# Orbit darkness time

The **orbit darkness time** determines the time a craft is staying in the shadow of an orbited object. The exact time depends on the current configuration of the orbit and the moons.

This page will give an expression for the worst-case scenario of how long a craft will be in darkness during an orbit. This information can be used to determine how many batteries are needed for a craft to remain powered during the dark portion of orbit.

## Contents

## General Result

Here is the result for calculating the longest amount of time spent in darkness (in units of seconds):

where is the semi-major axis, the semi-minor axis, the specific angular momentum, the eccentricity, and the radius of the planet or moon. For reference these terms can be calculated by knowing the apoapsis (Ap), periapsis (Pe) and body to orbit:

- , the apoapsis measured from the center of the body
- , the periapsis measured from the center of the body
- , the semi-major axis
- , the semi-minor axis
- , the eccentricity
- , the semi-latus rectum of the orbital ellipse
- , the gravitational parameter
- , the specific angular momentum

The following parameters are the minimum requirements to calculate the values from above:

- , planned apoapsis of the craft's orbit from the surface of the body
- : planned periapsis of the craft's orbit from the surface of the body
- : radius of body to orbit (a.k.a. equatorial radius)
- : mass of body to orbit

The radius and mass of a body can be obtained by visiting the page of the body (e.g. Kerbin) or by visiting Kerbol System/Table which contain the mass and radius of all celestial bodies. The actual page of a body also shows the gravitational parameter directly so it doesn't need to be multiplied by *G*.

When using kilometer for the orbital parameters the gravitational parameter needs to be divided by 1000^{3}. For example Kerbin has an equatorial radius 600 000 m or 600 km. Its gravitational parameter is 3.5316000×10^{12} m^{3}/s^{2} or 3.5316000×10^{3} km^{3}/s^{2}.

### Simplified Result

For (almost) circular orbits where the eccentricity is very small this equation becomes rather simple:

This simpler result only works when the apoapsis and periapsis are very close to the same altitude.

## Limitations

This method assumes the orbit is an ellipse/circle which is not changing or being altered by other bodies. It also assumes the sun's rays are parallel across the orbiting planet, although all bodies are small enough and far enough from the sun for this to be nearly true. The method does not take into account darkness caused by eclipses of a different body than the orbited body, for example, orbiting Laythe but Jool blocks the sun.

The method gives the longest amount of time spent in darkness, which for some orbits (e.g. polar orbits), will only be experienced periodically (see beta angle). However, it is a good idea to plan on the worst-case amount of time in darkness.

## Application

By entering the equations into a program such as Microsoft Excel or going to this calculator page, the darkness time can be calculated for various orbits around any planet or moon. Knowing the amount of time spent in darkness and the energy drain, the battery storage can be calculated by

- is the battery storage required in the electricity unit (⚡)
- is the rate of energy use in units of (⚡/sec)
- is the darkness time in seconds.

The darkness time along with the total orbital period can also be used to calculate how many solar panels are needed to recharge the batteries before the next cycle of darkness. However, the time spent in sunlight is usually quite long, and even a single small solar panel will recharge most any craft's batteries before the next darkness cycle.

## Examples

### Orbiting Kerbin with 100 km circular orbit

*T _{d} = 642 s (10 minutes 42 seconds)*

Using 0.05 ⚡/s for the RC-001S and 0.04 ⚡/s for the Illuminator Mk1 for a total drain of 0.09 ⚡/s:

*B = Ė _{d} 0.09 ⚡/s · 642 s = 57.8 ⚡*

So at least 57.8 ⚡ of electricity is needed in storage to make it through the darkness for this orbit around Kerbin, using 0.09 ⚡ per second. The battery with the next largest amount of storage (as of 0.90^{[outdated]}) is the Z-100 with 100 ⚡ of storage. The RC-001S can also store 15 ⚡ so in theory only 42.8 ⚡ are required.

### Orbiting Jool with apoapsis 700 km and periapsis 200 km

*T _{d} = 2403 s (40 minutes 3 seconds)*

Using 0.05 ⚡/s for the RC-001S and 0.16 ⚡/s for 4x Illuminator Mk1 for a total drain of 0.21 ⚡/s:

*B = Ė _{d} 0.21 ⚡/s · 2403 s ≈ 505 ⚡*

So at least 505 ⚡ of electricity is needed in storage to make it through the darkness for this orbit around Jool, using 0.21 ⚡ per second.

### Circular orbits around all planets and moons

The following table contains the darkness times for the planets and moons. The orbits are circular and have an altitude of 20% of the body's radius.

Body | Darkness time | Orbital period | Altitude above sea level | |
---|---|---|---|---|

Moho | 788 s | 13 m 8.4 s | 41 m 54.3 s | 50 000 m |

Eve | 531 s | 8 m 50.6 s | 28 m 12.2 s | 140 000 m |

Gilly | 1 333 s | 22 m 13.3 s | 1 h 10 m 52.1 s | 2 600 m |

Kerbin | 641 s | 10 m 40.5 s | 34 m 2.6 s | 120 000 m |

Mun | 908 s | 15 m 7.6 s | 48 m 14.5 s | 40 000 m |

Minmus | 906 s | 15 m 5.8 s | 48 m 8.7 s | 12 000 m |

Duna | 854 s | 14 m 14 s | 45 m 23.5 s | 64 000 m |

Ike | 891 s | 14 m 50.9 s | 47 m 21.1 s | 26 000 m |

Dres | 906 s | 15 m 5.8 s | 48 m 8.7 s | 27 600 m |

Jool | 2 265 s | 37 m 44.6 s | 2 h 0 m 21.8 s | 1 200 000 m |

Laythe | 654 s | 10 m 53.7 s | 34 m 44.8 s | 100 000 m |

Vall | 934 s | 15 m 34.3 s | 49 m 39.5 s | 60 000 m |

Tylo | 716 s | 11 m 56.1 s | 38 m 3.7 s | 120 000 m |

Pol | 890 s | 14 m 49.8 s | 47 m 17.6 s | 8 800 m |

Bop | 861 s | 14 m 20.7 s | 45 m 44.7 s | 13 000 m |

Eeloo | 914 s | 15 m 13.7 s | 48 m 33.8 s | 42 000 m |