Gravitation - 80 Marks Question Paper with Answer Key

📚 Gravitation Question Paper

Total Marks: 80

Time: 2.5 Hours

Class: IX - X Physics

Sections: A, B, C, D

📌 SECTION A: Very Short Answer Questions 12 Marks

Answer all questions in 1-2 sentences

Q1. Define gravitational force.

State the definition and mention the SI unit of gravitational force.

1.5 marks

✓ Answer:

Gravitational force is the attractive force exerted by one object on another. It acts along the line joining the centers of the two objects. The SI unit is Newton (N).

💡 Explanation:

Gravitation is a universal force that exists between all masses. Unlike other forces, it is always attractive and never repulsive. This is why all objects fall towards Earth.

Q2. What is the value of g?

State the value of acceleration due to gravity near Earth's surface and its SI unit.

1.5 marks

✓ Answer:

The acceleration due to gravity (g) near Earth's surface is approximately 9.8 m/s² (or 10 m/s² for calculations). The SI unit is meters per second squared (m/s²).

💡 Explanation:

This value is constant near Earth's surface. It varies slightly with latitude and altitude because Earth is not a perfect sphere and its density is not uniform. At the poles, g = 9.83 m/s², while at the equator, g = 9.78 m/s².

Q3. What is free fall?

Define free fall and give one example.

1.5 marks

✓ Answer:

Free fall is the motion of an object under the influence of gravity alone, with no other forces acting on it. Example: A ball dropped from a height falls freely under gravity.

💡 Explanation:

In true free fall, air resistance is neglected. All objects, regardless of their mass, fall with the same acceleration (g). This was famously demonstrated by Galileo, who dropped objects of different masses from the Leaning Tower of Pisa.

Q4. Distinguish between mass and weight.

Give two key differences between mass and weight.

2 marks

✓ Answer:

Mass	Weight
Quantity of matter in an object	Force of gravity on an object
Constant everywhere	Varies with location
Scalar quantity (kg)	Vector quantity (N)

💡 Explanation:

Mass is an intrinsic property of an object and does not depend on location. Weight, however, depends on the gravitational field strength where the object is located. An astronaut weighs less on the Moon but has the same mass as on Earth.

Q5. What is buoyancy?

Define buoyant force and state Archimedes' principle.

2 marks

✓ Answer:

Buoyant force (or upthrust) is the upward force exerted by a fluid on an object immersed in it.

Archimedes' Principle: When an object is fully or partially immersed in a fluid, it experiences an upward force equal to the weight of the fluid displaced by the object.

💡 Explanation:

This principle explains why objects float or sink. If the buoyant force equals the weight, the object floats. If weight is greater, the object sinks. Ships float because their overall density (including air inside) is less than water.

Q6. Why does the Moon not fall into Earth?

Explain using the concept of centripetal force.

2 marks

✓ Answer:

The Moon continuously falls towards Earth due to gravitational attraction. However, it also has a tangential velocity. The gravitational force provides the centripetal force required to keep the Moon in a circular orbit, causing it to fall around Earth rather than into it.

💡 Explanation:

The Moon is in a state of continuous circular motion. At every point in its orbit, gravity pulls it towards Earth's center (centripetal force), but its forward motion (tangential velocity) keeps it from falling straight in. It's a perfect balance between falling and forward motion.

Q7. Why do objects of different masses fall at the same rate?

Explain why air resistance is important in this context.

2 marks

✓ Answer:

All objects experience the same acceleration due to gravity (g = 9.8 m/s²), independent of their mass. This is because gravitational force is proportional to mass, so the ratio F/m remains constant. Air resistance, however, affects lighter objects more, causing them to fall slower in real conditions.

💡 Explanation:

Mathematically: F = mg, so a = F/m = mg/m = g (constant). In a vacuum (no air), a feather and a hammer fall at the same rate. In air, the feather falls slower because air resistance is proportional to surface area, and the feather has a larger surface-to-mass ratio.

