Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 2·0 uses

Tidal Force Calculator

Work out the tidal stretch a moon, star or planet exerts across a body: the 2GMr/d³ term, exact near- and far-side values, the transverse squeeze and gradient.

Tidal Force Calculator

Which component
Default is the Moon: GM_Moon = 4902.800118 km³/s² (JPL DE440) divided by G = 6.67430 × 10⁻¹¹ (CODATA 2022). For the Sun use 1.98841 × 10³⁰ kg.
kg
Default is the Moon's mean distance, 384 400 km (NASA). For the Sun use 1.495978707 × 10¹¹ m. Must be larger than the body radius below.
m
Default is Earth's volumetric mean radius, 6371.0084 km (JPL). For a person, use half their height.
m
Only converts the acceleration into a force in newtons. It does not affect the acceleration itself.
kg
Tidal acceleration (selected component)
1.100 × 10⁻⁶
The tidal acceleration of the component you selected, in m/s², as a positive magnitude.
Tidal force on the test mass
7.699 × 10⁻⁵ N
Sign convention, regime and limits
SIGN CONVENTION: every figure is a positive magnitude. The RADIAL components point AWAY from the body's centre on BOTH the near and the far side — the tide stretches the body along the line to the perturber, which is why there are two bulges, not one. The TRANSVERSE component points INWARD: it squeezes the body at 90° to that line, and it is exactly half the linearised radial term. This calculator gives the FORCING only. It is not a tide height and not a tide time: real tides depend on ocean depth, basin resonance, coastline shape and the Earth's own elastic response, none of which is modelled here.
Radial, linearised (2GMr/d³)
1.100 × 10⁻⁶ m/s²
Radial, exact near side
1.128 × 10⁻⁶ m/s²
Radial, exact far side
1.073 × 10⁻⁶ m/s²
Transverse (compressive)
5.499 × 10⁻⁷ m/s²
Tidal gradient (2GM/d³)
1.726 × 10⁻¹³ s⁻²
Linearisation error
2.54

Background.

Most coastlines get two high tides a day, roughly twelve and a half hours apart, even though the Moon is overhead only once. The explanation lies in what a tide actually responds to. The Moon accelerates the whole Earth toward itself at about 3.3 × 10⁻⁵ m/s², but a uniform pull would move everything together and raise no tide at all. What matters is the unevenness: water on the side facing the Moon is pulled a little harder than Earth's centre, and water on the far side a little less. Measured from the centre, both patches are displaced outward, so the ocean is drawn into two bulges and the planet turns beneath them.

That difference, the tidal acceleration, is about 3 % of the Moon's pull and roughly one ten-millionth of the gravity you stand in. Because it is a difference between neighbouring points, it falls off with the cube of distance rather than the square. This one extra power is why the Moon, which pulls on Earth about 179 times more weakly than the Sun, still raises the larger tide.

Enter a perturbing mass, a separation and the radius of the body being stretched, and the calculator returns the textbook linearised value 2GMr ⁄ d³ together with the exact near-side and far-side figures, the inward squeeze at right angles to the line between the bodies, and the gradient per metre. It defaults to the Moon acting on Earth, but the same equations describe Jupiter flexing Io, a star distorting a close planet, or an astronaut falling toward a black hole. The output is the forcing that drives tides; it is not a tide height or a time of high water.

What is tidal force calculator?

Tidal acceleration is the difference between the gravitational acceleration a distant mass produces at one point of an extended body and the acceleration it produces at the body's centre. Subtracting the centre's value removes the part of the pull that moves the whole body, leaving a field that stretches the body along the line to the perturber and compresses it across that line. The stretch is twice the squeeze. The tidal gradient, 2GM ⁄ d³, is the rate at which that difference grows per metre, so any body's tidal acceleration is the gradient times its radius.

How to use this calculator.

