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Radiation Pressure From Luminosity and Temperature

Estimate pressure from isotropic radiation at a surface and the isotropic thermal radiation pressure.

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

At 300 K, the retained radiation constant gives thermal radiation pressure about 2.04276 µPa.

psurface = s L cos²θ/(4π r² c); pthermal = a T⁴/3.

Worked example

Enter these known values and leave the other values blank.

Source luminosity (total emitted power)
1 L☉
Distance
1 au
Temperature
300 K
Pressure
4.54 μPa
Isotropic thermal radiation pressure
2.043e-15 GPa

Assumptions and limitations

  • Luminosity is total source power, despite the retained internal field name solar_constant. Irradiance follows inverse-square dilution.
  • The surface factor is one for ideal absorption and two for ideal specular reflection; pressure denotes the normal momentum transfer.
  • The retained radiation constant is 7.565767 × 10⁻¹⁶ J/(m³·K⁴). Temperature is absolute; surface angle is restricted to 0–90°.
  • Enter known quantities and clear the unknown to use available inverse branches. Locks retain supplied values. Consistency messages cover fully visible equations; identities with hidden dependencies may require a manual consistency check. Preset values are illustrative approximations, and unit selectors preserve the underlying quantity.

Common questions

How do I solve for a different quantity?

Clear that quantity and supply enough independent known values. Some outputs are one-way or tied to a selected preset, so they cannot determine every input.

References

Explicit physical domains for retained model.

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