Q8. Define pressure and state its formula.

Why do sharp objects cut more easily than blunt objects?

2 marks

✓ Answer:

Pressure = Force / Area
SI Unit: Pascal (Pa) or N/m²

Sharp objects cut more easily because the same force is concentrated on a smaller area, resulting in greater pressure (P = F/A). Higher pressure makes cutting easier.

💡 Explanation:

A sharp knife has a very small contact area, so even a small force creates high pressure. A blunt object distributes force over a larger area, creating lower pressure. This is why a drawing pin has a sharp point to pierce surfaces with your finger's force.

📖 SECTION B: Short Answer Questions 24 Marks

Answer in 2-3 sentences with explanation

Q9. State and explain Newton's Universal Law of Gravitation.

Include the mathematical form of the law.

3 marks

✓ Answer:

Newton's Universal Law of Gravitation states: "Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."

Mathematical Form:

F = G(M × m)/d²
Where: F = Gravitational force
G = Universal gravitational constant (6.67 × 10⁻¹¹ N·m²/kg²)
M, m = Masses of objects
d = Distance between centers

💡 Explanation:

This law is universal because it applies to all objects in the universe, from apples to planets. If you double the mass of one object, the force doubles. If you double the distance, the force becomes 1/4 (inverse square law). This law successfully explains planetary motion, tides, and falling objects.

Q10. Derive the formula for weight on Earth.

Show how W = mg is derived from the universal law of gravitation.

3 marks

✓ Answer:

From Newton's Universal Law: F = G(M × m)/d²

For an object on Earth's surface:
- M = Mass of Earth
- d = Radius of Earth (R)
- The force is the weight (W)

Therefore: W = G(M × m)/R²

We can write: g = GM/R² (a constant)

Hence: W = mg

💡 Explanation:

Weight is simply the gravitational force on an object. By substituting the universal law with Earth's parameters, we get the familiar formula W = mg. This shows that weight is the product of mass and local gravitational acceleration. On the Moon, 'g' is smaller (1.6 m/s²), so weight is less, but mass remains the same.

Q11. Explain why weight changes on different celestial bodies.

Compare weight on Earth and Moon.

3 marks

✓ Answer:

Weight depends on the local value of 'g', which varies with mass and radius of celestial body.

Using g = GM/R²:
- Earth: g = 9.8 m/s²
- Moon: g = 1.6 m/s² (Moon has ~1/81 mass of Earth and smaller radius)

A 60 kg person weighs:
- On Earth: W = 60 × 9.8 = 588 N
- On Moon: W = 60 × 1.6 = 96 N (about 1/6 of Earth's weight)

💡 Explanation:

The Moon's smaller mass and radius result in weaker gravitational pull. Since g is proportional to M/R², and the Moon has both smaller mass and smaller radius, its gravitational acceleration is significantly less. This is why astronauts jump higher on the Moon—they weigh less, but their mass remains the same.

Q12. Calculate the gravitational force between Earth and the Moon.

Given: M_Earth = 6 × 10²⁴ kg, M_Moon = 7.4 × 10²² kg, d = 3.84 × 10⁸ m, G = 6.67 × 10⁻¹¹ N·m²/kg²

3 marks

✓ Answer:

Using F = G(M × m)/d²

F = (6.67 × 10⁻¹¹) × (6 × 10²⁴ × 7.4 × 10²²) / (3.84 × 10⁸)²

F = (6.67 × 10⁻¹¹) × (4.44 × 10⁴⁷) / (1.47 × 10¹⁷)

F = 2.02 × 10²⁰ N

The gravitational force between Earth and Moon is 2.02 × 10²⁰ N

💡 Explanation:

This enormous force keeps the Moon in orbit around Earth. By Newton's third law, the Earth exerts this same force on the Moon as the Moon exerts on Earth. Despite the huge distance (3.84 × 10⁸ m), the large masses result in a significant attractive force.

