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Gravity & Universal Gravitation in Physics: Kepler's Laws & Satellites Guide
From an apple falling from a tree to Earth orbiting the Sun and galaxies clustering across cosmic filaments, **Gravity** is the invisible fundamental force of attraction governing the large-scale structure of the universe.
This 4,000+ word comprehensive exam guide covers **Newton's Universal Law of Gravitation ($F = G \frac{m_1 m_2}{r^2}$)**, **Universal Constant $G$ vs Acceleration $g$**, **Variations in $g$ (Height, Depth, Latitude, Earth's Shape & Rotation)**, **Mass vs Weight**, **Kepler's Three Laws of Planetary Motion**, **Orbital Velocity ($v_o$)**, **Escape Velocity ($v_e = 11.2\text{ km/s}$)**, **Geostationary vs Polar Satellites**, **Weightlessness**, and solved numerical problems for SSC CGL, RRB NTPC, and UPSC Prelims.
Table of Contents
- 1. Newton's Universal Law of Gravitation & Gravitational Constant ($G$)
- 2. Acceleration Due to Gravity ($g$) & Relation with $G$
- 3. Factors Causing Variations in $g$ (Height, Depth & Latitude)
- 4. Mass vs Weight: Definitions, Differences & Lift Mechanics
- 5. Kepler's Three Laws of Planetary Motion
- 6. Orbital Velocity ($v_o$) & Escape Velocity ($v_e = 11.2\text{ km/s}$)
- 7. Satellites: Geostationary ($36,000\text{ km}$) vs Polar Satellites
- 8. Weightlessness in Space & Free Fall Physics
- 9. Solved Numerical Examples for Competitive Exams
- 10. Must Remember Points for Quick Revision
- 11. Frequently Asked Questions (FAQ)
Key Takeaways & Core Highlights
- Newton's Law: $F = G \frac{m_1 m_2}{r^2}$. Inverse square law. Universal Constant $G = 6.674 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$.
- Acceleration due to Gravity ($g$): Standard value $g = 9.81\text{ m/s}^2$. $g = \frac{G M}{R^2}$.
- Variations in $g$:
- With Altitude: $g_h = g \left(1 - \frac{2h}{R}\right)$. Decreases above surface.
- With Depth: $g_d = g \left(1 - \frac{d}{R}\right)$. Decreases with depth; **$g = 0$ at Earth's center**!
- Latitude Variation: Max at **Poles** ($\approx 9.83\text{ m/s}^2$); Min at **Equator** ($\approx 9.78\text{ m/s}^2$).
- Escape Velocity Formula: $v_e = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g R} \approx \mathbf{11.2\text{ km/s}}$ on Earth ($\approx 2.38\text{ km/s}$ on Moon).
- Orbital Velocity Formula: $v_o = \sqrt{g R} \approx 7.92\text{ km/s}$. Relationship: $v_e = \sqrt{2} \cdot v_o \approx 1.414 \cdot v_o$.
- Kepler's 3rd Law (Law of Periods): $T^2 \propto r^3$.
- Geostationary Satellite: Height $\approx 35,786\text{ km}$, Time period $= 24\text{ hours}$, orbits west-to-east above equator.
1. Newton's Universal Law of Gravitation & Gravitational Constant ($G$)
Formulated by Sir Isaac Newton in his 1687 masterwork Philosophiæ Naturalis Principia Mathematica, the **Universal Law of Gravitation** states:
Newton's Universal Law Statement:
"Every point mass attracts every other point mass in the universe with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers."
Where $G = 6.674 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$ is the **Universal Gravitational Constant**, first accurately measured experimentally by Henry Cavendish in 1798 using a torsion balance.
2. Acceleration Due to Gravity ($g$) & Relation with $G$
The acceleration produced in a freely falling body under the sole action of Earth's gravitational pull is called **Acceleration Due to Gravity ($g$)**:
$$F = m \cdot g = G \frac{M \cdot m}{R^2} \implies g = \frac{G M}{R^2}$$Where $M = 5.972 \times 10^{24}\text{ kg}$ is Earth's mass and $R = 6371\text{ km}$ is Earth's mean radius.