  1. Choose the headline component: the linearised radial term for the textbook figure, an exact side when the body is large compared with the distance, or the transverse squeeze. All four are shown regardless.
  2. Enter the perturber's mass in kilograms. The default is the Moon, 7.3458 × 10²² kg; for the Sun use 1.98841 × 10³⁰ kg.
  3. Enter the centre-to-centre distance in metres (384,400 km for the Moon's mean distance, 1.495978707 × 10¹¹ m for the Sun) and the radius of the body being stretched. For a person, use half their height. The distance must exceed the radius.
  4. Enter a test mass only if you want a force in newtons. It scales the force output and leaves the accelerations untouched.
  5. Look at the linearisation error before relying on 2GMr ⁄ d³. If it exceeds 5 % the calculator warns you; quote the exact near-side and far-side values instead.

The formula.

a_t = 2GMr ⁄ d³ · a_near = GM[1⁄(d−r)² − 1⁄d²] · a_far = GM[1⁄d² − 1⁄(d+r)²] · a_⊥ = GMr ⁄ d³

Newton's law gives the acceleration toward a mass M at distance x as GM ⁄ x², with G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² (CODATA 2022). A body of radius r whose centre is at distance d has its near face at d − r and its far face at d + r. Subtracting the centre's acceleration gives the exact near-side value, a_near = GM[1 ⁄ (d − r)² − 1 ⁄ d²], and the exact far-side value, a_far = GM[1 ⁄ d² − 1 ⁄ (d + r)²]. Both point away from the body's centre.

Expanding either for small r ⁄ d gives the same leading term, a_t = 2GMr ⁄ d³. Differentiating an inverse-square law always adds one power of distance, which is where the cube comes from. The next term in the expansion makes the near side stronger and the far side weaker than a_t, so the linearised value sits between them and falls short of the near side by about 1.5 × r ⁄ d. The error output reports that shortfall exactly, as a_near ⁄ a_t − 1. The 5 % warning level, reached at r ⁄ d ≈ 0.032, is a convention chosen for this calculator, not a published threshold.

At 90° to the line of centres the perturber pulls at almost the same strength as at the centre but from a slightly different direction, and the leftover points inward with magnitude a_⊥ = GMr ⁄ d³, exactly half the radial stretch. All results are positive magnitudes; the directions are given in words beside them.

The model treats the perturber as a point mass and the stretched body as rigid, with no ocean dynamics, elastic response or relativity.

A worked example.

Example

Take the Moon acting on Earth. With G = 6.67430 × 10⁻¹¹ and M = 7.3458 × 10²² kg, GM = 4.9028 × 10¹² m³/s². Earth's mean radius is r = 6,371,008.4 m and the Moon's mean distance is d = 384,400,000 m. The numerator 2 × 4.9028 × 10¹² × 6.371 × 10⁶ comes to 6.2472 × 10¹⁹, and d³ is 5.6800 × 10²⁵. Dividing gives a linearised tidal acceleration of 1.0998 × 10⁻⁶ m/s². On a 70 kg test mass that is a force of 7.6989 × 10⁻⁵ N, the weight of about 7.9 milligrams. The exact forms bracket it. The near face, at 378,028,991.6 m, gives 1.1278 × 10⁻⁶ m/s²; the far face, at 390,771,008.4 m, gives 1.0731 × 10⁻⁶ m/s². The linearised figure is 2.5422 % below the near-side value, and the two bulges differ by about 5 % of the linearised value. The transverse squeeze is 5.4992 × 10⁻⁷ m/s², half the stretch, and the gradient is 1.7263 × 10⁻¹³ s⁻². Repeating the sum for the Sun, at 1.98841 × 10³⁰ kg and 1.495978707 × 10¹¹ m, gives 5.0510 × 10⁻⁷ m/s², or 45.92 % of the lunar figure. The Moon's tidal forcing is 2.1775 times the Sun's.

test Mass Kg70
componentradialApproximate
perturbing Mass Kg73,457,892,483,100,000,000,000
body Radius M6,371,008.4
separation M384,400,000

Frequently asked questions.