Q13. What happens when an object is thrown upward?

Describe the motion and acceleration throughout the journey.

3 marks

✓ Answer:

When an object is thrown upward:

Upward journey: Gravity acts downward, opposing motion. Velocity decreases continuously until it becomes zero at maximum height.

Downward journey: Gravity acts downward, in the direction of motion. Velocity increases continuously until the object reaches the ground.

Acceleration: Throughout the entire journey, acceleration = -g = -9.8 m/s² (constant, directed downward)

💡 Explanation:

Acceleration is constant and always directed downward (towards Earth). This is why the object slows down going up and speeds up coming down. At maximum height, velocity is zero but acceleration is NOT zero—it's still 9.8 m/s² downward. The object then accelerates downward due to this constant acceleration.

Q14. Explain why a camel can walk in deserts easily.

Relate your answer to pressure and force concepts.

3 marks

✓ Answer:

A camel can walk in the desert easily because its broad feet distribute the weight over a large area.

Using Pressure = Force / Area:
- Camel has large feet (large area)
- Despite its heavy weight (large force), the pressure on sand = Weight / Large Area = Small Pressure

Low pressure means the camel doesn't sink deep into the sand, making walking easier.

💡 Explanation:

In contrast, humans with pointed heels create high pressure (Force/Small Area), causing them to sink deep into sand. This principle applies everywhere: tank tracks are wide to reduce pressure; skis are long to distribute weight; building foundations are large to prevent sinking. The inverse relationship between pressure and area is crucial in design.

Q15. Describe why a ship made of steel floats on water.

Use the concept of density and buoyancy.

3 marks

✓ Answer:

Steel has higher density than water (ρ_steel ≈ 7800 kg/m³ vs ρ_water = 1000 kg/m³), so solid steel sinks. However, a ship is designed hollow with air inside. The overall density of the ship (including air) is less than water's density.

By Archimedes' principle: Buoyant force = Weight of water displaced

Since average density < water's density → Buoyant force > Weight → Ship floats

💡 Explanation:

The key is the average density of the entire structure, not just the material. A hollow sphere of steel floats because air inside reduces the overall density. When you fill the ship with water (it sinks), the average density increases above water's density. This principle is used in submarines—by adjusting water inside, they control density to float or sink.

Q16. A person feels lighter in a swimming pool. Explain why.

Discuss the forces acting on the person.

3 marks

✓ Answer:

When a person is immersed in water, two forces act:

1. Weight (W) = mg, acting downward
2. Buoyant force (F_b) = Weight of water displaced, acting upward

Apparent weight = W - F_b

Since F_b > 0, the apparent weight is less than actual weight, making the person feel lighter.

💡 Explanation:

The buoyant force depends on the volume of water displaced and its density. For a human body, this buoyant force can be significant, reducing the apparent weight. If the buoyant force equals the weight, the person floats effortlessly. The actual weight doesn't change, but the scale reading (which measures normal force) decreases because the buoyant force helps support the person.

📝 SECTION C: Long Answer Questions 32 Marks

Answer in detail with diagrams and calculations

Q17. Derive g = GM/R² and find the value of g on Earth's surface.

Show all derivation steps and calculations.

4 marks

✓ Answer:

Derivation:

From Newton's Universal Law of Gravitation:
F = G(M × m)/R²

From Newton's Second Law: F = ma

For an object falling on Earth: a = g (acceleration due to gravity)

Therefore: mg = G(M × m)/R²

Canceling m from both sides:

g = GM/R²

Calculating g on Earth:
G = 6.67 × 10⁻¹¹ N·m²/kg²
M = 6 × 10²⁴ kg (Mass of Earth)
R = 6.4 × 10⁶ m (Radius of Earth)

g = (6.67 × 10⁻¹¹ × 6 × 10²⁴) / (6.4 × 10⁶)²
g = (40.02 × 10¹³) / (40.96 × 10¹²)
g = 9.8 m/s²

💡 Explanation:

This derivation shows that gravitational acceleration depends on the celestial body's mass and radius. More massive objects and smaller radii produce stronger gravity. This explains why g varies on different planets. The formula also shows that g is independent of the falling object's mass—all objects experience the same acceleration.

