Prerequisites
Before studying this topic, make sure you understand:
- Newton’s Law of Universal Gravitation — The foundation for understanding g
- Circular Motion — For understanding Earth’s rotation effects
The Hook: Why Astronauts Train Underwater
Ever wondered why ISRO astronauts for Gaganyaan mission train in swimming pools?
Because underwater, their effective weight reduces due to buoyancy — simulating conditions where gravity is weaker (like on the Moon or at high altitudes).
But here’s the fascinating part: Gravity itself changes with location!
- Climb Mount Everest? You’re slightly lighter (g is less)
- Dive to the ocean floor? You’re heavier (g is more)
- Stand at the equator vs poles? Different weights!
Why does g vary? Let’s find out.
The Core Concept
What is ‘g’?
Acceleration due to gravity (g) is the acceleration experienced by a freely falling object near Earth’s surface.
At Earth’s surface, using Newton’s law:
$$mg = G\frac{Mm}{R^2}$$ $$\boxed{g = \frac{GM}{R^2}}$$Standard value: $g = 9.8$ m/s² (or 10 m/s² for quick calculations)
In simple terms: “g tells you how fast things accelerate when falling. It’s what gives you your weight: W = mg”
Interactive Demo: Visualize Gravitational Formulas
Explore how g changes with mass and radius of a planet.
| Property | $G$ (Universal Constant) | $g$ (Acceleration) |
|---|---|---|
| Value | $6.67 \times 10^{-11}$ N·m²/kg² | 9.8 m/s² |
| Nature | Fundamental constant | Derived quantity |
| Varies? | NO — same everywhere | YES — depends on location |
| Depends on | Nothing (universal) | Mass and radius of planet |
Formula connecting them: $g = \frac{GM}{R^2}$
Variation of g with Height
Above Earth’s Surface
For an object at height $h$ above Earth’s surface:
Distance from center: $r = R + h$
$$g_h = \frac{GM}{(R+h)^2}$$Comparing with surface value $g = \frac{GM}{R^2}$:
$$\boxed{g_h = g\left(\frac{R}{R+h}\right)^2}$$Approximation for Small Heights (h « R)
For satellites close to Earth or tall mountains:
$$g_h = g\left(1 + \frac{h}{R}\right)^{-2}$$Using binomial expansion $(1+x)^n \approx 1 + nx$ for small x:
$$\boxed{g_h \approx g\left(1 - \frac{2h}{R}\right)}$$Valid when: $h < 0.1R$ (i.e., h < 640 km)
At Mount Everest summit (h = 8.85 km):
$$g_h = g\left(1 - \frac{2 \times 8850}{6.4 \times 10^6}\right) = g(1 - 0.00276)$$ $$g_h \approx 0.9973g \approx 9.773 \text{ m/s}^2$$You’re about 0.27% lighter at the summit! A 60 kg person “loses” about 160 grams.
Variation of g with Depth
Below Earth’s Surface
Assumption: Earth has uniform density $\rho$
At depth $d$ below surface, only the mass of the sphere of radius $(R-d)$ contributes (shell theorem!).
$$M' = \frac{4}{3}\pi(R-d)^3 \rho$$Since $M = \frac{4}{3}\pi R^3 \rho$, we get $\frac{M'}{M} = \frac{(R-d)^3}{R^3}$
$$g_d = \frac{GM'}{(R-d)^2} = \frac{G \cdot M \cdot (R-d)^3}{R^3 \cdot (R-d)^2}$$ $$g_d = \frac{GM}{R^2} \cdot \frac{R-d}{R}$$ $$\boxed{g_d = g\left(1 - \frac{d}{R}\right)}$$Linear decrease! Unlike the 1/r² decrease with height, g decreases linearly with depth.
At the Center of Earth
When $d = R$:
$$g_{center} = g(1 - 1) = 0$$At Earth’s center, you’re pulled equally in all directions by the surrounding mass. All gravitational forces cancel out — you’d be weightless!
Movie Connection: In Total Recall and similar sci-fi movies, elevators through Earth’s core would experience zero gravity at the center.
Comparing Height vs Depth Variation
| Property | Height ($h$) | Depth ($d$) |
|---|---|---|
| Formula | $g_h = g\left(\frac{R}{R+h}\right)^2$ | $g_d = g\left(1 - \frac{d}{R}\right)$ |
| Approximation | $g_h \approx g\left(1 - \frac{2h}{R}\right)$ | No approximation needed |
| Variation type | Inverse square (1/r²) | Linear |
| At equal distance | $g_h < g_d$ | $g_d > g_h$ |
| Maximum decrease | Approaches 0 at $r \to \infty$ | Becomes 0 at center |
Key Insight: For same distance, depth has less effect than height!
