Complete mechanics guide • Step-by-step solutions
\( F = G \frac{m_1 m_2}{r^2} \)
Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The gravitational constant G ≈ 6.674 × 10⁻¹¹ N·m²/kg². This fundamental force governs planetary orbits, galaxy formation, and the structure of the universe.
Key properties of gravitational force:
Use this formula whenever analyzing celestial mechanics, orbital motion, or any system involving massive objects. It's essential for understanding planetary motion, satellite trajectories, and cosmic phenomena.
Gravitational force is one of the four fundamental forces of nature and describes the attractive force between any two objects with mass. Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This fundamental law governs everything from falling apples to planetary orbits and galactic structures.
Newton's law of universal gravitation is expressed as:
Where:
Key characteristics of gravitational force:
F = G(m₁m₂)/r², where F is force, G is gravitational constant, m₁ and m₂ are masses, r is distance.
g = GM/r², U = -Gm₁m₂/r, v_escape = √(2GM/r)
Where g is gravitational field strength, U is potential energy.
v = √(GM/r), where M is mass of central body.
Two objects of masses 1000 kg and 2000 kg are separated by 5 meters. What is the gravitational force between them? (G = 6.674 × 10⁻¹¹ N·m²/kg²)
Using the gravitational force formula F = G(m₁m₂)/r²:
F = (6.674 × 10⁻¹¹) × (1000 × 2000) / (5²)
F = (6.674 × 10⁻¹¹) × (2,000,000) / 25
F = (1.3348 × 10⁻⁴) / 25 = 5.34 × 10⁻⁶ N
The answer is A) 5.34 × 10⁻⁶ N.
This calculation demonstrates the weakness of gravitational force between relatively small masses at everyday distances. Even with masses of 1000 kg and 2000 kg, the force is extremely small (microNewtons). This is why we don't notice gravitational attraction between ordinary objects around us.
Gravitational Force: Attractive force between masses
Inverse Square Law: Force ∝ 1/r²
Gravitational Constant: G = 6.674 × 10⁻¹¹ N·m²/kg²
• F = G(m₁m₂)/r² (universal gravitation)
• Force is always attractive
• Very weak force compared to other fundamental forces
• Remember: force is always attractive
• Use scientific notation for large/small numbers
• Distance is center-to-center separation
• Forgetting to square the distance
• Using surface-to-surface distance instead of center-to-center
• Mixing up units in calculations
If the distance between two masses is tripled, by what factor does the gravitational force change? Explain the inverse square law relationship.
Step 1: Original force at distance r
F₁ = G(m₁m₂)/r²
Step 2: New force at distance 3r
F₂ = G(m₁m₂)/(3r)² = G(m₁m₂)/(9r²)
Step 3: Ratio of forces
F₂/F₁ = [G(m₁m₂)/(9r²)] / [G(m₁m₂)/r²] = 1/9
The force decreases by a factor of 9 when the distance is tripled.
The inverse square law means that when distance increases by a factor of n, the force decreases by a factor of n². This is because the force is proportional to 1/r².
The inverse square law is fundamental to gravitational force and many other physical phenomena (electric force, light intensity, etc.). This relationship means that gravitational force becomes significantly weaker with distance. For example, if Earth were twice as far from the Sun, the gravitational force would decrease by a factor of 4.
Inverse Square Law: F ∝ 1/r²
Proportional: Mathematical relationship
Center-to-Center: Distance between mass centers
• F ∝ 1/r² (inverse square relationship)
• Doubling distance reduces force by factor of 4
• Tripling distance reduces force by factor of 9
• Square the distance change factor
• Larger distance = smaller force
• Remember: r² in denominator
• Forgetting to square the distance factor
• Thinking force decreases linearly with distance
• Confusing direct vs inverse relationship
Calculate the gravitational force between Earth (mass = 5.972 × 10²⁴ kg) and the Moon (mass = 7.342 × 10²² kg) when they are 3.844 × 10⁸ m apart. Then calculate the acceleration each body experiences due to this force.
Step 1: Calculate gravitational force
F = G(m₁m₂)/r² = (6.674 × 10⁻¹¹) × (5.972 × 10²⁴) × (7.342 × 10²²) / (3.844 × 10⁸)²
F = (6.674 × 10⁻¹¹) × (4.385 × 10⁴⁷) / (1.478 × 10¹⁷)
F = 1.982 × 10²⁰ N
Step 2: Calculate acceleration of Earth
a_Earth = F/m_Earth = (1.982 × 10²⁰) / (5.972 × 10²⁴) = 3.32 × 10⁻⁵ m/s²
Step 3: Calculate acceleration of Moon
a_Moon = F/m_Moon = (1.982 × 10²⁰) / (7.342 × 10²²) = 2.70 × 10⁻³ m/s²
The gravitational force is 1.982 × 10²⁰ N, Earth's acceleration is 3.32 × 10⁻⁵ m/s², and Moon's acceleration is 2.70 × 10⁻³ m/s².
