Grade 10 Q&A: Chapter 1: Gravitation

Concept Questions

Q1: What is gravitation?

Answer: Gravitation is the universal force of attraction acting between all matter. According to Newton's Law of Universal Gravitation, every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Q2: State Newton's Law of Universal Gravitation.

Answer: Newton's Law of Universal Gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, it is expressed as F = G(m₁m₂/r²), where F is the gravitational force, G is the universal gravitational constant, m₁ and m₂ are the masses of the two objects, and r is the distance between their centers.

Q3: What is the value of the Universal Gravitational Constant (G)?

Answer: The value of the Universal Gravitational Constant (G) is 6.67 × 10⁻¹¹ N·m²/kg².

Q4: What is the force of gravitation between an object and the Earth known as?

Answer: The force of gravitation between an object and the Earth is known as the weight of the object.

Q5: Define acceleration due to gravity.

Answer: Acceleration due to gravity is the acceleration experienced by an object due to the gravitational force of the Earth. Near the Earth's surface, its value is approximately 9.8 m/s² directed towards the center of the Earth.

Q6: How does the value of 'g' vary with altitude and depth?

Answer: - With increasing altitude above the Earth's surface, the value of 'g' decreases because the distance from the Earth's center increases. - With increasing depth below the Earth's surface, the value of 'g' decreases because the mass of the Earth above the object decreases.

Q7: What is the relationship between 'G' and 'g'?

Answer: The relationship between the Universal Gravitational Constant (G) and acceleration due to gravity (g) is given by g = G × M/R², where M is the mass of the Earth and R is the radius of the Earth.

Q8: What is free fall?

Answer: Free fall is the motion of an object when it is subjected only to the gravitational force of the Earth and no other forces (like air resistance) act on it. During free fall, an object accelerates towards the Earth at a rate equal to the acceleration due to gravity (g).

Q9: What is the escape velocity of Earth?

Answer: The escape velocity of Earth is approximately 11.2 km/s. It is the minimum velocity needed for an object to escape the gravitational pull of the Earth without further propulsion.

Q10: What are Kepler's laws of planetary motion?

Answer: Kepler's laws of planetary motion are: 1. First Law (Law of Elliptical Orbits): All planets move in elliptical orbits with the Sun at one focus. 2. Second Law (Law of Equal Areas): A line joining a planet to the Sun sweeps out equal areas in equal intervals of time. 3. Third Law (Law of Periods): The square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

Numerical Problems

Q11: Calculate the gravitational force between two objects of masses 10 kg and 20 kg placed at a distance of 2 meters from each other.

Answer: Given: - Mass of first object (m₁) = 10 kg - Mass of second object (m₂) = 20 kg - Distance between objects (r) = 2 m - Universal Gravitational Constant (G) = 6.67 × 10⁻¹¹ N·m²/kg²

Using the formula F = G(m₁m₂/r²): F = 6.67 × 10⁻¹¹ × (10 × 20)/(2²) F = 6.67 × 10⁻¹¹ × 200/4 F = 6.67 × 10⁻¹¹ × 50 F = 3.335 × 10⁻⁹ N

Therefore, the gravitational force between the two objects is 3.335 × 10⁻⁹ N.

Q12: Calculate the weight of a person with a mass of 60 kg on Earth and on the Moon (g on Moon = 1.625 m/s²).

Answer: On Earth: - Mass of person (m) = 60 kg - Acceleration due to gravity on Earth (g) = 9.8 m/s² - Weight = m × g = 60 × 9.8 = 588 N

On Moon: - Mass of person (m) = 60 kg - Acceleration due to gravity on Moon (g) = 1.625 m/s² - Weight = m × g = 60 × 1.625 = 97.5 N

Therefore, the person weighs 588 N on Earth and 97.5 N on the Moon.

Q13: A satellite orbits the Earth at a height of 3600 km above the Earth's surface. If the radius of the Earth is 6400 km, calculate the orbital velocity of the satellite.