3. Factors Causing Variations in $g$ (Height, Depth & Latitude)
| Variation Parameter | Mathematical Formula | Physical Effect on Value of $g$ |
|---|---|---|
| Altitude (Height $h$) | $g_h = g \left(1 - \frac{2h}{R}\right) = \frac{G M}{(R+h)^2}$ | $g$ decreases with increasing height above Earth's surface. |
| Depth (Distance $d$ inside Earth) | $g_d = g \left(1 - \frac{d}{R}\right)$ | $g$ decreases linearly with depth. At Earth's center ($d=R$), **$g = 0$**! |
| Shape of Earth (Equatorial Bulge) | $g \propto \frac{1}{R^2}$ ($R_{\text{equator}} > R_{\text{pole}}$ by $21\text{ km}$) | $g$ is Maximum at Poles ($\sim 9.83\text{ m/s}^2$) and Minimum at Equator ($\sim 9.78\text{ m/s}^2$). |
| Earth Rotation ($\omega$) | $g' = g - \omega^2 R \cos^2 \lambda$ | Centrifugal force reduces effective $g$. Max reduction at equator ($\lambda=0^\circ$); Zero reduction at poles ($\lambda=90^\circ$). |
4. Mass vs Weight: Definitions, Differences & Lift Mechanics
- Mass ($m$): Quantity of matter contained in a body. Constant everywhere in the universe. Scalar quantity ($\text{kg}$).
- Weight ($W = mg$): Gravitational force pulling a body toward Earth. Vector quantity ($\text{Newton, N}$). Varies with $g$.
Apparent Weight in an Elevating Lift:
- Lift accelerating upward ($+a$): $W_{\text{apparent}} = m(g + a)$ (Feels heavier).
- Lift accelerating downward ($-a$): $W_{\text{apparent}} = m(g - a)$ (Feels lighter).
- Lift cable snaps (Free fall $a=g$): $W_{\text{apparent}} = m(g - g) = 0$ (State of **Weightlessness**!).
5. Kepler's Three Laws of Planetary Motion
Formulated by Johannes Kepler between 1609 and 1619:
- 1st Law (Law of Orbits): All planets move in elliptical orbits with the Sun located at one of the two foci.
- 2nd Law (Law of Areas): A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time ($\frac{dA}{dt} = \text{constant}$). Direct consequence of **Conservation of Angular Momentum**.
- 3rd Law (Law of Periods): The square of the orbital period ($T$) of a planet is directly proportional to the cube of the semi-major axis ($r$) of its orbit: $$T^2 \propto r^3 \implies \frac{T^2}{r^3} = \text{constant}$$
6. Orbital Velocity ($v_o$) & Escape Velocity ($v_e = 11.2\text{ km/s}$)
1. Orbital Velocity ($v_o$):
The horizontal velocity required for a satellite to maintain a stable circular orbit around Earth:
$$v_o = \sqrt{\frac{G M}{R}} = \sqrt{g R} \approx 7.92 \text{ km/s}$$2. Escape Velocity ($v_e$):
The minimum speed required for an unpropelled body to break free from Earth's gravitational field forever:
$$\frac{1}{2} m v_e^2 = \frac{G M m}{R} \implies v_e = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g R} \approx \mathbf{11.2 \text{ km/s}}$$Relationship Between Escape and Orbital Velocity:
$$v_e = \sqrt{2} \cdot v_o \approx 1.414 \cdot v_o$$Increasing a satellite's orbital speed by $41.4\%$ ($+41.4\%$) causes it to escape Earth's orbit into deep space!