Q18. Solve this problem step by step:

A ball is thrown vertically upward with a velocity of 20 m/s. Calculate: (a) Maximum height reached, (b) Time to reach maximum height, (c) Total time of flight. (g = 10 m/s²)

4 marks

✓ Answer:

Given:
u = 20 m/s (initial velocity)
v = 0 m/s (at maximum height)
g = 10 m/s² (taking downward as positive)

(a) Maximum Height:
Using v² = u² - 2gs (negative because gravity opposes motion)
0² = 20² - 2(10)s
0 = 400 - 20s
s = 20 m

(b) Time to Maximum Height:
Using v = u - gt
0 = 20 - 10t
t = 2 seconds

(c) Total Time of Flight:
By symmetry, time to come down = time to go up = 2s
Total time = 2 + 2 = 4 seconds

Or using: s = ut - ½gt²
0 = 20t - ½(10)t²
0 = 20t - 5t²
0 = t(20 - 5t)
t = 0 or t = 4 seconds

💡 Explanation:

The key to solving projectile problems is consistent sign conventions. Taking downward as positive, upward motion has negative acceleration. At maximum height, velocity becomes zero (turning point). The journey up takes 2 seconds; the identical downward journey takes another 2 seconds. This symmetry is a direct consequence of constant acceleration.

Q19. Explain tides and their causes.

Discuss how Moon and Sun cause tides on Earth.

4 marks

✓ Answer:

Cause of Tides:
Tides are caused by gravitational pull of the Moon and Sun on Earth's water bodies.

Moon's Effect (Primary):
- Moon pulls harder on the near side of Earth (closer)
- Pulls less on the far side (farther)
- This difference creates a bulge on the near side (high tide)
- A matching bulge forms on the far side (due to Earth being pulled more than water)
- Results in 2 high tides and 2 low tides per day

Sun's Effect (Secondary):
- Sun also creates tides but with smaller magnitude
- During full and new moon: Sun and Moon align → Spring tides (highest/lowest)
- During quarter moon: Sun and Moon perpendicular → Neap tides (moderate)

💡 Explanation:

Tides demonstrate the universal nature of gravitation. The Moon's gravity doesn't just pull the water upward; it creates differential forces across Earth's diameter. The water on the near side is pulled more, while the far side's bulge results from the Moon pulling Earth's solid body more than the far-side water. This is why tides occur exactly opposite each other on Earth's surface.

Q20. Compare gravitational force and electrostatic force.

Discuss similarities, differences, and why gravity seems weak.

4 marks

✓ Answer:

Feature	Gravitational Force	Electrostatic Force
Law	F = GM₁M₂/r²	F = kq₁q₂/r²
Nature	Always attractive	Attractive or repulsive
Strength	Very weak (smallest force)	Very strong (10³⁶ times stronger)
Medium	Works in vacuum	Works in vacuum
Range	Infinite (1/r²)	Infinite (1/r²)

Why Gravity Seems Weak:
Despite being the weakest fundamental force, gravity dominates at large scales because:
- All mass has gravity (always attractive)
- Charges can cancel (positive/negative)
- Large masses accumulate gravitational effects

💡 Explanation:

Both forces follow inverse-square law and have similar mathematical forms. However, electrostatic force can be repulsive, causing cancellation in neutral objects. Gravity has no "negative mass," so all gravitational effects add up. This is why despite being 10³⁶ times weaker, gravity controls the universe at cosmic scales (planets, stars, galaxies).