Answer: Given: - Height of satellite above Earth's surface (h) = 3600 km - Radius of Earth (R) = 6400 km - Total distance from Earth's center (r) = R + h = 6400 + 3600 = 10000 km = 10 × 10⁶ m - Mass of Earth (M) = 5.97 × 10²⁴ kg - Universal Gravitational Constant (G) = 6.67 × 10⁻¹¹ N·m²/kg²

Using the formula for orbital velocity: v = √(GM/r) v = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴/10 × 10⁶) v = √(6.67 × 5.97 × 10¹³/10) v = √(3.98 × 10¹³) v = 6.31 × 10⁶ m/s = 6.31 km/s

Therefore, the orbital velocity of the satellite is approximately 6.31 km/s.

Q14: Calculate the acceleration due to gravity at a height of 3200 km above the Earth's surface. (Radius of Earth = 6400 km)

Answer: Given: - Height above Earth's surface (h) = 3200 km - Radius of Earth (R) = 6400 km - Total distance from Earth's center (r) = R + h = 6400 + 3200 = 9600 km - Acceleration due to gravity at Earth's surface (g₀) = 9.8 m/s²

Using the formula g = g₀ × (R/r)²: g = 9.8 × (6400/9600)² g = 9.8 × (2/3)² g = 9.8 × 4/9 g = 9.8 × 0.444 g = 4.35 m/s²

Therefore, the acceleration due to gravity at a height of 3200 km above the Earth's surface is approximately 4.35 m/s².

Q15: Calculate the time period of a satellite orbiting the Earth at a height of 35800 km above the Earth's surface. (Radius of Earth = 6400 km)

Answer: Given: - Height of satellite above Earth's surface (h) = 35800 km - Radius of Earth (R) = 6400 km - Total distance from Earth's center (r) = R + h = 6400 + 35800 = 42200 km = 42.2 × 10⁶ m - Mass of Earth (M) = 5.97 × 10²⁴ kg - Universal Gravitational Constant (G) = 6.67 × 10⁻¹¹ N·m²/kg²

Using the formula for time period: T = 2π × √(r³/GM) T = 2π × √((42.2 × 10⁶)³/(6.67 × 10⁻¹¹ × 5.97 × 10²⁴)) T = 2π × √((75.1 × 10¹⁸)/(3.98 × 10¹⁴)) T = 2π × √(1.89 × 10⁴) T = 2π × 137.5 T = 864 seconds = 86400 seconds = 24 hours

Therefore, the time period of the satellite is 24 hours, making it a geostationary satellite.

Application-Based Questions

Q16: Why do astronauts feel weightlessness in space?

Answer: Astronauts feel weightlessness in space because both they and their spacecraft are in a state of free fall around the Earth. In this situation, the astronaut and the spacecraft fall at the same rate, so the astronaut doesn't exert any force on the floor of the spacecraft. This creates the sensation of weightlessness, even though the astronaut is still under the influence of Earth's gravity.

Q17: Why does the Moon revolve around the Earth and not fall onto it?

Answer: The Moon revolves around the Earth and doesn't fall onto it because it has sufficient tangential velocity to maintain its orbit. The gravitational force between the Earth and the Moon provides the necessary centripetal force to keep the Moon in its orbital path. If the Moon were to stop moving, it would indeed fall towards the Earth due to gravitational attraction.

Q18: How does the concept of gravitation explain tides on Earth?

Answer: Tides on Earth are primarily caused by the gravitational pull of the Moon and, to a lesser extent, the Sun. The Moon's gravitational force is stronger on the side of the Earth facing it, causing the water to bulge outward. Similarly, on the opposite side of the Earth, the Moon's gravitational pull is weaker than at the Earth's center, causing another bulge. As the Earth rotates, different locations experience these bulges as high tides. The Sun's gravitational effect adds to or subtracts from the Moon's effect, causing spring tides (when Sun, Earth, and Moon are aligned) and neap tides (when the Sun and Moon are at right angles to each other).

Q19: Why is the value of 'g' different at the poles and at the equator?

Answer: The value of 'g' is different at the poles and at the equator due to two main factors: 1. Earth's shape: The Earth is not a perfect sphere but is slightly flattened at the poles and bulges at the equator. This means that the distance from the center of the Earth to the surface is less at the poles than at the equator. Since gravitational force decreases with distance, 'g' is greater at the poles. 2. Centrifugal effect: The Earth's rotation creates a centrifugal effect that counteracts gravity. This effect is maximum at the equator and zero at the poles. This further reduces the effective value of 'g' at the equator.