This example shows how the same force produces different accelerations depending on the mass of the object. The Moon experiences a much greater acceleration than Earth due to its much smaller mass. This is why the Moon orbits Earth rather than Earth orbiting the Moon.
Acceleration: Change in velocity per unit time
Newton's Second Law: F = ma, so a = F/m
Orbital Motion: Result of gravitational force
• F = ma (Newton's second law)
• Same force, different acceleration due to different masses
• Smaller mass experiences greater acceleration
• Use F = ma to find acceleration
• Smaller object has greater acceleration
• Acceleration determines orbital motion
• Forgetting that both objects accelerate
• Thinking force is different for each object
• Not applying Newton's second law correctly
A 1000 kg satellite orbits Earth at an altitude of 400 km above the surface. Calculate the gravitational force on the satellite and the orbital speed required to maintain circular orbit. (Earth radius = 6.371 × 10⁶ m, Earth mass = 5.972 × 10²⁴ kg)
Step 1: Calculate orbital radius
r = R_Earth + altitude = 6.371 × 10⁶ + 400 × 10³ = 6.771 × 10⁶ m
Step 2: Calculate gravitational force
F = G(m_satellite × m_Earth)/r² = (6.674 × 10⁻¹¹) × (1000 × 5.972 × 10²⁴) / (6.771 × 10⁶)²
F = (3.986 × 10¹⁷) / (4.585 × 10¹³) = 8.69 × 10³ N
Step 3: Calculate orbital speed
For circular orbit: F_gravity = F_centripetal
G(m_Earth)/r² = v²/r
v = √[G(m_Earth)/r] = √[(6.674 × 10⁻¹¹ × 5.972 × 10²⁴) / (6.771 × 10⁶)]
v = √[5.898 × 10⁷] = 7.68 × 10³ m/s = 7.68 km/s
The gravitational force is 8.69 × 10³ N and the orbital speed is 7.68 km/s.
This problem demonstrates how gravitational force provides the centripetal force needed for circular orbital motion. The satellite is constantly falling toward Earth but moving sideways fast enough to miss it, resulting in stable orbit. This principle applies to all satellites, moons, and planets.
Centripetal Force: Force directed toward center of circular motion
Orbital Velocity: Speed required for stable orbitAltitude: Height above surface
• F_gravity = F_centripetal for circular orbit
• v_orbit = √(GM/r)
• Gravitational force provides centripetal force
• Always use center-to-center distance
• Orbital speed depends on distance from center
• Higher orbit = slower speed
• Using surface distance instead of center distance
• Confusing orbital velocity with escape velocity
• Forgetting that gravity provides centripetal force
Which of the following statements about gravitational force is correct?
Gravitational force is always attractive, never repulsive. It follows the inverse square law (F ∝ 1/r²), not inverse cube. Gravitational force is much weaker than electromagnetic force at the atomic scale. These are fundamental properties of gravity as described by Newton's law of universal gravitation.
The answer is C) Gravitational force is always attractive.
Gravitational force has unique properties that distinguish it from other fundamental forces. Unlike electromagnetic force (which can be attractive or repulsive), gravity only attracts. This property is fundamental to how matter clusters together to form stars, planets, and galaxies.
Fundamental Forces: Gravity, electromagnetic, strong nuclear, weak nuclear
Inverse Square Law: F ∝ 1/r²
Attractive Force: Always pulls objects together
• Gravity is always attractive
• F ∝ 1/r² (inverse square law)
• Weakest of fundamental forces
• Remember: gravity only attracts
• Compare strengths of fundamental forces
• Know the inverse square relationship
• Thinking gravity can repel
• Confusing inverse square with inverse cube
• Misunderstanding relative strength of forces
Q: Why don't we feel the gravitational force between everyday objects?
A: The gravitational force between everyday objects is extremely weak because the gravitational constant G is very small (6.674 × 10⁻¹¹ N·m²/kg²). Using the formula F = G(m₁m₂)/r², even for relatively large masses like 1000 kg objects separated by 1 meter, the force is only about 6.67 × 10⁻⁵ N, which is imperceptible.
We only notice gravitational effects when at least one of the masses is enormous, like Earth. The force between you and Earth is significant because Earth's mass is 5.972 × 10²⁴ kg, making the gravitational force substantial enough to keep you on the ground.
Q: How does the inverse square law affect satellite operations?
A: The inverse square law means that gravitational force decreases rapidly with distance. For satellites, this has important implications: if a satellite's distance from Earth doubles, the gravitational force becomes 1/4 of its original value. This affects orbital velocity requirements, as v = √(GM/r).
As satellites move to higher orbits, they experience weaker gravity and need lower orbital velocities. For example, GPS satellites at ~20,000 km altitude travel at about 3.8 km/s, while low Earth orbit satellites at ~400 km travel at ~7.7 km/s. This relationship is critical for mission planning and fuel calculations.