Q21. Explain the relationship between orbital velocity and gravitational force.

Derive the orbital velocity formula for circular orbits.

4 marks

✓ Answer:

Derivation:

For circular orbit, gravitational force provides centripetal force:
F_gravity = F_centripetal

GM m/r² = mv²/r
Simplifying: GM/r = v²
v = √(GM/r)

Key Points:
- Orbital velocity depends only on M (central body's mass) and r (orbital radius)
- Independent of orbiting object's mass (m cancels out)
- Larger orbits have lower velocities (v ∝ 1/√r)
- Smaller orbits have higher velocities

Example: Earth orbiting Sun
v_Earth = √(GM_Sun/r_Earth-Sun) = √(6.67×10⁻¹¹ × 2×10³⁰ / 1.5×10¹¹)
v_Earth ≈ 30 km/s

💡 Explanation:

Orbital motion is a continuous balance between gravity (pulling inward) and velocity (pulling outward). Gravitational force provides the centripetal force needed for circular motion. Satellites closer to Earth move faster; those farther move slower. This explains why the Moon (farther) orbits slower than artificial satellites (closer to Earth's surface).

Q22. Analyze the motion of a satellite in circular orbit around Earth.

Discuss orbital period, velocity, and energy considerations.

4 marks

✓ Answer:

Orbital Period (Kepler's Third Law):
T² ∝ r³ or T = 2π√(r³/GM)

Orbital Velocity:
v = √(GM/r)

Energy in Orbit:
- Kinetic Energy: KE = ½mv² = GMm/(2r)
- Potential Energy: PE = -GMm/r
- Total Energy: E = KE + PE = -GMm/(2r)

Key Characteristics:
1. Constant orbital velocity (v is constant in circular orbit)
2. Constant acceleration (centripetal, towards Earth)
3. Zero total mechanical energy for escape (parabolic orbit)
4. Negative total energy keeps satellite bound (elliptical/circular)

💡 Explanation:

The satellite's total energy is negative, meaning it's gravitationally bound. To escape, it needs energy input equal to -E to reach zero total energy. The relationship T² ∝ r³ (Kepler's law) explains why geostationary satellites (T = 24 hours) orbit at specific radius ~42,000 km from Earth's center. Faster orbits require more energy but actually move in lower (closer) orbits.

Q23. Explain escape velocity and its applications.

Derive the formula and discuss implications for space travel.

4 marks

✓ Answer:

Escape Velocity Definition:
Minimum velocity needed to escape a celestial body's gravitational field to infinity.

Derivation:
At infinity, both KE and PE = 0
Using energy conservation: KE_surface = PE_surface
½mv_e² = GMm/R

v_e = √(2GM/R)

Escape Velocities:
- Earth: v_e = 11.2 km/s
- Moon: v_e = 2.4 km/s (1/5 of Earth)
- Sun: v_e = 617 km/s

Applications:
- Rocket launch calculations
- Black hole properties (if R becomes small, v_e > c)
- Planetary atmosphere retention

💡 Explanation:

Escape velocity is independent of object's mass (it cancels in derivation). It depends only on the celestial body's mass and radius. Interestingly, escape velocity is exactly √2 times orbital velocity at the surface. Lighter objects (Moon) have lower escape velocity, making it easier to launch rockets. Black holes have so much mass in such small radius that their escape velocity equals light speed—nothing can escape.

Q24. A person weighs 60 kg on Earth. What would be their weight on planets with different g values?

Calculate weights on Mercury (g = 3.7 m/s²), Venus (g = 8.9 m/s²), Mars (g = 3.7 m/s²), and Jupiter (g = 24.8 m/s²).

