Compute the gravitational attraction between two masses using F = G·m₁·m₂ / r². Visualize the interaction, understand the inverse-square law, and explore real astronomical examples.
Isaac Newton first formulated the law of universal gravitation in his 1687 work Philosophiæ Naturalis Principia Mathematica. The law states that every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This fundamental interaction governs planetary orbits, tides, star formation, and the large-scale structure of the universe.
F = G · (m₁ · m₂) / r²
Where G = 6.67430 × 10⁻¹¹ N·m²·kg⁻² (2018 CODATA recommended value)
Our calculator implements the exact Newtonian formula using double‑precision arithmetic. Given masses m₁ and m₂ (in kg) and center‑to‑center distance r (in meters), the force magnitude is F = G × m₁ × m₂ / r². The direction is always attractive along the line joining the two masses. The accelerations a₁ = F/m₁ and a₂ = F/m₂ are also provided, demonstrating that the lighter mass experiences a greater acceleration — key to understanding why the Moon orbits the Earth.
The gravitational parameter μ = G·m (standard gravitational parameter) is also displayed, a crucial quantity in orbital mechanics (used in calculating orbital periods and velocities). The calculator handles extremely large and small numbers gracefully, using scientific notation where appropriate.
| System | Mass 1 (kg) | Mass 2 (kg) | Distance (m) | Force (N) | Physical meaning |
|---|---|---|---|---|---|
| Earth + 70 kg person | 5.972×10²⁴ | 70 | 6.371×10⁶ | ~686.5 | Weight force (mg ≈ 686 N) |
| Earth – Moon | 5.972×10²⁴ | 7.342×10²² | 3.844×10⁸ | 1.98×10²⁰ | Keeps Moon in orbit |
| Sun – Earth | 1.989×10³⁰ | 5.972×10²⁴ | 1.496×10¹¹ | 3.52×10²² | Central gravitational binding |
| Two 1 kg masses | 1 | 1 | 1 | 6.674×10⁻¹¹ | Extremely weak, negligible in daily life |
For a satellite to remain fixed above a point on Earth's equator, the gravitational force provides exactly the centripetal acceleration needed. Using our calculator, set m₁ = Earth mass (5.972e24 kg), m₂ = satellite mass (say 500 kg), and adjust distance r until the gravitational force matches the required centripetal force for an orbital period of 24 hours. The correct geostationary radius is approximately 42,164 km from Earth's center. Our tool helps confirm this value and demonstrates the inverse-square relation at work in modern telecommunications.