Gravity Games Slides Gravity Games
The Law of Universal Gravitation
MISSION CONTROL: PHYSICS UNIT
The Invisible Thread
Gravity isn't just "down." It is a mutual attraction between every single piece of matter in the universe.
"Every object in the universe pulls on every other object."
You pull on the Earth.
The Earth pulls on you.
You pull on your pencil!
Newton's Secret Recipe
THE FORMULA
\[ F = G \frac{m_1 m_2}{r^2} \]
F
Force (N)
G
Constant
m
Mass (kg)
r
Distance (m)
1. The Mass Multiplier
Mass is directly proportional to force.
Double Mass = Double Force
Half Mass = Half Force
Think About It
Earth pulls harder than the Moon because it has significantly more Mass. Simple math!
2. The Distance Dilemma
1 / r²
The Inverse Square Law makes distance the true power-player of gravity.
Distance is inversely squared.
Double Distance (2x)
1/4 Force
Triple Distance (3x)
1/9 Force
Half Distance (1/2x)
4x Force
Orbital Mechanics
Orbits are a balancing act between Inertia and Gravity.
1
Gravity pulls the satellite toward the center.
2
The satellite's speed (inertia) wants it to go straight.
3
A perpetual "fall" that misses the target every time.
Velocity
Gravity
Cosmic Combat
Commanders: Defence grids are offline.
Calculate
Predict
Defend
Cosmic Combat Reference Sheet Cosmic Combat
Tactical Reference Sheet
Sector:
Stardate:
Universal Gravitation Formula
\[ F = G \frac{m_1 m_2}{r^2} \]
F Gravitational Force (Newtons)
m₁, m₂ Masses (Kilograms)
G \( 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2 \)
r Distance between centers (Meters)
1. The Mass Multiplier
Force is directly proportional to mass.
Mass Change Force Change 2x Mass 2x Force 1/2x Mass 1/2x Force Double BOTH Masses 4x Force
2. The Distance Dilemma
The Inverse Square Law: \( F \propto 1/r^2 \)
Distance Change Force Change 2x Distance 1/4 Force 3x Distance 1/9 Force 1/2x Distance 4x Force
Orbital Stability Check
Centripetal Force:
\[ F_c = F_g \]
To stay in orbit, gravitational pull must equal the centripetal force required for circular motion.
Increasing speed requires a smaller orbit or more mass to hold.
Higher mass planets have a stronger "grip" on their satellites.
Weightlessness in orbit is just Continuous Freefall!
Tactical Intel
Earth Mass (\( M_E \)) \( 5.97 \times 10^{24} \text{ kg} \)
Earth Radius (\( R_E \)) \( 6.37 \times 10^6 \text{ m} \)
Unit Conversion 1 km = 1,000 meters
Command Authority: Unified Planetary Defense Force • Starchart v4.2
Cosmic Combat Mission Log Mission Log
Operation: Gravity Shield
Commander:
Squadron:
Classified: Level 5 Access Required
01
Targeting: The Mass Multiplier
Intelligence reports a massive rogue comet entering your sector. We need to predict its gravitational influence on nearby moons.
Scenario Alpha:
The current gravitational pull between the Comet and Moon A is \( F \). If the Comet suddenly hits an asteroid and its mass doubles, what is the new gravitational force?
Scenario Beta:
The Comet moves toward Planet Titan. Titan has 3 times the mass of Moon A. If the distance remains the same, how many times stronger is the pull on Titan compared to Moon A?
Pro-Tip for Commanders:
Mass and Force are direct proportional partners. What you do to the top of the fraction happens to the Force!
02
Targeting: The Distance Dilemma
Your scout ships are attempting to hide in the "Shadow Zone" between two stars. Distance is critical for survival.
Scenario Gamma:
A scout ship is \( 1,000\text{ km} \) from a star. It jumps to a new position \( 2,000\text{ km} \) away (double the distance). By what factor does the gravitational pull decrease?