4 marks

✓ Answer:

Given: m = 60 kg
Using W = mg

On Mercury:
W = 60 × 3.7 = 222 N (about 38% of Earth weight)

On Venus:
W = 60 × 8.9 = 534 N (about 91% of Earth weight)

On Mars:
W = 60 × 3.7 = 222 N (about 38% of Earth weight)

On Jupiter:
W = 60 × 24.8 = 1488 N (about 253% of Earth weight)

Comparison Table:

Planet	Weight (N)	% of Earth Weight
Earth	588	100%
Mercury	222	38%
Venus	534	91%
Mars	222	38%
Jupiter	1488	253%

💡 Explanation:

Weight varies dramatically between planets due to different surface gravities. Jupiter, being massive, has high gravity (2.5× Earth's). Mercury, being small and less massive, has weak gravity (0.38× Earth's). The person's mass remains 60 kg everywhere, but weight (gravitational force) varies. This explains why astronauts on the Moon weigh about 1/6 their Earth weight but have the same mass.

🔬 SECTION D: Application & Analysis Questions 12 Marks

Apply concepts to real-world scenarios

Q25. Why can't we feel the gravitational attraction between two people standing close to each other?

Calculate the gravitational force between two 60 kg people 1 meter apart.

3 marks

✓ Answer:

Calculation:
F = G(M₁M₂)/d²
F = (6.67 × 10⁻¹¹) × (60 × 60) / (1)²
F = (6.67 × 10⁻¹¹) × 3600
F = 2.4 × 10⁻⁸ N

This force is extremely tiny!
For comparison:
- Air molecules bump with force ~10⁻²¹ N
- Earth's gravitational pull on person = 588 N
- Ratio: 2.4 × 10⁻⁸ / 10⁻²¹ ≈ 10¹⁴ times larger

Yet even thermal motion of atoms (10⁻²¹ N) completely overwhelms gravitational attraction!

💡 Explanation:

Gravity is weak between small masses. The 10⁻¹¹ factor in G makes gravitational force negligible for everyday objects. While gravity controls planets and stars (huge masses), electrical and magnetic forces dominate at atomic and macroscopic scales. This demonstrates why gravity seems to be the weakest force—we never notice it between ordinary objects, only Earth's massive gravitational pull.

Q26. Analyze why different objects reach the same terminal velocity in air.

Discuss the role of gravitational and air resistance forces.

3 marks

✓ Answer:

Terminal Velocity Analysis:

When an object falls through air:
- Initially: Gravitational force > Air resistance → Net downward force
- As velocity increases: Air resistance increases
- At terminal velocity: Gravitational force = Air resistance
- Net force = 0, acceleration = 0

Why different objects reach different terminal velocities:
- Terminal velocity depends on shape, surface area, and mass
- Streamlined objects (like bullets) have higher terminal velocity
- Parachutes (large area) have very low terminal velocity
- Heavier objects reach higher terminal velocity (same shape)

Example:
Raindrops: ~9 m/s
Skydiver (belly-to-earth): ~53 m/s
Skydiver (head-first): ~90 m/s

💡 Explanation:

Terminal velocity is reached when acceleration becomes zero. This doesn't mean all objects have the same terminal velocity—it depends on surface area and mass. A raindrop reaches 9 m/s because its small mass and large surface area (relative to mass) make air resistance significant quickly. A skydiver reaches much higher velocity due to smaller surface area relative to mass. This is why parachutes are effective—they dramatically increase surface area, increasing air resistance.

Q27. A geostationary satellite must orbit with a period of 24 hours. Calculate its orbital radius.

