Gravitation: Comprehensive NEET Physics Notes

1. Variation of 'g' (Acceleration due to Gravity)

1.1 Introduction to Acceleration due to Gravity

The acceleration due to gravity, denoted by $g$ , refers to the gravitational acceleration experienced by an object near the Earth's surface. The standard value is approximately 9.8 m/s² on Earth's surface. However, this value varies based on factors like height (altitude), depth, latitude, and local geological formations. Understanding these variations is essential for solving NEET questions involving gravitational concepts.

1.2 Factors Affecting Variation in 'g'

1.2.1 Variation with Height (Altitude)

As we move away from the Earth's surface, the value of $g$ decreases. The acceleration due to gravity at a height $h$ above the Earth's surface is given by:

$g_{h} = g_{0} (1 - \frac{2 h}{R _{E}})$

Where:

$g_{h}$ is the acceleration due to gravity at height $g$ .
$g_{0}$ is the acceleration due to gravity on Earth's surface.
$R_{E}$ is the radius of the Earth.

This equation shows that for small heights, the decrease in gravity is nearly linear. As altitude increases, gravitational force weakens due to the inverse square law.

NEET Problem-Solving Strategy:

When solving NEET questions that involve altitude changes, use the approximation that for small heights, $g$ decreases almost linearly.

Common Misconception:

Students often mistakenly assume that $g$ remains constant irrespective of height. Always apply the altitude variation formula to ensure accuracy.

Visual Aid Recommendation:

Include a diagram illustrating the decrease of gravity with altitude and the relationship between the distance from Earth's center and gravity.

1.2.2 Variation with Depth

When we move below the Earth's surface, $g$ also decreases. The value of $g$ at depth $d$ is:

$g_{d} = g_{0} (1 - \frac{d}{R _{E}})$

At greater depths, less of the Earth's mass contributes to the gravitational pull on the object, reducing the gravitational force. Near the Earth’s core, $g$ would approach zero.

Did You Know?

The concept of concentric spherical shells is used to explain why gravity decreases inside the Earth. At any point, only the mass within the radius of that point contributes to gravitational pull.

Real-life Application:

In mining engineering, the variation of $g$ with depth affects the weight of materials, which is crucial for designing tunnels and shafts.

Visual Aid Recommendation:

Add a diagram depicting a cross-section of Earth with gravity decreasing as depth increases.

1.2.3 Variation with Latitude

The value of $g$ changes slightly with latitude due to Earth's rotation and its slightly oblate shape. This effect is accounted for by the following formula:

$g_{latitude} = g_{0} - R_{E} ω^{2} cos^{2} λ$

Where:

$g_{latitude}$ is the acceleration due to gravity at latitude $λ$ .
$R_{E}$ is the Earth's radius.
$ω$ is the angular velocity of the Earth.
$λ$ is the latitude.

The centrifugal force caused by the Earth’s rotation reduces gravity at the equator and increases it at the poles, making gravity stronger at the poles and weaker at the equator.

NEET Tip:

For latitude-based questions, remember that gravity is weaker at the equator due to the centrifugal force. This is a key point often tested in NEET exams.

Visual Aid Recommendation:

Add a visual representation showing how centrifugal force affects gravity at different latitudes, with a comparison between equatorial and polar regions.

Quick Recap:

Altitude: Gravity decreases with height using the relation $g_{h} = g_{0} (1 - \frac{2 h}{R _{E}})$ .
Depth: Gravity decreases with depth as given by $g_{d} = g_{0} (1 - \frac{d}{R _{E}})$ .
Latitude: Gravity decreases towards the equator due to the Earth’s rotation and is greatest at the poles.

Concept Connection:

Biology Link: The effect of gravity variations at altitude is important in biology, particularly in understanding high-altitude sickness.
Chemistry Link: Gravity influences atmospheric pressure, affecting gas distribution, which is crucial in understanding gas laws in chemistry.

Practice Questions:

A satellite orbits the Earth at a height of 1200 km. Calculate the value of $g$ at that height.
Given: Earth's radius $R_{E} = 6.4 \times 1 0^{6} m$ and $g_{0} = 9.8 m / s^{2}$ .
Solution:
Using $g_{h} = g_{0} (1 - \frac{2 h}{R _{E}})$ :
$g_{h} = 9.8 (1 - \frac{2 \times 1200 \times 1 0 ^{3}}{6.4 \times 1 0 ^{6}}) = 7.1 m / s^{2}$ .
At what depth below Earth's surface will the value of $g$ be reduced to half of its surface value?
Solution:
Given $g_{d} = \frac{g _{0}}{2}$ , we find that:
$\frac{g _{0}}{2} = g_{0} (1 - \frac{d}{R _{E}})$ , which gives $d = \frac{R _{E}}{2} = 3200 km$ .
Determine the change in $g$ at the equator compared to the poles if Earth's angular velocity is $7.27 \times 1 0^{- 5} r a d / s$ .
Calculate the value of $g$ at an altitude of 8000 m above sea level if $g_{0}$