- Every object attracts every other with a gravitational force.
- Force increases with mass — more mass means a stronger pull.
- Force decreases with distance — doubling the distance halves the force.
- A force acts along the line joining the centres (or centres of mass) of the two bodies.
Topics
Gravitation
- Concept of Gravitation
- Force and Motion
- Centripetal Force
- Kepler’s Laws
- Law of Orbit or Kepler's First Law
- Law of Areas or Kepler's Second Law
- Law of Periods or Kepler's Third Law
- Newton's Universal Law of Gravitation
- Uniform Circular Motion (UCM)
- Earth’s Gravitational Force
- Earth’s Gravitational Acceleration
- Mass and Weight
- Gravitational Waves
- Free Fall
- Gravitational Potential Energy
- Escape Velocity
- Weightlessness in Space
Periodic Classification of Elements
- Classification of Elements
- Dobereiner’s Triads
- Newland's Law of Octaves
- Mendeleev’s Periodic Table
- Insights into Mendeleev’s Periodic Table
- Modern Periodic Law
- The Modern Periodic Table
- Structure of the Modern Periodic Table
- Modern Periodic Table and Electronic Configuration of Elements
- Groups and Electronic Configuration
- Periods and Electronic Configuration
- Periodic Trends in the Modern Periodic Table
- Atomic Size
- Metallic and Non-metallic Characters
- Gradation in Halogen Family
Chemical Reactions and Equations
- Chemical Reaction
- Chemical Equations
- Balancing Chemical Equation
- Types of Chemical Reactions > Combination Reaction
- Types of Chemical Reactions > Decomposition Reaction
- Types of Chemical Reactions > Single Displacement Reaction
- Types of Chemical Reactions > Double Displacement Reaction
- Endothermic and Exothermic Processes
- Rate of Chemical Reaction
- Factors Affecting the Rate of a Chemical Reaction
- Chemical Properties of Carbon Compounds > Oxidation
- Chemical Properties of Carbon Compounds > Reduction
- Corrosion of Metals
- Rancidity
Effects of Electric Current
- Electric Circuit
- Heating Effect of Electric Current
- Magnetic Effect of Electric Current
- Right-hand Thumb Rule
- Applications of Biot-Savart's Law > Magnetic Field at the Centre of a Circular Loop
- Applications of Ampere’s Circuital Law > Magnetic Field of a Long Straight Solenoid
- Force on a Current Carrying Conductor in a Magnetic Field
- Fleming’s Left Hand Rule
- Electric Motor
- Electromagnetic Induction
- Galvanometer
- Faraday's Laws of Electromagnetic Induction
- Fleming’s Right Hand Rule
- Alternating current (AC) and Direct Current (DC)
- Electric Generator
Heat
Refraction of Light
Lenses
- Concept of Lenses
- Images Formed by Convex Lenses
- Images Formed by Concave Lenses
- Sign Convention
- Lens Formula
- Magnification
- Power of a Lens
- Combination of Lenses
- The Human Eye
- Defects of Vision and Their Corrections > Myopia
- Defects of Vision and Their Corrections > Hypermetropia
- Defects of Vision and Their Corrections > Presbyopia
- Apparent Size of an Object
- Use of Concave Lenses
- Use of Convex Lenses
- Persistence of Vision
Metallurgy
- Physical Properties of Metals
- Physical Properties of Non-metal
- Chemical Properties of Metal
- Reactions of Metals
- Reactivity Series of Metals
- Chemical Properties of Non-metal
- Ionic Compounds
- Metallurgy
- Basic Principles of Metallurgy > Concentration of Ores
- Basic Principles of Metallurgy > Extraction of Metals
- Basic Principles of Metallurgy > Refining of Metals
- Corrosion of Metals
- Prevention of Corrosion
Carbon Compounds
- Bonds in Carbon Compounds
- Carbon: A Versatile Element
- Hydrocarbons
- Straight chains, Branched chains, and Rings of Carbon atoms
- Functional Groups in Carbon Compounds
- Homologous Series
- Nomenclature
- Chemical Properties of Carbon Compounds > Combustion
- Chemical Properties of Carbon Compounds > Oxidation
