Tidal Heating Power Calculator
Estimate tidal heating power in an orbiting moon from its radius, orbital eccentricity and period, rigidity, and the mass of the primary body.
Tidal Heating Power Calculator
Calculator
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Formula: Ė = (21/2) × (k₂/Q) × R⁵ n⁵ e² / G, with n = 2π/P
Worked example — Tidal heating: 9.3e13 W (93 TW) | Surface heat flux: 2.24 W/m2
Formula
Ė = (21/2) × (k₂/Q) × R⁵ n⁵ e² / G, with n = 2π/P
Leading-order tidal dissipation in a synchronously rotating satellite on a slightly eccentric orbit. R is the body's radius, n its mean motion (2π divided by the orbital period), e the orbital eccentricity and G the gravitational constant. The ratio k₂/Q combines the degree-two Love number, which says how easily the body deforms, with the quality factor Q, which says how lossy that deformation is; this tool assumes k₂/Q = 0.015, a typical silicate value. The equivalent form (21/2)(k₂/Q)·G·M_p²·n·R⁵·e²/a⁶ reduces to this one through Kepler's third law, n²a³ = GM_p. Because power scales as e² and as the fifth power of both radius and mean motion, a large body on a short-period, slightly eccentric orbit dominates.
Worked Examples
Example 1: Io: The Most Volcanically Active Body in the Solar System
Problem:Io has a radius of 1,821 km, a forced eccentricity of 0.0041 and an orbital period of 1.769 days (152,853 s). Using k2/Q = 0.015, find the tidal heating power and the resulting surface heat flux.
Solution:Mean motion n = 2 pi / P = 6.2832 / 152,853 = 4.111e-5 rad/s E_dot = (21/2) x (k2/Q) x R^5 x n^5 x e^2 / G R^5 = (1.821e6)^5 = 2.003e31 m^5 n^5 = (4.111e-5)^5 = 1.174e-22 s^-5 e^2 = 0.0041^2 = 1.681e-5 E_dot = 10.5 x 0.015 x 2.003e31 x 1.174e-22 x 1.681e-5 / 6.674e-11 = 9.32e13 W = 93 TW Surface heat flux = E_dot / (4 pi R^2) = 9.32e13 / 4.168e13 = 2.24 W/m2, about 26 times Earth's mean geothermal flux of 0.087 W/m2.
Result:Tidal heating: 9.3e13 W (93 TW) | Surface heat flux: 2.24 W/m2
Example 2: Europa: Enough Heat for a Subsurface Ocean
Problem:Europa has a radius of 1,561 km, eccentricity 0.0094 and an orbital period of 3.551 days (306,822 s). Compare its tidal heating with Io's under the same k2/Q = 0.015.
Solution:n = 6.2832 / 306,822 = 2.048e-5 rad/s, half Io's mean motion E_dot = 10.5 x 0.015 x (1.561e6)^5 x (2.048e-5)^5 x 0.0094^2 / 6.674e-11 = 6.96e12 W = 7.0 TW Surface heat flux = 6.96e12 / (4 pi x (1.561e6)^2) = 0.23 W/m2 Europa's eccentricity is more than twice Io's, which alone would raise the heating five-fold, but the n^5 term costs a factor of 2.01^5 = 32.5 and the smaller radius costs another factor of (1821/1561)^5 = 2.2, so the net is roughly one thirteenth of Io.
Result:Tidal heating: 7.0 TW | Surface heat flux: 0.23 W/m2 | About 1/13 of Io
Example 3: Enceladus: A Small Moon With a Large Plume
Problem:Enceladus has a radius of only 252 km, eccentricity 0.0047 and an orbital period of 1.370 days (118,386 s). What does the leading-order model predict, and how does it compare with observation?
Solution:n = 6.2832 / 118,386 = 5.307e-5 rad/s, slightly faster than Io E_dot = 10.5 x 0.015 x (2.521e5)^5 x (5.307e-5)^5 x 0.0047^2 / 6.674e-11 = 2.24e10 W = 0.022 TW = 22 GW Surface heat flux = 2.24e10 / (4 pi x (2.521e5)^2) = 0.028 W/m2 The R^5 term is decisive: Enceladus is 7.2 times smaller in radius than Io, which costs 7.2^5 = 19,000 in power. Cassini measured roughly 10 GW emerging from the south polar terrain alone, so a homogeneous-body estimate of this order is reasonable, though the real dissipation is concentrated in a thin ice shell over a global ocean rather than spread uniformly.