If the Sun pulls on Earth far harder than the Moon does, why is the Moon's tide bigger?
Direct pull scales as M ⁄ d², and by that measure the Sun wins by a factor of about 179. Tidal stretch scales as M ⁄ d³, and the Sun's distance, about 389 times the Moon's, costs it that extra factor. The result is a solar tide 45.92 % of the lunar one, a ratio of 2.18.
Some tide references say the Moon's tide is 2.5 times the Sun's. Why does this give 2.18?
NOAA's tide primer, Our Restless Tides, gives the ratio as approximately 2.5. Working it out from the gravitational constants and mean distances gives 2.1775, which matches the commonly quoted statement that the solar tide is about 46 % of the lunar one. The calculator reports what the inverse-cube law and those constants produce.
What happens at spring and neap tides?
At new and full Moon the Sun and Moon line up and their forcing adds, to about 1.46 times the lunar figure alone. At the quarter Moons they act at right angles and partly cancel, leaving about 0.54 times. Actual tidal ranges also depend on the ocean's response, so coastal tables differ from this ratio.
Can I use these numbers to predict high tide at my beach?
No. Real tides are the ocean's response to this forcing, shaped by water depth, basin resonance, coastline geometry, friction and weather. They can lag the Moon by hours, and the range varies enormously: the Bay of Fundy exceeds 15 metres. Tide tables are built from long records at each gauge, not from this equation.
Does the Moon's tide stretch my body?
In principle, yes, but immeasurably. For a person 2 metres tall the head-to-toe difference is 3.45 × 10⁻¹³ m/s². Earth's own gravity weakens by about 3.1 × 10⁻⁶ m/s² for every metre you climb, a gradient roughly 18 million times the Moon's, so the Moon's stretch on you is lost completely.
Is spaghettification near a black hole the same effect?
Yes, in its Newtonian form. The stretch grows as 1 ⁄ d³, so close to a compact mass the difference across even a human body can exceed the strength holding it together. Very near a black hole general relativity is needed for accurate numbers, but the picture of radial stretching and transverse squeezing carries over.

References& sources.

  1. [1]NIST, CODATA 2022 recommended value of the Newtonian constant of gravitation: G = 6.674 30(15) × 10⁻¹¹ m³ kg⁻¹ s⁻², relative standard uncertainty 2.2 × 10⁻⁵. Independent, open access. Retrieved 2026-07-29.
  2. [2]NASA JPL Solar System Dynamics, 'Planetary Satellite Physical Parameters': Earth's Moon, GM = 4902.800 ± 0.001 km³/s² and mean radius 1737.4 km, from ephemeris DE440. Source of the default perturbing mass. Independent, open access. Retrieved 2026-07-29.
  3. [3]NASA JPL Solar System Dynamics, 'Astrodynamic Constants': astronomical unit 149 597 870 700 m; heliocentric gravitational constant GM_Sun = 1.327 124 400 412 794 19 × 10²⁰ m³ s⁻² (DE440). Source of the solar comparison figures. Independent, open access. Retrieved 2026-07-29.
  4. [4]NASA JPL Solar System Dynamics, 'Planetary Physical Parameters': Earth volumetric mean radius 6371.0084 ± 0.0001 km. Source of the default body radius. Independent, open access. Retrieved 2026-07-29.
  5. [5]NOAA, Center for Operational Oceanographic Products and Services, 'Our Restless Tides', chapter 4: states that the tide-producing force varies 'in inverse proportion to the third power of the distance' between the bodies. Chapter 3 of the same document states that 'the tide producing force of the moon is approximately 2.5 times that of the sun' — a figure this page does NOT adopt; see the FAQ on the Moon-to-Sun ratio for the recorded disagreement. Independent, open access. Retrieved 2026-07-29.
  6. [6]NASA Science, 'Moon Facts' (page updated 12 February 2026): 'The Moon is an average of 238,855 miles (384,400 kilometers) away.' Source of the default separation. Independent, open access. Retrieved 2026-07-29.

How this page was produced

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Quanta Calculator
Primary sources
6 cited below
Method
a_t = 2GMr ⁄ d³ · a_near = GM[1⁄(d−r)² − 1⁄d²] · a_far = GM[1⁄d² − 1⁄(d+r)²] · a_⊥ = GMr ⁄ d³
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