- Chemical Properties of Carbon Compounds > Addition Reaction
- Chemical Properties of Carbon Compounds > Substitution Reaction
- Ethanol
- Ethanoic Acid
- Macromolecules and Polymers
Space Missions
School of Elements
The Magic of Chemical Reactions
- Chemical Equations
- Types of Chemical Reactions > Combination Reaction
- Types of Chemical Reactions > Decomposition Reaction
- Types of Chemical Reactions > Single Displacement Reaction
- Types of Chemical Reactions > Double Displacement Reaction
- Chemical Properties of Carbon Compounds > Oxidation
- Types of Double Displacement: Neutralization Reaction
The Acid Base Chemistry
- Properties of Acids > Physical Properties
- The pH Scale
- Acids, Bases and Their Reactivity
- Acid or a Base in a Water Solution
- Preparation and Uses of Baking Soda
- Preparation and Uses of Bleaching Powder
- Preparation and Uses of Washing Soda
- Preparation and Uses of Plaster of Paris
- Chemicals from Common Salt - Soap as a Salt
The Electric Spark
All about Electromagnetism
- Magnetic force
- The Bar Magnet
- Right-hand Thumb Rule
- Applications of Biot-Savart's Law > Magnetic Field at the Centre of a Circular Loop
- Applications of Ampere’s Circuital Law > Magnetic Field of a Long Straight Solenoid
- Force on a Current Carrying Conductor in a Magnetic Field
- Electric Motor
- Electromagnetic Induction
- A.C. Generator
- Simple D.C. Motor
- Household Electrical Circuits
Wonders of Light 1
Wonders of Light 2
Striving for better Environment 1
- Abatement of Pollution
- Sustainable Use of Resources
- Introduction
- History/Origin
- Definition: Universal Law of Gravitation
- Formula: Universal Law of Gravitation
- Key Points: Newton's Universal Law of Gravitation
- Characteristics
- Relationship to the Acceleration of the Moon
- Generalisation to Force
- Force Due to the Collection of Masses
- Special Cases for Extended Objects
- Significance
- Example 1
- Example 2
- Real-Life Examples
Maharashtra State Board: Class 10, 11
Definition: Universal Law of Gravitation
"Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."
Maharashtra State Board: Class 10, 11
Formula: Universal Law of Gravitation
The gravitational force of attraction (F) between two bodies of mass m1 and m2 separated by a distance r is:
\[\mathbf{F} = \mathbf{G}\frac{m_1 m_2}{r^2}\]
-
F: Gravitational Force of attraction (in Newtons, N).
-
\[m_1, m_2\]: Masses of the two objects (in kilograms, kg).
-
r (or d in the first part): Distance between the two objects (in meters, m).
-
G: The constant of proportionality, called the Universal gravitational constant.
-
Value in SI units: \[G=6.67\times10^{-11}\mathrm{N}\cdot\mathrm{m}^2/\mathrm{kg}^2\]
-
Dimensions: \[[G]=[\mathrm{L}^3\mathrm{M}^{-1}\mathrm{T}^{-2}]\]
-
Maharashtra State Board: Class 10
Key Points: Newton's Universal Law of Gravitation
Maharashtra State Board: Class 11
Introduction
Newton's Universal Law of Gravitation is one of the most important laws in physics. Every object in the Universe attracts every other object with a definite force. This force depends on two things: the masses of the objects and the distance between them. The law explains why objects fall to Earth and why planets move around the Sun. This law was developed by Newton after studying the work of famous scientists like Kepler and Galileo. Understanding this law helps us understand how gravity works everywhere in the universe.
Maharashtra State Board: Class 11
History/Origin
- 1564-1642: Galileo discovered that heavy and light objects fall to Earth with the same acceleration when released from the same height.
- 1571-1630: Johannes Kepler, a German astronomer and mathematician, worked as a helper to Tycho Brahe in Prague starting in 1600. After Brahe's sudden death in 1601, Kepler became the Royal mathematician.
