Welcome to our detailed guide on Class 9 Science Chapter 10: Gravitation! Whether you’re a Class 9 student gearing up for your 2025 CBSE exams, a parent helping your child understand Physics, or a teacher seeking in-depth resources, this blog post is your go-to solution. We’ll explore the in-text questions and answers, exercise solutions, and detailed notes for Chapter 10, breaking down the concepts of gravity, free fall, and weight into clear, comprehensive explanations. By the end, you’ll have a thorough grasp of how gravitation shapes our world—equipping you to excel in your exams and solidify your Physics knowledge!
Chapter 10 delves into the invisible force of gravitation, explaining why objects fall, planets orbit, and tides rise. Let’s dive into the Class 9 Science Chapter 10 questions, answers, and notes!
Class 9 Science Chapter 10 Notes
1. Gravitation
- Definition: The force of attraction between any two masses in the universe.
- Universal Law (Newton): Every object attracts every other object with a force proportional to their masses and inversely proportional to the square of the distance between them:
- F = G × (m₁m₂/r²), where G = gravitational constant (6.67 × 10⁻¹¹ Nm²/kg²).
2. Gravity
- Definition: Gravitation specific to Earth, pulling objects toward its center.
- Acceleration due to Gravity (g): 9.8 m/s² on Earth’s surface, varies with altitude and depth.
3. Free Fall
- Definition: Motion under gravity alone, no other forces (e.g., air resistance negligible).
- Equations of Motion: Same as uniform acceleration, with a = g (e.g., v = u + gt).
4. Mass and Weight
- Mass: Amount of matter in an object (scalar, kg).
- Weight: Force due to gravity (W = mg, vector, N).
- Variation: Weight changes with g (e.g., less on Moon, g = 1/6th Earth’s).
5. Thrust and Pressure
- Thrust: Force perpendicular to a surface (e.g., pushing a nail).
- Pressure: Force per unit area (P = F/A, unit: Pascal, Pa).
- Buoyancy: Upward force by fluids on submerged objects, depends on displaced fluid’s weight.
6. Archimedes’ Principle
- An object in a fluid experiences an upward force equal to the weight of the fluid displaced.
- Applications: Floating ships, submarines.
These notes encapsulate the chapter’s essentials—use them as a foundation for the Q&As!
In-Text Questions and Answers: Page 141
Question 1: Provide a comprehensive explanation of why the value of the acceleration due to gravity varies across different locations on Earth, including the factors that influence this variation and examples of how it differs in specific regions or conditions.
Answer:
The value of the acceleration due to gravity (g), typically 9.8 m/s² on Earth’s surface, varies across different locations due to several factors tied to Earth’s shape, rotation, and composition, affecting how strongly gravity pulls objects downward. Firstly, Earth isn’t a perfect sphere but an oblate spheroid, flattened at the poles and bulging at the equator. This means the distance from the center of Earth (radius, r) is greater at the equator than at the poles. Since gravitational force follows Newton’s law (F = G × m₁m₂/r²), a larger r at the equator reduces g slightly—about 9.78 m/s²—compared to 9.83 m/s² at the poles, where r is smaller.
Secondly, Earth’s rotation introduces a centrifugal effect, strongest at the equator, pushing objects outward and effectively reducing the net gravitational pull, further lowering g there. At the poles, with no rotational effect, g is higher.
Thirdly, altitude impacts g: as you climb a mountain (e.g., Mount Everest), r increases, so g drops marginally (e.g., 9.77 m/s² at high altitudes). Depth also matters—inside a mine, closer to Earth’s dense core, g slightly increases, but beyond a point, it decreases due to less mass below. Local geology, like dense rock formations, can also tweak g minutely. For instance, g on the Moon is 1.62 m/s² (1/6th Earth’s) due to its smaller mass and radius. These variations—though small on Earth—matter in precise measurements, like satellite orbits or sports at different altitudes, showing how gravity adapts to Earth’s dynamic structure and motion.
Question 2: Discuss in detail how the gravitational force between two objects changes when the distance between them is altered, including the mathematical relationship, practical implications, and examples from both terrestrial and astronomical contexts.