Example: At h = d = R/2:
- $g_h = g \cdot \frac{4}{9} \approx 0.44g$
- $g_d = g \cdot \frac{1}{2} = 0.5g$
$g_d > g_h$ — You’re heavier deep underground than high in the sky!
Variation of g with Latitude
Effect of Earth’s Rotation
Earth rotates with angular velocity $\omega = \frac{2\pi}{T}$ where $T = 24$ hours.
An object at latitude $\lambda$ experiences:
- True gravitational force: $mg$ (towards center)
- Centrifugal force: $m\omega^2 R\cos\lambda$ (outward)
Effective $g'$ (apparent gravity):
$$\boxed{g' = g - \omega^2 R\cos^2\lambda}$$At Different Latitudes
| Location | Latitude $\lambda$ | $\cos\lambda$ | Effective g |
|---|---|---|---|
| Poles | 90° | 0 | $g_{pole} = g$ (maximum) |
| Equator | 0° | 1 | $g_{eq} = g - \omega^2 R$ (minimum) |
| 45° latitude | 45° | $1/\sqrt{2}$ | $g_{45} = g - \frac{\omega^2 R}{2}$ |
Numerical value: $\omega^2 R \approx 0.034$ m/s²
So: $g_{pole} - g_{eq} \approx 0.034$ m/s² (about 0.35% difference)
Why does ISRO launch rockets from Sriharikota (near equator)?
At equator:
- $g$ is minimum — less fuel needed to escape gravity
- Earth’s rotational speed is maximum (about 460 m/s eastward) — free boost to orbital velocity!
For eastward launches: Effective launch velocity = rocket velocity + Earth’s rotation
This saves fuel and increases payload capacity!
Combined Variation with Height and Latitude
At height $h$ and latitude $\lambda$:
$$g' = g\left(\frac{R}{R+h}\right)^2 - \omega^2(R+h)\cos^2\lambda$$For small heights:
$$g' \approx g\left(1 - \frac{2h}{R}\right) - \omega^2 R\cos^2\lambda$$Memory Tricks & Patterns
Mnemonic for Variation Formulas
“Height Hurts Heavily, Depth Decreases Directly”
- Height: Square law → $\left(\frac{R}{R+h}\right)^2$ or $1 - \frac{2h}{R}$
- Depth: Linear → $1 - \frac{d}{R}$
The 1/2 vs 2 Pattern
Height approximation: Factor is 2h
$$g_h \approx g\left(1 - \frac{2h}{R}\right)$$Depth formula: Factor is just d
$$g_d = g\left(1 - \frac{d}{R}\right)$$Remember: Height has “2” because it follows inverse square law!
Quick Comparison Trick
Same distance h = d:
For both: decrease = $\frac{2h}{R}$ and $\frac{d}{R}$
Since $\frac{2h}{R} > \frac{d}{R}$ for same distance:
Height decreases g MORE than depth → $g_h < g_d$
When to Use Which Formula
At surface (no height/depth):
- Simple problems: $g = 9.8$ m/s² or $g = 10$ m/s²
- Deriving g: $g = \frac{GM}{R^2}$
Above surface (height h):
- General: $g_h = g\left(\frac{R}{R+h}\right)^2$
- Small heights (h < 640 km): $g_h \approx g\left(1 - \frac{2h}{R}\right)$
- Far from Earth (satellites): Use $g = \frac{GM}{r^2}$ directly
Below surface (depth d):
- Always: $g_d = g\left(1 - \frac{d}{R}\right)$
- At center: $g = 0$
Rotation effects:
- Add: $g' = g - \omega^2 R\cos^2\lambda$
Common Mistakes to Avoid
Wrong: $g_h = \frac{GM}{h^2}$
Correct: $g_h = \frac{GM}{(R+h)^2}$
Distance is measured from Earth’s center, not surface!
For small h:
Wrong: $g_h \approx g\left(1 - \frac{h}{R}\right)$ ❌
Correct: $g_h \approx g\left(1 - \frac{2h}{R}\right)$ ✓
Don’t forget the factor of 2!
Height and depth formulas are NOT the same!
Height: $g_h = g\left(1 - \frac{2h}{R}\right)$ (approximation)
Depth: $g_d = g\left(1 - \frac{d}{R}\right)$ (exact)
Different factors: 2 vs 1
The depth formula assumes uniform density. Real Earth has denser core, so:
- Formula is approximate
- In mines, measured g is slightly higher than predicted
For JEE, assume uniform density unless stated otherwise.