7. Satellites: Geostationary ($36,000\text{ km}$) vs Polar Satellites
| Satellite Category | Orbital Altitude ($h$) | Time Period ($T$) | Primary Operational Uses |
|---|---|---|---|
| Geostationary Satellite (GEO) | $\approx 35,786 \text{ km} \ (\sim 36,000 \text{ km})$ | 24 Hours (Synchronized with Earth rotation) | TV broadcasting, telecommunications, weather monitoring (INSAT, GSAT) |
| Polar Satellite (LEO / Sun-Synchronous) | $500 \text{ to } 800 \text{ km}$ | $\approx 100 \text{ minutes}$ ($1.5 \text{ hours}$) | Remote sensing, earth observation, military intelligence, mapping (IRS) |
8. Weightlessness in Space & Free Fall Physics
Astronauts inside the International Space Station (ISS) float not because gravity is zero ($g \approx 8.7\text{ m/s}^2$ at $400\text{ km}$ altitude!), but because the station and astronauts are in a continuous state of **Free Fall around Earth**.
9. Solved Numerical Examples for Competitive Exams
Numerical Problem 1 (Kepler's 3rd Law Calculation):
Question: The orbital radius of Planet A is 4 times that of Planet B ($r_A = 4 r_B$). If Planet B takes 1 year to complete an orbit, calculate the time period of Planet A.
Solution:
$$\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_A}{r_B}\right)^3 = (4)^3 = 64 \implies \frac{T_A}{T_B} = \sqrt{64} = 8$$ $$T_A = 8 \times T_B = 8 \times 1 \text{ year} = 8 \text{ years}$$Answer: Planet A takes $8\text{ years}$ to complete one orbit.
Numerical Problem 2 (Weight on Moon):
Question: A person weighs $600\text{ N}$ on Earth. What will be their mass and weight on the Moon? ($g_{\text{moon}} = g_{\text{earth}}/6$).
Solution:
$$\text{Mass on Earth } m = \frac{W}{g} = \frac{600}{10} = 60\text{ kg}$$ $$\text{Mass on Moon} = 60\text{ kg} \quad (\text{Mass never changes!})$$ $$\text{Weight on Moon } W_{\text{moon}} = \frac{600}{6} = 100\text{ N}$$Answer: Mass is $60\text{ kg}$ and weight on Moon is $100\text{ N}$.
10. Must Remember Points for Quick Revision
Exam Revision Cheat Sheet:
- Newton's Formula: $F = G \frac{m_1 m_2}{r^2}$. $G = 6.674 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$.
- Earth Acceleration: $g = \frac{GM}{R^2} \approx 9.81\text{ m/s}^2$.
- $g$ Variations: Max at Poles, Min at Equator; Zero at Earth Center ($g_{\text{center}} = 0$).
- Escape Velocity ($v_e$): $11.2\text{ km/s}$ on Earth; $v_e = \sqrt{2} v_o$.
- Kepler's 3rd Law: $T^2 \propto r^3$. 2nd Law $\rightarrow$ Angular Momentum Conservation.
- Geostationary Satellite: Height $\approx 36,000\text{ km}$, Period $= 24\text{ hours}$.
- Weight on Moon: $1/6^{\text{th}}$ of weight on Earth ($W_{\text{moon}} = W_{\text{earth}}/6$).
11. Frequently Asked Questions (FAQ)
What is Newton's Universal Law of Gravitation and value of G?
Newton's Law states that every point mass attracts every other point mass in the universe with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them: F = G (m₁ m₂)/r². The Universal Gravitational Constant G = 6.674 × 10⁻¹¹ N·m²/kg².
How does acceleration due to gravity (g) vary across Earth's surface?
Acceleration due to gravity g is maximum at the Poles (~9.83 m/s²) and minimum at the Equator (~9.78 m/s²). This variation is due to Earth's equatorial bulge (R_equator > R_pole) and Earth's rotational centrifugal force.
What is the escape velocity from Earth's surface?
Escape velocity is the minimum speed required for an unpropelled body to break free from Earth's gravitational attraction forever. From Earth's surface, v_e = √(2 g R) ≈ 11.2 km/s.
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