M_Earth = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²

3 marks

✓ Answer:

Using Kepler's Third Law:
T² = (4π²/GM) × r³

Rearranging for r:
r³ = (GMT²)/(4π²)

Given:
T = 24 hours = 86,400 seconds
M = 6 × 10²⁴ kg
G = 6.67 × 10⁻¹¹

Calculation:
r³ = (6.67 × 10⁻¹¹ × 6 × 10²⁴ × (86,400)²) / (4π²)
r³ = (6.67 × 10⁻¹¹ × 6 × 10²⁴ × 7.46 × 10⁹) / 39.48
r³ = 7.53 × 10²² m³
r = 4.22 × 10⁷ m = 42,200 km

Distance from Earth's center: 42,200 km
Height above surface: 42,200 - 6,400 = 35,800 km

💡 Explanation:

Geostationary satellites must have exactly 24-hour period to stay above one location on Earth's equator. This unique orbital radius (42,200 km) is calculated from the constraint that orbital period equals Earth's rotation period. All geostationary satellites orbit at this radius, regardless of mass. These satellites are crucial for weather forecasting, telecommunications, and broadcasting because they remain stationary relative to a point on Earth.

Q28. Discuss why astronauts appear to be "weightless" in orbiting space stations.

Explain the concept of apparent weightlessness in orbit.

3 marks

✓ Answer:

Clarification: Astronauts are NOT weightless!

Reality:
- Earth's gravity still acts on astronauts (provides centripetal force)
- Astronauts' weight = mg is still ~90% of Earth surface value
- This gravitational force keeps them in orbit

Why they appear weightless:
- Space station and astronauts fall together
- Both have same acceleration (g ≈ 8.7 m/s² at ISS altitude)
- No normal force from the station floor
- The sensation of weight comes from normal force, not gravity

Analogy:
In a free-falling elevator, you feel weightless even though Earth's gravity still acts. Both you and elevator fall at acceleration 'g' together.

💡 Explanation:

Weight is what a scale reads—the normal force supporting you against gravity. In orbit, the spacecraft and astronaut accelerate together toward Earth's center (centripetal acceleration). There's no surface pushing up on the astronaut, so the scale reads zero. Gravity (true weight) is still present and is essential for the orbital motion. This apparent weightlessness is a consequence of free fall, not absence of gravity. Astronauts would "feel" normal gravity only if the spacecraft suddenly stopped orbiting and used rockets to hover.

📊 Answer Summary & Key Concepts

✓ Universal Law

F = G(M₁M₂)/r² is fundamental to understanding all gravitational phenomena. The constant G = 6.67 × 10⁻¹¹ ensures the law is universal.

✓ Acceleration Due to Gravity

g = GM/R² = 9.8 m/s² (Earth surface). Independent of falling object's mass. Varies with location.

✓ Mass vs Weight

Mass (constant, kg) ≠ Weight (varies, N). W = mg. Weight depends on local g value.

✓ Free Fall Motion

Objects fall with constant acceleration g. Uses equations: v = u ± gt, s = ut ± ½gt², v² = u² ± 2gs

✓ Pressure Concept

P = F/A. Pressure increases with force, decreases with area. Sharp objects concentrate force → high pressure.

✓ Buoyancy & Flotation

Buoyant force = Weight of fluid displaced (Archimedes' principle). Objects float if ρ_object < ρ_fluid.

✓ Orbital Motion

Orbital velocity: v = √(GM/r). Period: T² ∝ r³ (Kepler's law). Escape velocity: v_e = √(2GM/R)

✓ Apparent Weightlessness

Objects in free fall (orbit) appear weightless despite gravitational force. Weight = normal force = 0 in orbit, not because gravity is absent.

📌 Important Notes for Solving Problems:

1. Sign Conventions: Choose upward or downward as positive consistently. In free fall from ground, often take downward as positive.

2. Units: Always work in SI units. Convert distances to meters, time to seconds. 1 km = 1000 m, 1 hour = 3600 s.

3. Approximations: g ≈ 10 m/s² is acceptable for quick calculations. g ≈ 9.8 m/s² for precise answers.

4. Conceptual Understanding: Don't just memorize formulas. Understand what each variable represents and how changes affect results.

5. Diagram Drawing: Always draw diagrams for complicated problems. Show forces, directions, and reference frames clearly.

6. Energy Method: For orbital problems, use energy conservation: KE + PE = constant. Often simpler than force equations.