Practice Problems
Level 1: Foundation (NCERT/Basics)
At what height above Earth’s surface will g be 4.9 m/s² (half of 9.8 m/s²)? (Earth’s radius = 6400 km)
Solution:
$$g_h = g\left(\frac{R}{R+h}\right)^2$$ $$\frac{g}{2} = g\left(\frac{R}{R+h}\right)^2$$ $$\frac{1}{2} = \left(\frac{R}{R+h}\right)^2$$ $$\frac{R}{R+h} = \frac{1}{\sqrt{2}}$$ $$R+h = R\sqrt{2}$$ $$h = R(\sqrt{2} - 1) = R(1.414 - 1) = 0.414R$$ $$h = 0.414 \times 6400 = 2650 \text{ km}$$Answer: At height 2650 km above Earth’s surface
At what depth below Earth’s surface is g reduced to 60% of its surface value? (R = 6400 km)
Solution:
$$g_d = g\left(1 - \frac{d}{R}\right)$$ $$0.6g = g\left(1 - \frac{d}{R}\right)$$ $$0.6 = 1 - \frac{d}{R}$$ $$\frac{d}{R} = 0.4$$ $$d = 0.4R = 0.4 \times 6400 = 2560 \text{ km}$$Answer: At depth 2560 km below surface
Level 2: JEE Main Type
Compare the values of g at height h and depth h below Earth’s surface, where h « R.
Solution:
At height h:
$$g_h = g\left(1 - \frac{2h}{R}\right)$$At depth h:
$$g_d = g\left(1 - \frac{h}{R}\right)$$Ratio:
$$\frac{g_h}{g_d} = \frac{1 - \frac{2h}{R}}{1 - \frac{h}{R}}$$For small h/R, approximate:
$$\frac{g_h}{g_d} \approx \left(1 - \frac{2h}{R}\right)\left(1 + \frac{h}{R}\right)$$ $$\approx 1 - \frac{2h}{R} + \frac{h}{R} = 1 - \frac{h}{R}$$Conclusion: $g_h < g_d$
The decrease is more at height than at depth!
Numerical example: If h = R/100 (64 km):
- $g_h = g(1 - 0.02) = 0.98g$
- $g_d = g(1 - 0.01) = 0.99g$
Ratio: $\frac{g_h}{g_d} = \frac{0.98}{0.99} \approx 0.99$
If g at Earth’s surface is 9.8 m/s², find g at height equal to Earth’s radius.
Solution:
Given: $h = R$
$$g_h = g\left(\frac{R}{R+h}\right)^2 = g\left(\frac{R}{R+R}\right)^2$$ $$g_h = g\left(\frac{R}{2R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$$ $$g_h = \frac{9.8}{4} = 2.45 \text{ m/s}^2$$Answer: 2.45 m/s² (one-fourth of surface value)
Note: At h = R, distance from center is 2R, so by inverse square law, g becomes (1/2)² = 1/4.
A body weighs 63 N on Earth’s surface. What will it weigh at a height of R/2? (g = 10 m/s²)
Solution:
Weight on surface: $W = mg = 63$ N → $m = 6.3$ kg
At height h = R/2:
$$g_h = g\left(\frac{R}{R + R/2}\right)^2 = g\left(\frac{R}{3R/2}\right)^2 = g\left(\frac{2}{3}\right)^2 = \frac{4g}{9}$$New weight:
$$W_h = mg_h = m \cdot \frac{4g}{9} = \frac{4mg}{9} = \frac{4 \times 63}{9} = 28 \text{ N}$$Answer: 28 N
Shortcut: $W_h = W \cdot \left(\frac{R}{R+h}\right)^2 = 63 \times \frac{4}{9} = 28$ N
Level 3: JEE Advanced Type
At what distance from Earth’s center is the value of g same as at a depth of R/2 below surface?
Solution:
At depth d = R/2:
$$g_d = g\left(1 - \frac{R/2}{R}\right) = g\left(\frac{1}{2}\right) = \frac{g}{2}$$At distance r from center (height h = r - R):
$$g_r = \frac{GM}{r^2} = \frac{g}{2}$$Since $g = \frac{GM}{R^2}$:
$$\frac{GM}{r^2} = \frac{1}{2} \cdot \frac{GM}{R^2}$$ $$\frac{1}{r^2} = \frac{1}{2R^2}$$ $$r^2 = 2R^2$$ $$r = R\sqrt{2}$$Answer: At distance $R\sqrt{2}$ from Earth’s center
Height above surface: $h = R\sqrt{2} - R = R(\sqrt{2} - 1) \approx 0.414R$
Find the ratio of heights above Earth’s surface where g reduces to (a) 36% of surface value, and (b) where weight reduces to 36% of surface weight.
