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Force Of Attraction Formula Calculator

\[ F = \frac{G m_1 m_2}{r^2} \]

1. What is the Force Of Attraction Formula Calculator?

Definition: This calculator computes the gravitational force of attraction (\(F\)) between two masses using the formula \( F = \frac{G m_1 m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between their centers.

Purpose: It is used in physics to determine the gravitational force between objects, applicable in astronomy (e.g., planetary orbits), engineering (e.g., satellite dynamics), and educational contexts.

2. How Does the Calculator Work?

The calculator uses Newton’s law of universal gravitation:

Formula: \[ F = \frac{G m_1 m_2}{r^2} \] where:

  • \(F\): Force of attraction (N, lb-force)
  • \(G\): Gravitational constant (m³/(kg·s²), ft³/(lb·s²), lb·ft/(slug²·s²))
  • \(m_1, m_2\): Masses (kg, lb, slug)
  • \(r\): Distance (m, ft, mi)

Unit Conversions:

  • Gravitational Constant (\(G\)):
    • 1 m³/(kg·s²) = 1 m³/(kg·s²)
    • 1 ft³/(lb·s²) = \( \frac{1}{0.028316846592 / 0.45359237} \) m³/(kg·s²) \(\approx 4.169 \times 10^{-9} \, \text{ft}^3/(\text{lb·s}^2)\)
    • 1 lb·ft/(slug²·s²) = \( \frac{1}{0.028316846592 / 14.593903} \) m³/(kg·s²) \(\approx 8.683 \times 10^{-11} \, \text{lb·ft}/(\text{slug}^2·\text{s}^2)\)
  • Masses (\(m_1, m_2\)):
    • 1 kg = 1 kg
    • 1 lb = 0.45359237 kg
    • 1 slug = 14.593903 kg
  • Distance (\(r\)):
    • 1 m = 1 m
    • 1 ft = 0.3048 m
    • 1 mi = 1609.344 m
  • Force (Output):
    • 1 N = 1 N
    • 1 lb-force = 4.4482216152605 N
The force is calculated in newtons (N) and can be converted to the selected output unit (N, lb-force). Results greater than 10,000 or less than 0.001 are displayed in scientific notation; otherwise, they are shown with 4 decimal places.

Steps:

  • Select a scenario (e.g., Earth and Object) or enter custom values for the gravitational constant (\(G\)), masses (\(m_1, m_2\)), and distance (\(r\)) with their units (default: Earth and Object scenario, \(G = 6.67430 \times 10^{-11} \, \text{m}^3/(\text{kg·s}^2)\), \(m_1 = 5.972 \times 10^{24} \, \text{kg}\), \(m_2 = 1 \, \text{kg}\), \(r = 6371000 \, \text{m}\)).
  • Convert inputs to SI units (m³/(kg·s²), kg, m).
  • Validate that gravitational constant, masses, and distance are greater than 0.
  • Calculate the force in newtons using the formula.
  • Convert the force to the selected output unit.
  • Display the result, using scientific notation if the value is greater than 10,000 or less than 0.001, otherwise rounded to 4 decimal places.

3. Importance of Force of Attraction Calculation

Calculating the gravitational force of attraction is crucial for:

  • Astronomy: Determining the forces between celestial bodies, such as the attraction between the Earth and the Moon, which governs orbits and tides.
  • Engineering: Designing spacecraft and satellites, where gravitational forces affect trajectories and orbital dynamics.
  • Education: Teaching Newton’s law of universal gravitation and the principles of gravitational interactions in physics.

4. Using the Calculator

Examples:

  • Example 1: Calculate the gravitational force between Earth (\( m_1 = 5.972 \times 10^{24} \, \text{kg} \)) and a 1 kg object (\( m_2 = 1 \, \text{kg} \)) at Earth’s surface (\( r = 6371000 \, \text{m} \)), with \( G = 6.67430 \times 10^{-11} \, \text{m}^3/(\text{kg·s}^2) \), output in N:
    • Enter \( G = 6.67430 \times 10^{-11} \, \text{m}^3/(\text{kg·s}^2) \), \( m_1 = 5.972 \times 10^{24} \, \text{kg} \), \( m_2 = 1 \, \text{kg} \), \( r = 6371000 \, \text{m} \).
    • Numerator: \( G m_1 m_2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times 1 \approx 3.986 \times 10^{14} \, \text{N·m}^2 \).
    • Denominator: \( r^2 = (6371000)^2 \approx 4.061 \times 10^{13} \, \text{m}^2 \).
    • Force: \( F = \frac{3.986 \times 10^{14}}{4.061 \times 10^{13}} \approx 9.817 \, \text{N} \).
    • Output unit: N (no conversion needed).
    • Result: \( \text{Force of Attraction} = 9.8170 \, \text{N} \).
  • Example 2: Calculate the gravitational force between two objects (\( m_1 = 2204.62 \, \text{lb} \), \( m_2 = 2204.62 \, \text{lb} \)) separated by \( r = 32.8084 \, \text{ft} \), with \( G = 4.169 \times 10^{-9} \, \text{ft}^3/(\text{lb·s}^2) \), output in lb-force:
    • Enter \( G = 4.169 \times 10^{-9} \, \text{ft}^3/(\text{lb·s}^2) \), \( m_1 = 2204.62 \, \text{lb} \), \( m_2 = 2204.62 \, \text{lb} \), \( r = 32.8084 \, \text{ft} \).
    • Convert: \( G = 4.169 \times 10^{-9} \times \frac{0.028316846592}{0.45359237} \approx 6.67430 \times 10^{-11} \, \text{m}^3/(\text{kg·s}^2) \), \( m_1 = 2204.62 \times 0.45359237 = 1000 \, \text{kg} \), \( m_2 = 2204.62 \times 0.45359237 = 1000 \, \text{kg} \), \( r = 32.8084 \times 0.3048 = 10 \, \text{m} \).
    • Numerator: \( G m_1 m_2 = 6.67430 \times 10^{-11} \times 1000 \times 1000 = 6.67430 \times 10^{-5} \, \text{N·m}^2 \).
    • Denominator: \( r^2 = 10^2 = 100 \, \text{m}^2 \).
    • Force in N: \( F = \frac{6.67430 \times 10^{-5}}{100} = 6.67430 \times 10^{-7} \, \text{N} \).
    • Convert to output unit (lb-force): \( 6.67430 \times 10^{-7} \times \frac{1}{4.4482216152605} \approx 1.500 \times 10^{-7} \, \text{lb-force} \).
    • Result: \( \text{Force of Attraction} = 1.5000 \times 10^{-7} \, \text{lb-force} \).

5. Frequently Asked Questions (FAQ)

Q: What is the force of attraction?
A: The force of attraction (\(F\)) is the gravitational force between two masses, given by Newton’s law of universal gravitation \( F = \frac{G m_1 m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between their centers. It is always attractive and acts along the line joining the masses.

Q: Why must the distance be greater than zero?
A: The distance (\(r\)) must be greater than zero to avoid division by zero in the formula and to represent a physically meaningful separation between the masses. If \( r = 0 \), the objects would occupy the same space, which is not practical for this calculation.

Q: How is the gravitational constant determined?
A: The gravitational constant (\(G\)) is determined experimentally, most famously through the Cavendish experiment, which measures the weak gravitational attraction between lead spheres using a torsion balance. Its value is approximately \( 6.67430 \times 10^{-11} \, \text{m}^3/(\text{kg·s}^2) \), with ongoing refinements in precision.

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