{3}pi R^3}$.
  
  
  2. At depth d, only the mass of the sphere of radius (R-d) contributes to gravity. Let this mass be $M' =
     ho cdot frac{4}{3}pi (R-d)^3$.
  
  
  3. Substitute M' into $g_d = frac{GM'}{(R-d)^2}$.
  
  
  4. After simplification, using $g = frac{GM}{R^2}$:
  
  
  5. Resulting formula: $g_d = gleft(1 - frac{d}{R}
     ight)$.
  
  
  
  


• Key Point: Acceleration due to gravity decreases with increasing depth, becoming zero at the Earth's center.

4. Variation of 'g' with Latitude (Rotation of Earth)

Qualitative understanding and the formula are important for CBSE, though detailed derivation is less common.

• Due to Earth's rotation, a pseudo-force (centrifugal force) acts outwards, effectively reducing the gravitational pull.


• Formula: $g' = g - Romega^2 cos^2lambda$, where $omega$ is the angular velocity of Earth and $lambda$ is the latitude.


• Implications:
  
  
  
  • At Poles ($lambda = 90^circ$, $coslambda = 0$): $g_p = g$ (maximum 'g').
  
  
  • At Equator ($lambda = 0^circ$, $coslambda = 1$): $g_e = g - Romega^2$ (minimum 'g').
  
  
  
  


• Key Point: 'g' is maximum at the poles and minimum at the equator due to Earth's rotation.

5. Variation of 'g' with Shape of Earth

• Earth is not a perfect sphere; it's an oblate spheroid, flattened at the poles and bulging at the equator.


• Implication: Radius at poles (R_p) is less than radius at equator (R_e).


• Since $g propto frac{1}{R^2}$, a smaller radius means larger 'g'.


• This also contributes to 'g' being greater at poles than at the equator.

CBSE Exam Tip

Be prepared to derive the expressions for variation of 'g' with height and depth. Practice solving numerical problems based on these formulas, especially those comparing 'g' at different locations or finding the height/depth for a given percentage change in 'g'.

Welcome, future engineers! This section on 'Acceleration due to gravity and its variation' is a recurring favorite in JEE Main. Mastering its nuances, especially the conditions for using approximations, is key to scoring well. Let's focus on the essential aspects.

JEE Main Focus Areas: Acceleration due to Gravity and its Variation

The acceleration due to gravity (g) is a fundamental concept. While its standard value is approximately 9.8 m/s², JEE problems often test its variations under different conditions.

1. Definition and Standard Value

• The acceleration experienced by a body due to Earth's gravitational pull is called acceleration due to gravity (g).


• Its value on the surface of Earth is given by $g = frac{GM}{R^2}$, where M is Earth's mass and R is Earth's radius.


• Standard value: $g approx 9.8 ext{ m/s}^2$ (or sometimes 10 m/s² for calculation ease in problems, unless specified).

2. Variation with Altitude (Height 'h')

• 
  General Formula: At a height 'h' above the Earth's surface, the acceleration due to gravity is:
  $g_h = frac{GM}{(R+h)^2} = g left( frac{R}{R+h}
  ight)^2$
  


• 
  Approximation for h << R: If 'h' is very small compared to 'R' (typically h < 5% of R), we can use the binomial approximation $(1+x)^n approx 1+nx$ for $x ll 1$.
  $g_h approx g left( 1 - frac{2h}{R}
  ight)$
  
  
  JEE Tip: Always check if $h ll R$ before using the approximation. Using the general formula is safer if unsure or if 'h' is comparable to 'R'. Problems often ask for percentage change, for which the approximation is useful.
  

3. Variation with Depth (Depth 'd')

• 
  At a depth 'd' below the Earth's surface, the acceleration due to gravity is:
  $g_d = g left( 1 - frac{d}{R}
  ight)$
  


• 
  At the Earth's center: When $d = R$, then $g_d = 0$. This is an important conceptual point.
  


• 
  JEE Tip: For depth, g decreases linearly from the surface to the center. For height, g decreases quadratically. Be prepared for graphs of g vs. distance from the center.
  

4. Variation due to Earth's Rotation (Latitude 'λ')

• 
  Due to the Earth's rotation with angular speed 'ω', the effective acceleration due to gravity at a latitude 'λ' is:
  $g_lambda = g - Romega^2 cos^2lambda$
  


• 
  At Equator ($λ = 0^circ$, $coslambda = 1$): $g_{equator} = g - Romega^2$ (minimum value).
  


• 
  At Poles ($λ = 90^circ$, $coslambda = 0$): $g_{poles} = g$ (maximum value).
  


• 
  JEE Tip: This variation is often combined with scenarios involving a change in Earth's angular speed (e.g., "What if Earth stops rotating?"). Note that $Romega^2$ is a small quantity ($approx 0.034$ m/s²).
  

5. Variation due to Shape of Earth

• Earth is not a perfect sphere; it's an oblate spheroid (flattened at poles, bulged at equator).


• The equatorial radius ($R_e$) is greater than the polar radius ($R_p$).


• Since $g propto 1/R^2$, $g_{pole} > g_{equator}$. This effect works in conjunction with rotational effects to make gravity strongest at the poles and weakest at the equator.

JEE Main Problem-Solving Strategies:

1. 
   Identify the variation: Is it height, depth, or rotation? Sometimes problems combine these.
   


2. 
   Choose the correct formula: For height, decide if 'h' is small enough for the approximation.
   


3. 
   Percentage Change: Problems often ask for the percentage change in 'g'.
   $ ext{Percentage change} = frac{Delta g}{g} imes 100\% = frac{|g_{final} - g_{initial}|}{g_{initial}} imes 100\%$
   


4. 
   Graphical Analysis: Be prepared to interpret or sketch graphs of 'g' vs. distance from Earth's center (r), showing linear variation inside and inverse square variation outside.
   

Focus on understanding the conditions for each formula and approximation. Consistent practice with problems involving these variations will solidify your understanding and boost your confidence. You've got this!