Solution:
Case (a): g reduces to 36%
$$g_h = 0.36g$$ $$g\left(\frac{R}{R+h}\right)^2 = 0.36g$$ $$\frac{R}{R+h} = 0.6$$ $$R = 0.6R + 0.6h$$ $$0.4R = 0.6h$$ $$h = \frac{2R}{3}$$Case (b): Weight reduces to 36%
Weight = mg, so if weight reduces to 36%, then g ALSO reduces to 36% (mass doesn’t change!).
$$h = \frac{2R}{3}$$(same as case a)
Answer: Both are the same = $\frac{2R}{3}$
Key Insight: Reduction in weight is ALWAYS equal to reduction in g (since W = mg and m is constant).
A tunnel is dug along a diameter of Earth. A particle is dropped into it. Show that it will execute SHM and find the time period. (Assume uniform density, neglect air resistance)
Solution:
At distance $x$ from center:
$$g(x) = g \cdot \frac{x}{R}$$(This is the depth formula rearranged: at depth d, $g_d = g(1 - d/R) = g \cdot (R-d)/R = g \cdot x/R$ where x = R-d)
Force on particle of mass m:
$$F = -mg(x) = -mg\frac{x}{R}$$Negative sign because force is towards center (restoring).
$$F = -\frac{mg}{R} \cdot x$$This is of the form $F = -kx$ where $k = \frac{mg}{R}$
This is SHM!
Acceleration:
$$a = \frac{F}{m} = -\frac{g}{R}x$$Comparing with $a = -\omega^2 x$:
$$\omega^2 = \frac{g}{R}$$ $$\omega = \sqrt{\frac{g}{R}}$$Time period:
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}}$$Numerical value:
$$T = 2\pi\sqrt{\frac{6.4 \times 10^6}{10}} = 2\pi\sqrt{640000} = 2\pi \times 800$$ $$T \approx 5024 \text{ seconds} \approx 84 \text{ minutes}$$Answer: Time period = $2\pi\sqrt{\frac{R}{g}} \approx 84$ minutes
Amazing fact: This is independent of the mass of the particle! Also, this is the same as the time period of a satellite orbiting very close to Earth’s surface!
Quick Revision Box
| Situation | Formula | Key Point |
|---|---|---|
| At height h (general) | $g_h = g\left(\frac{R}{R+h}\right)^2$ | Inverse square law |
| At height h (h « R) | $g_h \approx g\left(1 - \frac{2h}{R}\right)$ | Note the factor of 2 |
| At depth d | $g_d = g\left(1 - \frac{d}{R}\right)$ | Linear decrease |
| At Earth’s center | $g = 0$ | All forces cancel |
| Due to rotation | $g' = g - \omega^2R\cos^2\lambda$ | Minimum at equator |
| At poles | $g' = g$ | Maximum value |
| At equator | $g' = g - \omega^2R$ | Minimum value |
| Same distance h = d | $g_h < g_d$ | Height affects more |
Critical Values:
- $g = 9.8$ m/s² (or 10 m/s²)
- $R = 6.4 \times 10^6$ m
- $\omega^2R \approx 0.034$ m/s²
Related Topics
Within Gravitation Chapter
- Universal Gravitation — Foundation for deriving g
- Gravitational Field and Potential — Energy aspects
- Satellites — Uses variation of g with height
Connected Chapters
- Kinematics — Free fall uses constant g
- Circular Motion — Earth’s rotation effects
- Simple Harmonic Motion — Tunnel through Earth problem
Experimental
- Experimental Skills — Measuring g using pendulum or free fall
Teacher’s Summary
g is Not Constant: It varies with height, depth, and latitude — only approximately constant near Earth’s surface
Height vs Depth: For same distance, height reduces g MORE than depth ($g_h < g_d$) because of inverse square vs linear variation
Two Key Formulas: Height → $g\left(1 - \frac{2h}{R}\right)$, Depth → $g\left(1 - \frac{d}{R}\right)$ — note the factor of 2 difference!
Rotation Matters: Earth’s spin makes you lighter at the equator — that’s why rockets launch from there
Center is Special: At Earth’s center, g = 0 despite enormous mass around you — symmetry cancels all forces
“From mountain peaks to ocean depths, from poles to equator — gravity’s strength varies. Master these variations, and you’ll understand why satellites orbit, why ISRO launches from Sriharikota, and why your weight certificate from sea level won’t match at Shimla!”