Surface Gravity
Gravitational acceleration at a body's surface.
Formula
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Example
The Moon’s smaller mass and radius give it surface gravity of about 1.6 m/s².
How it works
This formula gives the acceleration an object would experience near the surface of a spherical body such as a planet, moon, or star. It comes from Newton’s law of gravitation, where the force on a test mass is GMm/r^2; dividing by the test mass m leaves the acceleration g. The stronger the mass M, the greater the pull, but the effect drops rapidly with distance because it follows an inverse-square law, so larger radius means weaker surface gravity for the same mass.
When to use it
Use this when you need the gravity at the surface of a planet, moon, or other roughly spherical body, or to compare how heavy objects would feel on different worlds. It is also useful in astronomy problems involving escape speed, orbital motion near a body, or estimating weights on another planet.
Worked example, step by step
Find the surface gravity of Mars. Given: G = 6.67 × 10^-11 N·m^2/kg^2, M = 6.42 × 10^23 kg, r = 3.39 × 10^6 m Step 1: Substitute into g = GM/r^2 g = (6.67 × 10^-11 × 6.42 × 10^23) / (3.39 × 10^6)^2 Step 2: Multiply the numerator 6.67 × 10^-11 × 6.42 × 10^23 ≈ 4.28 × 10^13 Step 3: Square the radius (3.39 × 10^6)^2 ≈ 1.15 × 10^13 Step 4: Divide g ≈ (4.28 × 10^13) / (1.15 × 10^13) ≈ 3.7 m/s^2 Final answer: Mars’s surface gravity is about 3.7 m/s^2.
Common mistakes to avoid
- Using the diameter instead of the radius in the formula
- Forgetting to square the radius
- Mixing up surface gravity g with gravitational force, which also depends on the object’s mass
Did You Know?
You would weigh nearly 2.5 times more standing on Jupiter's cloud tops than on Earth.
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