Answer:
The gravitational force between two objects changes with distance according to Newton’s Universal Law of Gravitation: F = G × (m₁m₂/r²), where F is the force, m₁ and m₂ are the masses, r is the distance between their centers, and G is the gravitational constant. This inverse-square relationship means that if the distance doubles (r becomes 2r), the force drops to 1/4th (1/(2²)) of its original value; if r triples, F falls to 1/9th (1/(3²)). Conversely, halving the distance increases F fourfold. The masses remain constant, so distance is the key variable.
Practically, this has profound implications. On Earth, consider a 1 kg rock held 1 m above the ground—its weight (gravitational force) is mg ≈ 9.8 N. Lift it to 2 m, and the force barely changes because r (Earth’s radius, ~6,371 km) is so large that adding 1 m is negligible.
But in space, it’s dramatic: Earth and Moon, about 384,400 km apart, exert a force keeping the Moon in orbit. If this distance halved, the force would quadruple, potentially disrupting orbits. Astronomically, the Sun’s pull on Earth (150 million km away) weakens as distance squares—doubling it would slash the force, altering Earth’s orbit and climate. In daily life, this explains why gravity feels constant near Earth’s surface but weakens significantly for satellites farther out. This law governs everything from falling apples to planetary motion, showing distance’s outsized role in gravitational interactions.
In-Text Questions and Answers: Page 148
Question 1: Explain extensively why the weight of an object on the Moon is significantly less than its weight on Earth, incorporating the differences in gravitational acceleration, the resulting calculations, and real-world examples of this effect.
Answer:
The weight of an object on the Moon is significantly less than on Earth because weight (W = mg) depends on the acceleration due to gravity (g), which is much weaker on the Moon due to its smaller mass and size. Earth’s g is 9.8 m/s², driven by its mass (5.97 × 10²⁴ kg) and radius (~6,371 km). The Moon, with a mass of 7.34 × 10²² kg (1/81st of Earth’s) and radius of ~1,738 km, has a g of about 1.62 m/s²—roughly 1/6th of Earth’s.
This stems from Newton’s law (F = G × m₁m₂/r²): the Moon’s lower mass reduces the gravitational pull, and its smaller radius increases it slightly, but the mass effect dominates. For a 60 kg person, weight on Earth is 60 × 9.8 = 588 N.
On the Moon, it’s 60 × 1.62 = 97.2 N—about 1/6th as much. This calculation holds for any object: a 1 kg rock weighs 9.8 N on Earth but 1.62 N on the Moon. Real-world examples include astronauts during the Apollo missions—they bounced easily, lifting heavy gear with less effort, because their weight (not mass) dropped. A lunar rover, weighing 210 kg on Earth (2,058 N), weighs just 343 N on the Moon, simplifying mobility. This reduced weight affects design—Moon vehicles need less power than Earth’s. The difference highlights how gravity, tied to celestial body properties, dictates weight, making lunar exploration distinct from terrestrial experiences.
Question 2: Provide a detailed analysis of why a sheet of paper falls more slowly than a coin under normal conditions, and explain how this behavior changes when air resistance is minimized, with examples and underlying principles.
Answer:
A sheet of paper falls more slowly than a coin under normal conditions because of air resistance, which opposes motion and depends on an object’s shape, size, and surface area, while gravity pulls both equally. Both objects experience the same gravitational acceleration (g = 9.8 m/s²), so in a vacuum, they’d fall at the same rate (free fall). However, in air, the paper’s larger surface area encounters more air molecules, creating greater drag that counteracts gravity. The coin, being small and dense, cuts through air with minimal resistance, falling faster.
For a 0.01 kg paper and 0.01 kg coin, weight is 0.098 N each, but the paper’s flat shape increases air friction, slowing its descent—say, taking 5 seconds to fall 5 m versus the coin’s 1 second. This drag is proportional to speed and area, not mass, so the lightweight paper is more affected.
