Gravitational force in astronomy is the attractive force between any two masses, governed by Newton's Law of Universal Gravitation, which states that the force is proportional to the product of the masses and inversely proportional to the square of the distance between them. This force is responsible for holding planets in orbit around the Sun, governing the motion of moons, shaping the structure of galaxies, and dictating the trajectories of spacecraft. It is the dominant long-range force at astronomical scales and underlies phenomena from tidal locking to the formation of planetary systems.
F = G × (m₁ × m₂) / r²
LaTeX: F = G \frac{m_1 m_2}{r^2}
| Symbol | Meaning | Unit |
|---|---|---|
| F | Gravitational force between the two masses | Newtons (N) |
| G | Universal gravitational constant (6.674 × 10⁻¹¹) | N·m²·kg⁻² |
| m₁ | Mass of the first object | kilograms (kg) |
| m₂ | Mass of the second object | kilograms (kg) |
| r | Distance between the centres of the two masses | metres (m) |
Problem
Calculate the gravitational force between the Earth (mass = 5.972 × 10²⁴ kg) and the Moon (mass = 7.342 × 10²² kg), separated by a distance of 3.844 × 10⁸ m.
Solution
Step 1: Write the formula: F = G × m₁ × m₂ / r². Step 2: Substitute values: F = (6.674 × 10⁻¹¹) × (5.972 × 10²⁴) × (7.342 × 10²²) / (3.844 × 10⁸)². Step 3: Numerator = 6.674 × 10⁻¹¹ × 4.384 × 10⁴⁷ = 2.924 × 10³⁷. Step 4: Denominator = (3.844 × 10⁸)² = 1.478 × 10¹⁷. Step 5: F = 2.924 × 10³⁷ / 1.478 × 10¹⁷ = 1.978 × 10²⁰ N.
Answer
F ≈ 1.98 × 10²⁰ N (approximately 198 quintillion Newtons).
| Body | Mass (kg) | Radius (km) | Surface Gravity (m/s²) | Relative to Earth |
|---|---|---|---|---|
| Sun | 1.989 × 10³⁰ | 695,700 | 274.0 | 27.9× |
| Earth | 5.972 × 10²⁴ | 6,371 | 9.81 | 1.0× |
| Moon | 7.342 × 10²² | 1,737 | 1.62 | 0.165× |
| Mars | 6.39 × 10²³ | 3,390 | 3.72 | 0.379× |
| Jupiter | 1.898 × 10²⁷ | 69,911 | 24.79 | 2.53× |
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