- Kepler's Contribution: Kepler used observations of planetary positions made by Brahe to discover the laws of planetary motion. His work was later used by Newton in postulating the law of gravitation.
- 1665: Newton studied the motion of the Moon around the Earth. He knew the Moon completes one revolution about the Earth in 27.3 days and is at a distance of 3.85 × 10⁵ km.
- Newton's Achievement: Newton combined the ideas of Galileo and Kepler with his own laws of motion to formulate the Universal Law of Gravitation.
Maharashtra State Board: Class 11
Characteristics
- Nature of Force: It is an attractive force.
- Action-Reaction Pair: The forces exerted by the two bodies on each other have the same magnitude but opposite directions (obeying Newton's Third Law).
- Line of Action: The force always acts along the line joining the centres (or centres of mass) of the two bodies.
- Universality: The law applies to all material objects in the universe.
- Impact of Mass: If the mass of one object is doubled, the force between the two objects doubles.
- Impact of Distance: If the distance between the two objects is doubled, the force decreases by a factor of 4 (due to the inverse square relationship).
Maharashtra State Board: Class 11
Relationship to Acceleration of the Moon
1. Moon's Centripetal Acceleration: Newton studied the Moon's motion around the Earth, which is almost circular. The centripetal force required is F = mrω², leading to an acceleration of a = rω².
- Using the period T (27.3 days) and distance r (\[3.85 \times 10^5\] km), the Moon's acceleration (\[a_{moon}\]) towards the Earth was calculated:
\[a = r \left(\frac{2\pi}{T}\right)^2 \approx 0.0027 \, \text{m/s}^2\]
2. Comparison of Accelerations: This acceleration (0.0027 m/s²) is much smaller than the acceleration of objects near the Earth's surface (aobject ≈ 9.8 m/s²).
3. Inverse Square Relationship: By comparing the ratio of accelerations and the ratio of distances:
\[\frac{a_{object}}{a_{moon}} \approx 3600 \quad \text{and} \quad \frac{\text{distance of moon}}{\text{distance of object}} \approx 60\]
It was found that:
\[\frac{a_{object}}{a_{moon}} = \left[\frac{\text{distance of moon}}{\text{distance of object}}\right]^2\]
4. Conclusion: Newton concluded that the acceleration of an object towards the Earth is inversely proportional to the square of its distance from the center of the Earth (\[\mathbf{a}\propto1/\mathbf{r}^2\]).
Maharashtra State Board: Class 11
Generalization to Force
- Force on an Object: Since F = ma, the force exerted by the Earth on an object of mass m at a distance r is F ∝ m/r².
- Newton's Third Law: The object also exerts an equal and opposite force on the Earth (FE ∝ M/r²).
- Final Generalization: By combining these, Newton concluded that the gravitational force between the Earth (Mass M) and an object (Mass m) is F ∝ Mm/r².
- Universal Law: This was generalized to the force between any two objects (m1 and m2) as F ∝ m1m2/r².
Maharashtra State Board: Class 11
Force Due to Collection of Masses (Superposition Principle)
- For a collection of point masses, the force on any one mass is the vector sum of the gravitational forces exerted by all the other point masses.
- For n particles, the force on the ith mass (\[\vec F_i\]) is:
\[F_i=\sum_{
\begin{array}
{l}j=1 \\
j\neq i
\end{array}}^nF_{ij}\]
Maharashtra State Board: Class 11
Special Cases for Extended Objects
The source provides two special cases for gravitational force involving extended objects:
- Force inside a hollow spherical shell: The gravitational force of attraction due to a hollow, thin spherical shell of uniform density, on a point mass situated inside it is zero.
- Force outside a hollow shell or solid sphere: The gravitational force between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside is just as if the entire mass is concentrated at the center of the shell or sphere.
Maharashtra State Board: Class 11
Significance
- It led to an explanation of terrestrial gravitation (why things fall on Earth).
- It explains Kepler’s laws of planetary motion.
- It provides the reason behind the observed motion of planets around the Sun.
- It explains the force that keeps the Moon in its circular orbit around the Earth.
