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Circular Motion & Gravitation

Master circular motion dynamics, Kepler's laws, and universal gravitation through interactive problem-solving and celestial visualizations.

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Topic Overview

This topic covers the laws of dynamics of objects in circular motion and the application of the law of gravity to the motion of celestial bodies. The combined use of the centripetal force formula, Kepler's laws, and gravitational calculations are often tested in exams in relation to specific situations (e.g., satellite orbits, planetary motions), requiring strong physical modeling skills.

Key Points

Sources and formulas for centripetal force in uniform circular motion
Kepler's three laws for the description of the orbital motion of celestial bodies
Application of the formula for the law of gravity and the derivation of celestial parameters
Analysis of energy and velocity changes in the satellite re-orbiting problem
Relationship between orbital period, radius, and velocity
Gravitational potential energy and escape velocity
Comparison of circular and elliptical orbits
Geosynchronous and low Earth orbits

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Quick Tip

Remember: Centripetal force is not a new force—it's the net force causing circular motion. For orbits, gravitational force provides the centripetal force.

1 Basic Centripetal Force Calculation

Easy

A 2 kg object moves in a horizontal circle of radius 3 m with a constant speed of 4 m/s. What is the magnitude of the centripetal force acting on the object?

Circular Motion Visualization

2 kg

r = 3 m

Velocity is tangential to the circle

Centripetal force points toward the center

Centripetal Force Calculation

Centripetal force formula: F_c = mv²/r

F_c = (2 kg) × (4 m/s)² / (3 m)

F_c = (2 × 16) / 3 = 32 / 3

F_c = 10.67 N

Key properties of centripetal force:

• Always directed toward center of circle

• Perpendicular to velocity (does no work)

• Causes change in direction, not speed

• Not a new force—it's the net radial force

10.67 N

6.0 N

24.0 N

2.67 N

Hint: Use F_c = mv²/r. Centripetal force depends on mass, speed squared, and inversely on radius.

Step-by-Step Solution

1. Centripetal force formula: F_c = mv²/r

2. Given: m = 2 kg, v = 4 m/s, r = 3 m

3. Calculate v²: (4 m/s)² = 16 m²/s²

4. Calculate mv²: 2 kg × 16 m²/s² = 32 kg·m²/s² = 32 N·m

5. Divide by radius: 32 N·m / 3 m = 10.67 N

6. Units: kg·m/s² = N (Newton)

7. Interpretation: This force is required to keep the object moving in a circle of radius 3 m at 4 m/s

Answer: A

2 Orbital Velocity of Satellite

Medium

A satellite orbits Earth at an altitude of 500 km above the surface. Earth's radius is 6370 km and mass is 5.97 × 10²⁴ kg. What is the orbital velocity of the satellite? (G = 6.67 × 10⁻¹¹ N·m²/kg²)

Satellite Orbit Visualization

Earth

Sat

R = 6370 km

h = 500 km

Orbital Velocity Calculation

Orbital velocity formula: v = √(GM/r)

where r = R_earth + h (distance from Earth's center)

Given:

• Earth radius: R = 6370 km = 6.37 × 10⁶ m

• Altitude: h = 500 km = 5.00 × 10⁵ m

• Orbital radius: r = R + h = 6.87 × 10⁶ m

• Earth mass: M = 5.97 × 10²⁴ kg

• G = 6.67 × 10⁻¹¹ N·m²/kg²

Calculation:

GM = (6.67×10⁻¹¹) × (5.97×10²⁴) = 3.98×10¹⁴ m³/s²

v = √(3.98×10¹⁴ / 6.87×10⁶)

v = √(5.79×10⁷)

v = 7610 m/s ≈ 7.61 km/s

7.61 km/s

3.07 km/s

11.2 km/s

5.42 km/s

Hint: For circular orbits, gravitational force provides centripetal force: GMm/r² = mv²/r → v = √(GM/r). Remember r = Earth radius + altitude.

Step-by-Step Solution

1. For a satellite in circular orbit: gravitational force = centripetal force

2. GMm/r² = mv²/r (m cancels)

3. v² = GM/r → v = √(GM/r)

4. Given: M = 5.97 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²

5. Earth radius: R = 6370 km = 6.37 × 10⁶ m

6. Altitude: h = 500 km = 5.00 × 10⁵ m

7. Orbital radius: r = R + h = 6.87 × 10⁶ m

8. Calculate GM: 6.67×10⁻¹¹ × 5.97×10²⁴ = 3.98×10¹⁴ m³/s²

9. v = √(3.98×10¹⁴ / 6.87×10⁶) = √(5.79×10⁷)

10. v = 7610 m/s = 7.61 km/s

11. Note: This is typical for low Earth orbit satellites

Answer: A

3 Kepler's Third Law Application

Hard

Planet X orbits its star with a period of 8 Earth years. If Earth's orbital radius around the Sun is 1 AU, what is the orbital radius of planet X in AU? (Assume circular orbits and same star mass for both)

Kepler's Third Law Visualization

Star

Earth

Planet X

1 AU

R_x = ?

Kepler's Third Law: T² ∝ r³

The square of the orbital period is proportional to the cube of the semi-major axis

Kepler's Law Calculation

Kepler's Third Law: T² ∝ r³ or T²/r³ = constant

For two planets orbiting same star: T₁²/T₂² = r₁³/r₂³

Given:

• Earth: T₁ = 1 year, r₁ = 1 AU

• Planet X: T₂ = 8 years, r₂ = ?

Apply Kepler's Third Law:

T₁²/r₁³ = T₂²/r₂³

1²/1³ = 8²/r₂³

1 = 64/r₂³

r₂³ = 64

r₂ = ∛64 = 4 AU

Memory aid: T² ∝ r³ means if period increases by factor of 8, radius increases by factor of 8^(2/3) = 4.

4 AU

2 AU

8 AU

64 AU

Hint: Use Kepler's Third Law: T² ∝ r³. Set up ratio: (T_X/T_Earth)² = (r_X/r_Earth)³.

Step-by-Step Solution

1. Kepler's Third Law: For planets orbiting same star, T² ∝ r³

2. In equation form: T₁²/r₁³ = T₂²/r₂³ (constant for given star)

3. Let Earth: T₁ = 1 year, r₁ = 1 AU

4. Planet X: T₂ = 8 years, r₂ = ?

5. Set up proportion: T₁²/r₁³ = T₂²/r₂³

6. 1²/1³ = 8²/r₂³

7. 1 = 64/r₂³

8. r₂³ = 64

9. r₂ = ∛64 = 4 AU

10. Alternative: (T_X/T_Earth)² = (r_X/r_Earth)³ → (8/1)² = (r_X/1)³ → 64 = r_X³ → r_X = 4 AU

11. Interpretation: Planet X is 4 times farther from its star than Earth is from Sun

Answer: A

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