Tidally locked planets keep one hemisphere facing their star (permanent day) while the opposite hemisphere remains in perpetual night. The strongest heating occurs at the substellar point (the point directly under the star). Moving away from that point toward the terminator (the day–night boundary), incoming starlight arrives at a lower angle and the absorbed energy drops. The result is often imagined as a circumplanetary “twilight” band where temperatures may fall in a comfortable range for liquid water, agriculture, or settlement.
This calculator estimates the angular width, surface distance, and surface area of the region on the dayside where the surface temperature lies between your chosen minimum and maximum comfortable temperatures. It uses a deliberately simplified energy-balance relation intended for quick exploration and worldbuilding rather than climate prediction.
Inputs:
Geometry:
cos(θ) ranges from 1 down to 0.A common first-order approximation for radiative equilibrium on the illuminated hemisphere is that absorbed flux scales with cos(θ), while equilibrium temperature scales with the fourth root of flux. This yields:
Solving for angle at a chosen temperature T:
cos(θ) = (T / T_s)^4
Then:
θ_hot = arccos((Tmax/Ts)^4)θcold = arccos((Tmin/Ts)^4)The ring’s angular thickness on the dayside is Δθ = θcold - θhot (in radians or degrees, depending on how you report it). The approximate surface width along the ground is:
width_km = R × Δθ (radians)
On a sphere, the area between two zenith angles measured from the substellar point is:
A = 2π R^2 (cos(θ_hot) - cos(θcold))
This is the area of the dayside band whose temperatures fall between Tmax and Tmin under the model assumptions.
Suppose a tidally locked rocky planet has:
Compute cosine values:
cos(θ_hot) = (310/400)^4 ≈ 0.361cos(θcold) = (270/400)^4 ≈ 0.208Angles (degrees):
θ_hot ≈ arccos(0.361) ≈ 68.8°θcold ≈ arccos(0.208) ≈ 78.0°Angular thickness: Δθ ≈ 9.2° ≈ 0.161 rad. Surface width: width ≈ 6371 × 0.161 ≈ 1030 km.
Area:
A = 2π R^2 (0.361 - 0.208) ≈ 2π (6371^2) (0.153) ≈ 39 million km^2 (order-of-magnitude).
Interpretation: under this simplified model, a sizable belt near the terminator stays within 270–310 K. In a worldbuilding context, that could support a broad “ring civilization” with large agricultural area—if the atmospheric and circulation assumptions below are reasonably satisfied.
| Quantity | What it measures | Why it matters |
|---|---|---|
| Hot-edge angle (θhot) | How far from the substellar point you must go before it cools to Tmax | Defines the inner boundary where overheating becomes a problem |
| Cold-edge angle (θcold) | How far you can go before dropping below Tmin | Defines the outer boundary where freezing becomes a problem |
| Ring width (km) | Surface distance between the two boundaries | Useful for planning travel, infrastructure, and biome size |
| Ring area (km²) | Total dayside band area within the target temperatures | Proxy for total habitable/usable real estate |
cos(θ). Real planets include greenhouse effects and wavelength-dependent absorption.(T/Ts)^4 > 1 or < 0 are physically invalid for this model and should be treated as “no solution” or clamped boundaries.The cosine-based insolation scaling and T ∝ F1/4 radiative equilibrium relationship are standard tools in introductory planetary energy balance modeling and are commonly used as first approximations in exoplanet climate discussions.