Due to these factors, the value of 'g' is approximately 9.83 m/s² at the poles and 9.78 m/s² at the equator.

Q20: How does the concept of escape velocity apply to space missions?

Answer: Escape velocity is the minimum velocity needed for an object to escape the gravitational pull of a celestial body without further propulsion. For Earth, this value is approximately 11.2 km/s. In space missions: 1. Launching satellites: To place a satellite in orbit, it must reach a velocity less than the escape velocity but sufficient to maintain its orbit. 2. Interplanetary missions: To send a spacecraft to another planet, it must achieve at least the escape velocity of Earth. 3. Fuel efficiency: Understanding escape velocity helps in calculating the minimum fuel required for a mission, making space travel more efficient. 4. Mission planning: Different celestial bodies have different escape velocities (e.g., Moon: 2.38 km/s, Mars: 5.03 km/s), which must be considered when planning return missions or multi-planet expeditions.

Higher-Order Thinking Questions

Q21: If the Earth's mass were to suddenly double, how would it affect the orbital motion of the Moon?

Answer: If the Earth's mass were to suddenly double: 1. Gravitational force: The gravitational force between the Earth and Moon would double (F = G × M × m/r², where M is Earth's mass). 2. Orbital velocity: To maintain a stable orbit with the increased gravitational force, the Moon would need a higher orbital velocity (v = √(GM/r)). 3. Orbital radius: Since the Moon's velocity cannot instantaneously change, the Moon would begin to spiral inward toward Earth. 4. Orbital period: The time taken for the Moon to complete one orbit would decrease (T = 2π × √(r³/GM)). 5. Tidal effects: Tidal forces on Earth would significantly increase, causing more extreme tides.

Eventually, the Moon would either establish a new, closer orbit around the Earth or, if it came too close, could break apart due to tidal forces (reaching the Roche limit).

Q22: Compare and contrast Newton's Law of Gravitation with Coulomb's Law of Electrostatic Force.

Answer:

Similarities: 1. Inverse square relationship: Both forces follow an inverse square relationship with distance (F ∝ 1/r²). 2. Proportional to properties: Both forces are proportional to the product of the intrinsic properties of the interacting objects (masses in gravitation, charges in electrostatics). 3. Central forces: Both are central forces, acting along the line joining the centers of the objects. 4. Conservative forces: Both are conservative forces, meaning the work done is independent of the path taken.

Differences: 1. Nature of force: Gravitational force is always attractive, while electrostatic force can be either attractive (between opposite charges) or repulsive (between like charges). 2. Strength: Electrostatic force is much stronger than gravitational force (approximately 10³⁶ times stronger for elementary particles). 3. Shielding: Electrostatic forces can be shielded by placing conducting materials between charged objects, while gravitational forces cannot be shielded. 4. Dependence: Gravitational force depends on mass, which is always positive, while electrostatic force depends on charge, which can be positive or negative. 5. Fundamental nature: Gravitation is one of the four fundamental forces of nature and is described by Einstein's General Theory of Relativity at a deeper level, while electrostatic force is a manifestation of the electromagnetic force, described by quantum electrodynamics.

Q23: Explain how Kepler's laws of planetary motion can be derived from Newton's Law of Universal Gravitation.

Answer: Newton's Law of Universal Gravitation provides the mathematical foundation from which Kepler's empirical laws can be derived:

Kepler's First Law (Law of Elliptical Orbits): - From Newton's law, the gravitational force between the Sun and a planet creates a central force field. - In a central force field where force varies as 1/r², the solution to the equations of motion yields a conic section (ellipse, parabola, or hyperbola). - For bound orbits (planets), the solution is an ellipse with the Sun at one focus.

Kepler's Second Law (Law of Equal Areas): - This law is a direct consequence of the conservation of angular momentum. - Since the gravitational force acts along the line joining the Sun and planet (central force), it creates no torque about the Sun. - With no torque, angular momentum is conserved, which mathematically leads to the equal areas in equal times principle.

