This MCQ module is based on: G Earth Altitude Depth
TOPIC 30 OF 33
G Earth Altitude Depth
🎓 Class 11
Physics
CBSE
Theory
Ch 7 – Gravitation
⏱ ~14 min
🌐 Language: [gtranslate]
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G Earth Altitude Depth
7.4 The Gravitational Constant — and What It Tells Us
We saw in Part 1 that Newton's law contains the constant G = 6.67×10⁻¹¹ N m² kg⁻². Knowing G allows us to "weigh" the Earth itself: by measuring the free-fall acceleration g of a falling apple, we can deduce the mass of our planet without ever stepping off it.
7.5 Acceleration due to Gravity of the Earth
Consider an object of mass m placed on Earth's surface at a distance \(R_E\) from Earth's centre. By Newton's law of gravitation:
\[F = G\frac{M_E\,m}{R_E^2}\]
The same force, by Newton's Second Law, equals \(m\cdot g\). Cancelling m gives the famous expression for acceleration due to gravity at Earth's surface:
Surface gravity:
\[g = \frac{G M_E}{R_E^2}\approx 9.8\,\text{m s}^{-2}\]
Using \(M_E = 6.0\times 10^{24}\) kg and \(R_E = 6.4\times 10^6\) m.
The Shell Theorem (Newton, 1687) lets us treat Earth as if all its mass were concentrated at its centre when computing the field outside Earth. Inside Earth, only the mass within the sphere of radius r below us contributes — the spherical shell above us cancels itself out (its inward and outward pulls perfectly balance).
7.6 Acceleration due to Gravity Below and Above the Earth's Surface
7.6.1 Above the Surface (Altitude h)
If we go to height h above the surface, the distance from Earth's centre becomes \(R_E + h\). So:
\[g(h) = \frac{G M_E}{(R_E+h)^2}\]
Dividing by surface g:
\[\frac{g(h)}{g} = \left(\frac{R_E}{R_E+h}\right)^2\]
For small heights \(h\ll R_E\), using binomial expansion:
\[g(h) \approx g\left(1-\frac{2h}{R_E}\right)\]
So g decreases linearly with altitude for small h. At the top of Mount Everest (h ≈ 8.85 km), g drops by about 0.28%. At the International Space Station (h ≈ 400 km), g is still about 89% of surface g — astronauts "float" not because gravity is absent but because they are in continuous free fall around Earth.
7.6.2 Below the Surface (Depth d)
Imagine descending to depth d below the surface. Only the mass inside a sphere of radius \((R_E - d)\) attracts you; the outer shell contributes zero net force (Shell Theorem). Assuming uniform density \(\rho\):
\[M_{\text{inside}} = \frac{4}{3}\pi(R_E-d)^3 \rho\]
So the gravity at depth d is:
\[g(d) = \frac{G\cdot \frac{4}{3}\pi(R_E-d)^3\rho}{(R_E-d)^2}=\frac{4}{3}\pi G\rho(R_E-d)\]
Variation with depth:
\[g(d) = g\left(1-\frac{d}{R_E}\right)\]
At the centre of the Earth (d = R_E), g = 0! Below the surface, gravity decreases linearly, while above the surface it decreases as the inverse square.
7.6.3 Variation Due to Latitude (Rotation Effect)
Earth rotates about its axis with angular velocity \(\omega = 7.27\times 10^{-5}\) rad/s. An object at latitude \(\lambda\) moves in a circle of radius \(R_E\cos\lambda\). The required centripetal force is provided partly by gravity, so the apparent g is reduced:
\[g'(\lambda) = g - R_E\,\omega^2\cos^2\lambda\]
This effect is maximum at the equator (\(\lambda = 0\)) where it reduces g by ≈0.034 m/s². At the poles, the effect vanishes. Together with Earth's equatorial bulge, this is why g is largest at the poles (9.83 m/s²) and smallest at the equator (9.78 m/s²).
| Location | Latitude | g (m/s²) |
|---|---|---|
| North Pole | 90° | 9.832 |
| Helsinki | 60° | 9.819 |
| New Delhi | 28° | 9.791 |
| Singapore | 1° | 9.781 |
| Equator (sea level) | 0° | 9.780 |
7.7 The Gravitational Field
The gravitational field at a point is the force experienced per unit test mass placed at that point:
\[\vec{E}_g = \frac{\vec{F}}{m_0} = -\frac{GM}{r^2}\hat{r}\]
Field is a vector pointing toward the source mass. Its SI unit is N/kg (= m/s²). The field due to several masses is the vector sum of individual fields (superposition principle).
Interactive Simulation: g at Different Altitudes
Slide the altitude to see how g changes above (and below) Earth's surface.
g = 9.80 m/s²
Negative altitude = depth below surface. h = −6400 km is Earth's centre (g = 0).
Worked Example 1: g at the top of Everest
Find g at the top of Mt. Everest (h = 8.85 km).
\[g(h)\approx g\left(1-\frac{2h}{R_E}\right) = 9.8\left(1-\frac{2\times 8.85}{6400}\right)\]
\[ = 9.8\times(1-0.00276) = \boxed{9.773\,\text{m/s}^2}\]
About 0.28% less than sea-level g.