⚠️ Critical Warning:
An enemy tractor beam is pulling your ship. To reduce the force to 1/16th of its current value, how many times further away must you move?
Visual Mapping
Sketch how Force changes as \( r \) increases
Inverse Square Rule
"Double the distance, quarter the force. Triple the distance, one-ninth the force."
03
Tactical Calculations
Now, provide the exact numbers for HQ. Use your reference sheet for the constant \( G \). Show all work for full credit.
Problem 1: The Earth-Moon Grip
Target ID: 399-301
Calculate the force of gravity between the Earth (\( 5.97 \times 10^{24}\text{ kg} \)) and the Moon (\( 7.35 \times 10^{22}\text{ kg} \)) given the average distance between their centers is \( 3.84 \times 10^8\text{ meters} \).
Show Tactical Workings Below
Problem 2: The Satellite Sync
Target ID: GPS-12
A \( 500\text{ kg} \) satellite orbits Earth at a distance of \( 4.2 \times 10^7\text{ meters} \) from the Earth's center. What is the gravitational force acting on the satellite?
Show Tactical Workings Below
04
Orbital Maneuvers
Final Phase: Analyze the stability of our defense grid.
Critical Reasoning Challenge
Cosmic Combat Answer Key Mission Log: Answer Key
Operation: Gravity Shield • Teacher Resource
Internal Use Only
Mission 01: The Mass Multiplier
Scenario Alpha:
Answer: \( 2F \)
Explanation: Force is directly proportional to mass. Doubling one mass doubles the force.
Scenario Beta:
Answer: 3 times stronger (\( 3F \))
Explanation: Tripling the mass of one object triples the total gravitational pull between the pair.
Mission 02: The Distance Dilemma
Scenario Gamma:
Answer: Decreases by a factor of 4 (1/4th the original force)
Explanation: \( F \propto 1/r^2 \). If distance doubles (2x), the force is \( 1/2^2 = 1/4 \).
Critical Warning Scenario:
Answer: 4 times further away
Explanation: To get \( 1/16 \) force, we need \( r^2 = 16 \). Therefore, \( r = \sqrt{16} = 4 \).
Mission 03: Tactical Calculations
Problem 1: Earth-Moon Grip
Variables:
\( m_1 = 5.97 \times 10^{24}\text{ kg} \)
\( m_2 = 7.35 \times 10^{22}\text{ kg} \)
\( r = 3.84 \times 10^8\text{ m} \)
Calculation Steps:
1. \( m_1 \times m_2 = 4.38795 \times 10^{47} \)
2. \( r^2 = 1.47456 \times 10^{17} \)
3. \( (m_1 \times m_2) / r^2 = 2.9757 \times 10^{30} \)
4. Multiply by \( G \)
Final Answer: \( 1.98 \times 10^{20}\text{ Newtons} \)
Problem 2: Satellite Sync
Variables:
\( m_1 = 5.97 \times 10^{24}\text{ kg} \) (Earth)
\( m_2 = 500\text{ kg} \)
\( r = 4.2 \times 10^7\text{ m} \)
Calculation Steps:
1. \( m_1 \times m_2 = 2.985 \times 10^{27} \)
2. \( r^2 = 1.764 \times 10^{15} \)
3. Multiply by \( G \)
Final Answer: \( 112.87\text{ Newtons} \)
Mission 04: Orbital Maneuvers
Conceptual Solution:
"The speed must decrease. As distance (\( r \)) increases, the gravitational force (\( F_g \)) weakens significantly due to the inverse square law. To stay in a stable orbit, the required centripetal force (\( F_c = mv^2/r \)) must match this weaker gravity. If the satellite kept its original high speed at a further distance, inertia would carry it away from Earth's pull entirely."
Verified for Newtonian Universe Sequence • Unified Planetary Defense Force Answer Protocol