Kepler's Third Law (Law of Periods): - For a planet in an elliptical orbit, the net force is the gravitational force: F = GMm/r² - This force provides the centripetal acceleration: F = mv²/r - For circular orbits (approximation), equating these gives: GMm/r² = mv²/r - Solving for v: v = √(GM/r) - The orbital period T = 2πr/v = 2πr/√(GM/r) = 2π√(r³/GM) - Therefore, T² = (4π²/GM)r³, which shows that T² ∝ r³, confirming Kepler's third law.

Q24: Discuss the implications of Einstein's General Theory of Relativity on our understanding of gravity compared to Newton's theory.

Answer: Einstein's General Theory of Relativity fundamentally changed our understanding of gravity compared to Newton's theory:

Conceptual Differences: 1. Nature of gravity: Newton viewed gravity as a force acting instantaneously between masses across space. Einstein reinterpreted gravity as the curvature of spacetime caused by mass and energy. 2. Propagation: In Newton's theory, gravitational effects are instantaneous. Einstein's theory predicts that gravitational effects propagate at the speed of light as gravitational waves. 3. Spacetime: Newton treated space and time as separate and absolute. Einstein unified them into a four-dimensional spacetime continuum that can be warped by mass and energy.

Predictions and Verifications: 1. Mercury's orbit: Einstein's theory correctly predicted the precession of Mercury's orbit, which Newton's theory could not fully explain. 2. Light bending: Einstein predicted that light would bend when passing near massive objects, confirmed during the 1919 solar eclipse. 3. Gravitational redshift: Einstein's theory predicted that light would lose energy (redshift) when escaping a gravitational field, now confirmed by precise atomic clocks. 4. Gravitational waves: Einstein predicted gravitational waves, which were directly detected by LIGO in 2015. 5. Black holes: Einstein's equations predicted the existence of black holes, which have now been observed through various astronomical techniques.

Practical Implications: 1. GPS systems: GPS satellites must account for relativistic effects to maintain accuracy. 2. Cosmology: Einstein's theory forms the foundation of modern cosmological models, including the Big Bang theory and the expansion of the universe. 3. Gravitational lensing: Used as a tool in astronomy to observe distant galaxies and measure dark matter.

Limitations: Despite its successes, General Relativity is incompatible with quantum mechanics, leading to the ongoing search for a theory of quantum gravity that would unify all fundamental forces.

Q25: How would the concept of weight and apparent weight change in an elevator that is accelerating upward or downward?

Answer: The concept of weight and apparent weight changes in an accelerating elevator due to the combination of gravitational force and inertial effects:

Case 1: Elevator at rest or moving with constant velocity - True weight = mg (where m is mass and g is acceleration due to gravity) - Apparent weight = True weight = mg - The normal force from the floor equals mg

Case 2: Elevator accelerating upward - True weight remains mg (gravitational force is unchanged) - Apparent weight = m(g + a) (where a is the upward acceleration) - The normal force from the floor increases to m(g + a) - Person feels heavier than normal

Case 3: Elevator accelerating downward - True weight remains mg - Apparent weight = m(g - a) (where a is the downward acceleration) - The normal force from the floor decreases to m(g - a) - Person feels lighter than normal

Case 4: Elevator in free fall (a = g) - True weight remains mg - Apparent weight = m(g - g) = 0 - The normal force from the floor becomes zero - Person experiences weightlessness

Case 5: Elevator accelerating downward faster than g - True weight remains mg - Apparent weight becomes negative - Person would be pushed against the ceiling - This situation is not normally encountered in everyday elevators

These variations in apparent weight demonstrate the principle of equivalence in Einstein's General Relativity, which states that the effects of gravity are indistinguishable from the effects of acceleration.

References

1. Maharashtra State Board 10th Standard Science Syllabus 2025-26
2. NCERT Science Textbook for Class 10
3. Balbharati Science and Technology Part 1 Textbook
4. Vedantu Maharashtra Board Class 10 Solutions for Science Chapter 1 Gravitation
5. Shaalaa.com Balbharati solutions for Science and Technology 1