Maharashtra State Board: Class 11
Example 1
Problem: The gravitational force between two bodies is 1 N. If the distance between them is doubled, what will be the gravitational force between them?
Solution:
Let the masses of the two bodies be m₁ and m₂, and the distance between them be r.
F₁ = \[\frac{Gm_1m_2}{r^2}\]
When distance is doubled (new distance = 2r): F₂ = \[\frac{Gm_1m_2}{(2r)^2}=\frac{Gm_1m_2}{4r^2}\].
\[\frac{F_1}{F_2}=\frac{\frac{Gm_1m_2}{r^2}}{\frac{Gm_1m_2}{4r^2}}=4\]
F₂ = \[\frac {F_1}{4}\]
Since F₁ = 1 N:
F₂ = \[\frac {1}{4}\] = 0.25 N
Answer: When the distance is doubled, the gravitational force becomes one-fourth (0.25 N).
Maharashtra State Board: Class 11
Example 2
Problem: Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = l. A fourth particle D is kept on the perpendicular bisector of AC at a distance l from B. Determine the gravitational force on D.
Solution:
First, we need to find the distances AD and CD:
CD = AD = \[\sqrt{(AB)^2+(BD)^2}=\sqrt{l^2+l^2}=\sqrt{2l^2}=\sqrt{2}\mathbf{l}\]
Force due to A on D:
\[F_{DA}=\frac{Gmm}{(\sqrt{2}l)^2}=\frac{Gm^2}{2l^2}\]
Force due to C on D:
\[\begin{aligned}
F_{DC}=\frac{Gmm}{(\sqrt{2}l)^2}=\frac{Gm^2}{2l^2}
\end{aligned}\]
Force due to B on D:
\[F_{DB}=\frac{Gmm}{(l)^2}=\frac{Gm^2}{l^2}\]
Let the unit vector along the horizontal direction (AC) be î and along the vertical direction (BD) be ĵ.
Horizontal Component (along î direction):
- The force \[\vec F_{DB}\] is purely vertical, so its horizontal component is zero (cos 90° = 0).
- The triangle ABD is a right isosceles triangle, so the angle ∠DAB = 45°.
- FDA and FDC have equal horizontal components, but they are in opposite directions (canceling each other out).
-
\[\text{Net Horizontal Force} = F_{DC} \cos 45^\circ - F_{DA} \cos 45^\circ = 0\]
Vertical Component (along ĵ direction):
From A: \[\frac{Gm^2}{2l^2}\times\frac{1}{\sqrt{2}}=\frac{Gm^2}{2\sqrt{2}l^2}\]
From B: \[\frac{Gm^{2}}{2l^{2}}\times\frac{1}{\sqrt{2}}=\frac{Gm^{2}}{2\sqrt{2}l^{2}}\]
From C: \[\frac{Gm^2}{l^2}\]
Net Vertical Force:
= \[\frac{Gm^{2}}{2\sqrt{2}l^{2}}+\frac{Gm^{2}}{l^{2}}+\frac{Gm^{2}}{2\sqrt{2}l^{2}}\]
= \[\frac{Gm^2}{l^2}\left[\frac{1}{2\sqrt{2}}+1+\frac{1}{2\sqrt{2}}\right]\]
= \[\frac{Gm^2}{l^2}\left[\frac{2}{2\sqrt{2}}+1\right]=\frac{Gm^2}{l^2}\left[\frac{1}{\sqrt{2}}+1\right]\]
Final Answer:
The net gravitational force on D is \[{\frac{Gm^{2}}{l^{2}}}\left(1+{\frac{1}{\sqrt{2}}}\right)\] directed along DB (toward B).
Maharashtra State Board: Class 11
Real-Life Examples
- Tides: The gravitational force exerted by the Moon and the Sun causes the ocean tides on Earth.
- Satellite Launch: This force is used to calculate the required escape velocity for launching rockets and satellites into Earth's orbit.
- The Solar System: Gravitation is the force that holds the planets, asteroids, and comets in their orbits around the Sun, structuring the entire solar system.
- Walking/Standing: It is the force that keeps us grounded and gives us weight.
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