Surface Gravity
Gravitational acceleration at a body surface.
Formula
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Example
Moon: 1.62 m/s²
How it works
This formula gives the strength of gravity at the surface of a spherical body, such as a planet or moon. It comes from Newton’s law of gravitation: the gravitational force on a small mass is F = GMm/r^2, and dividing by the mass m gives the acceleration g. The result shows that a body with more mass has stronger surface gravity, while a larger radius weakens surface gravity because the same mass is spread farther from the surface.
When to use it
Use this when you need the gravitational acceleration on a planet, moon, or star from its mass and radius. It is useful for comparing worlds, estimating weight changes, or checking orbital and landing calculations near a body’s surface.
Worked example, step by step
Find the surface gravity of Mars. Given: G = 6.67 × 10^-11 N·m²/kg², M = 6.42 × 10^23 kg, r = 3.39 × 10^6 m Step 1: Substitute into g = GM/r^2 Step 2: g = (6.67 × 10^-11 × 6.42 × 10^23) / (3.39 × 10^6)^2 Step 3: Calculate the numerator: 6.67 × 6.42 × 10^( -11 + 23 ) ≈ 42.8 × 10^12 = 4.28 × 10^13 Step 4: Calculate the denominator: (3.39 × 10^6)^2 ≈ 1.15 × 10^13 Step 5: Divide: g ≈ (4.28 × 10^13) / (1.15 × 10^13) ≈ 3.7 m/s² Final answer: Mars’ surface gravity is about 3.7 m/s².
Common mistakes to avoid
- using the diameter instead of the radius in the denominator
- forgetting to square the radius
- mixing up mass and weight, or using kilograms for g instead of m/s²
Did You Know?
Escape velocity from the Sun's surface is about 618 km/s, over fifty times Earth's.
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