Worked Example 2: g at a depth equal to R_E/2
Find g at depth d = R_E/2.
\[g(d) = g\left(1-\frac{d}{R_E}\right) = 9.8\left(1-\frac{1}{2}\right) = \boxed{4.9\,\text{m/s}^2}\]
Exactly half of surface g.
Worked Example 3: Mass of Earth from g
Using g = 9.8 m/s² and R_E = 6.4×10⁶ m, find Earth's mass.
\[M_E = \frac{g R_E^2}{G} = \frac{9.8\times(6.4\times 10^6)^2}{6.67\times 10^{-11}}\]
\[ = \frac{9.8\times 4.1\times 10^{13}}{6.67\times 10^{-11}} \approx \boxed{6.0\times 10^{24}\,\text{kg}}\]
We have "weighed" the Earth using only G and a falling object!
Activity 7.2 — Measuring g with a Simple PendulumL4 Analyse
Materials: Thread (≈1 m), bob, stand, stopwatch.
- Suspend the bob and measure thread length L (to bob's centre).
- Pull the bob aside (small angle <10°) and release.
- Time 20 oscillations; divide by 20 to get period T.
- Use \(T = 2\pi\sqrt{L/g}\) to compute g.
Predict: Will your value match 9.8 m/s² exactly? What sources of error could shift it?
Typical student results: 9.4–10.1 m/s². Errors arise from: large amplitude (use small angle), air drag, parallax in measuring length, reaction-time error in timing (mitigate by counting 20+ swings). The pendulum method was the standard for measuring g until the 20th century — and Foucault's pendulum still demonstrates Earth's rotation in many planetariums.
Competency-Based Questions
An Indian Air Force fighter jet flies at 25 km above sea level. A scientist on board carries a high-precision gravimeter to measure local g. Meanwhile, a deep-mining team in Karnataka works 3 km below the surface and notices their precision balance reads slightly differently than at the surface.
Q1. The percentage decrease in g at 25 km altitude (compared to surface) is closest to:L3 Apply
Answer: (b). Δg/g ≈ 2h/R_E = 2×25/6400 = 0.0078 = 0.78%.
Q2. At 3 km depth, by what fraction does g decrease?L3 Apply
Δg/g = d/R_E = 3/6400 ≈ 0.047%. So g_3km ≈ 9.795 m/s². Smaller change than at the same height above ground.
Q3. True or False: At Earth's centre, your weight would be zero but your mass would also be zero.L4 Analyse
FALSE. Weight = mg becomes zero (because g = 0 at centre), but mass is an invariant property of the object — it stays exactly the same. This distinction between mass and weight is crucial.
Q4. Why does g have its maximum value at the poles?L2 Understand
Two reasons: (i) Earth is slightly flattened at the poles, so polar radius is smaller — closer to the centre means stronger g. (ii) No rotational reduction (centripetal effect = R_Eω²cos²λ vanishes at λ = 90°). Together these make g_pole ≈ 9.83 m/s² vs g_eq ≈ 9.78 m/s².
Q5. HOT: A future colony lives inside a hollow spherical asteroid (no rotation). Predict the gravitational field they would feel inside, and design a clever way they could verify the prediction.L6 Create
Prediction: Inside a hollow shell, the gravitational field is exactly zero everywhere (Shell Theorem corollary). So colonists would float weightless no matter where they stood inside. Verification idea: Release a metallic ball at different points inside the shell. If it stays motionless wherever placed (no preferred direction), the field is null. They could also fire a laser between two points and check it travels in straight line — gravitational lensing absence further confirms zero field.
Assertion–Reason Questions
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
A: g decreases linearly with depth inside the Earth.
R: Only the mass within a sphere of radius (R_E − d) attracts a body at that depth.
(A). Both true and R correctly explains A. The outer shell's pull cancels out.
A: Astronauts in the ISS are weightless because gravity is absent at that altitude.
R: At 400 km altitude, g is still about 89% of its surface value.
(D). A is FALSE — astronauts float because they and the station are in free fall around Earth, not because gravity is zero. R is TRUE (correct physics).
A: The value of g is greater at the poles than at the equator.
R: Earth's rotation reduces the apparent g most at the equator and not at all at the poles.
(A). Both true; R explains A (along with Earth's oblateness — equatorial bulge).
Frequently Asked Questions - G Earth Altitude Depth
What is the main concept covered in G Earth Altitude Depth?
In NCERT Class 11 Physics Chapter 7 (Gravitation), "G Earth Altitude Depth" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is G Earth Altitude Depth useful in real-life applications?
Real-life applications of G Earth Altitude Depth from NCERT Class 11 Physics Chapter 7 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in G Earth Altitude Depth?
Key formulas in G Earth Altitude Depth (NCERT Class 11 Physics Chapter 7 Gravitation) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 7?
NCERT Class 11 Physics Chapter 7 (Gravitation) is structured so each part builds on the previous one. G Earth Altitude Depth connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from G Earth Altitude Depth?
CBSE board questions from G Earth Altitude Depth typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the G Earth Altitude Depth lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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