Result:Tidal heating: 2.2e10 W (22 GW) | Surface heat flux: 0.028 W/m2
Frequently Asked Questions
Why is Io so volcanic?
Jupiter's immense gravity and Io's forced orbital eccentricity (from resonance with Europa and Ganymede) generate ~100 TW of tidal heating — enough to drive intense volcanism.
Which other moons experience significant tidal heating?
Europa (Jupiter) has tidal heating of about 0.1-1 TW that likely maintains a liquid water ocean beneath its ice shell, making it a prime target for life. Enceladus (Saturn) has only about 16 GW of tidal heating but its small size and high porosity concentrate that heat enough to drive active geysers of water vapor and ice. Titan and Ganymede experience much weaker tidal heating due to their orbital parameters.
How does orbital eccentricity drive tidal heating?
A moon in a circular orbit experiences tidal forces that are constant in direction and magnitude, causing static deformation without energy dissipation. In an eccentric orbit, the tidal force varies throughout the orbit — stronger at periapsis, weaker at apoapsis — causing the moon to flex periodically. This flexing generates internal friction and heat, proportional to the square of eccentricity. The 4:2:1 Laplace resonance between Io, Europa, and Ganymede continuously replenishes Io's orbital eccentricity.
What is the quality factor Q in tidal heating and what does a low Q mean?
The tidal quality factor Q (similar to Q in damped oscillators) describes how efficiently a body dissipates tidal energy. A low Q means high dissipation and more tidal heating — rocky, partially molten bodies like Io have low Q. A high Q means little dissipation — rigid icy bodies or gas giants have high Q. The ratio k₂/Q (Love number over quality factor) combines both the deformability and dissipation efficiency, directly controlling tidal heating rate.
Why is tidal heating so sensitive to the moon's radius and mean motion?
The dissipation rate carries R⁵n⁵, so both terms are fifth powers. Doubling the radius raises the heating by a factor of 32, because a larger body presents a much bigger tidal bulge and a larger volume of flexing material. Mean motion n is 2π divided by the orbital period, and Kepler's third law makes n proportional to a^−3⁄2, so n⁵ varies as a^−7.5. Orbital distance is therefore the single most decisive parameter.
How does tidal heating compare with radiogenic heating inside a moon?
Radiogenic heat comes from the slow decay of uranium, thorium and potassium-40 in silicate rock and declines over billions of years. Tidal heating is set by the current orbit, so it can be far larger and can persist or even grow. Io radiates on the order of 2 W/m², more than twenty times Earth's mean geothermal flux of about 0.087 W/m² — a gap radioactive decay alone cannot explain for a body that small.
Why does a synchronously rotating moon still flex if it always shows the same face?
Synchronous rotation is uniform, but orbital motion on an eccentric path is not. The moon sweeps faster near periapsis than near apoapsis, so the sub-primary point swings back and forth in longitude — the optical and physical libration. At the same time the changing separation makes the bulge grow and shrink radially. Both the librational and radial components deform the interior once per orbit, and the friction from that deformation is the heat source.
What is the feedback between tidal dissipation and orbital eccentricity?
Dissipation inside the satellite removes orbital energy and damps eccentricity, so an isolated moon would circularise and its heating would fade. Mean-motion resonances counter this: the 4:2:1 Laplace resonance between Io, Europa and Ganymede repeatedly forces Io's eccentricity back up. The moon settles near an equilibrium eccentricity where resonant pumping balances tidal damping, which is why the heating in the e² term can be sustained over geological time.
How does tidal heating sustain a subsurface ocean beneath an ice shell?
Ice is a poor conductor, so modest dissipation trapped under a shell can hold a layer above the melting point. Heat is generated in two places: the silicate interior and the viscoelastic ice shell itself, where libration and obliquity forcing concentrate strain near the base and at fractures. If dissipation exceeds what conduction through the shell can remove, the ocean persists — the argument behind Europa's ocean and the Enceladus south-polar plumes.
Can this calculation be applied to close-in exoplanets?
Yes, with care. The same k₂/Q framework explains why short-period planets circularise quickly and become tidally locked, and why residual eccentricity in some systems implies ongoing internal heating. The caution is that k₂ and Q are poorly constrained for objects we cannot probe, and the e² expansion assumes small eccentricity. For a strongly eccentric planet the full frequency-dependent tidal model, not this leading-order expression, is required.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer · Editorial policy
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