Tidal Force
Differential gravitational force across a body.
Formula
Variables
Example
Causes ocean tides
How it works
Tidal force measures how strongly gravity changes from one side of an object to the other, rather than the total pull on its center. It scales with the object’s size because a larger body has a bigger distance difference across it, and it falls off as d^3 because the gradient of gravity gets much weaker very quickly with distance. The factor 2GMm r/d^3 comes from comparing the gravitational acceleration at the near and far sides of the body and taking the difference.
When to use it
Use this when you need the stretching or squeezing effect of gravity on an extended body, such as a moon near a planet, a satellite near a massive planet, or a star passing close to a black hole. It is especially useful in problems involving tidal disruption, Roche limits, and tidal heating.
Worked example, step by step
Find the tidal force on a 2.0×10^3 kg satellite of radius 1.5 m near Jupiter. Given: G = 6.67×10^-11 N·m^2/kg^2, M = 1.90×10^27 kg, m = 2.0×10^3 kg, r = 1.5 m, d = 7.0×10^7 m Use F_t ∝ 2GMmr/d^3 Plug in the values: F_t ≈ 2(6.67×10^-11)(1.90×10^27)(2.0×10^3)(1.5) / (7.0×10^7)^3 Calculate the numerator: 2 × 6.67×10^-11 × 1.90×10^27 × 2.0×10^3 × 1.5 ≈ 7.60×10^20 Calculate the denominator: (7.0×10^7)^3 = 3.43×10^23 Divide: F_t ≈ (7.60×10^20) / (3.43×10^23) ≈ 2.2×10^-3 N Final answer: the tidal force is about 2.2×10^-3 N.
Common mistakes to avoid
- Using the distance to the object’s surface instead of the center-to-center distance d
- Forgetting that tidal effects depend strongly on size r, so larger bodies experience larger differential force
- Treating tidal force as the same as ordinary gravitational force rather than the difference in gravity across the object
Did You Know?
Tidal forces stretch objects because gravity pulls their near side harder than their far side.
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