When air resistance is minimized (e.g., in a vacuum chamber), both fall identically, as shown in a famous experiment by astronaut David Scott on the Moon in 1971: a feather and hammer dropped together hit the surface simultaneously, with no air to interfere. On Earth, crumpling the paper into a ball reduces its area, making it fall nearly as fast as the coin (e.g., 1.2 s vs. 1 s for 5 m). This illustrates Galileo’s principle—all objects fall equally under gravity alone—but air resistance, a real-world factor, alters timing based on shape, not weight, until removed.
Exercise Questions and Answers: Page 163
Question 1: Discuss comprehensively why an object floats or sinks when placed in a liquid, including the role of buoyant force, density relationships, and practical examples from everyday life and engineering.
Answer:
An object floats or sinks in a liquid based on the balance between its weight (downward gravitational force, W = mg) and the buoyant force (upward force from the liquid), governed by Archimedes’ Principle: the buoyant force equals the weight of the liquid displaced. If buoyant force exceeds or equals weight, the object floats; if weight exceeds buoyant force, it sinks. This depends on density (mass/volume). An object denser than the liquid (e.g., a steel nail, density 7.8 g/cm³ vs. water’s 1 g/cm³) displaces less water by weight than its own (W > F_b), so it sinks. A less dense object (e.g., wood, 0.8 g/cm³) displaces water weighing more than itself, floating (W < F_b). For a 1 kg wooden block (volume 0.00125 m³), weight = 9.8 N; it displaces 0.00125 m³ of water (1.25 kg, 12.25 N), so F_b > W, and it floats.
A 1 kg steel block (volume 0.000128 m³) displaces only 0.128 kg of water (1.25 N), so W > F_b, and it sinks. Everyday examples: ice (0.92 g/cm³) floats on water, while a coin sinks. Engineering uses this—ships float by displacing vast water volumes with hollow hulls, while submarines adjust density with ballast tanks. Neutral buoyancy (W = F_b) occurs in fish with swim bladders. Density and displacement dictate flotation, making this principle vital for navigation, design, and nature.
Question 2: The volume of a 500 g sealed packet is 350 cm³. Will it float or sink in water, given that the density of water is 1 g/cm³? Calculate in detail and explain what will happen if the packet’s mass increases to 700 g, including all steps and reasoning.
Answer:
To determine if a 500 g sealed packet with a volume of 350 cm³ floats or sinks in water (density 1 g/cm³), we compare its density to water’s and analyze buoyant force versus weight.
- Density of packet: Density = mass/volume = 500 g / 350 cm³ = 1.428 g/cm³. Since 1.428 > 1, the packet is denser than water.
- Weight: W = mg = 0.5 kg × 9.8 m/s² = 4.9 N (convert 500 g to 0.5 kg).
- Buoyant force: Volume = 350 cm³ = 0.00035 m³. Water displaced = 0.00035 m³ × 1000 kg/m³ = 0.35 kg. F_b = 0.35 × 9.8 = 3.43 N.
- Comparison: W (4.9 N) > F_b (3.43 N), so the packet sinks.
Now, if mass increases to 700 g: - New density: 700 g / 350 cm³ = 2 g/cm³, still > 1 g/cm³.
- New weight: 0.7 kg × 9.8 = 6.86 N.
- Buoyant force: Remains 3.43 N (volume unchanged).
- Comparison: W (6.86 N) > F_b (3.43 N), so it still sinks, and faster, as the net downward force (6.86 – 3.43 = 3.43 N) is greater than before (4.9 – 3.43 = 1.47 N). The packet’s fixed volume can’t displace enough water to match its growing weight, confirming sinking in both cases, with increased mass amplifying the effect.
In The End
Class 9 Science Chapter 10: Gravitation unveils the force that binds our universe, from falling objects to orbiting planets, preparing you for your 2025 CBSE exams. With these detailed notes, extended questions, and solutions, you’re ready to master gravitation’s principles. This chapter blends theory with real-world applications, making it both fascinating and essential.
For more practice, explore NCERT solutions and sample papers online. Have questions? Leave them in the comments—we’re here to help! Keep studying and stay curious!
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