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The following information is from ISRO's webpage on "Detection of Argon-40 in the lunar exosphere" (Textual information relevant to my question is in bold face):

The Chandra’s Atmospheric Composition Explorer-2 (CHACE-2) payload aboard the Chandrayaan-2 orbiter, is a neutral mass spectrometer-based payload which can detect constituents in the lunar neutral exosphere in the range of 1-300 amu (atomic mass unit). As part of its early operation, it has detected $^{40}$Ar in the lunar exosphere from an altitude of ~100 km, capturing the day-night variations of concentration. $^{40}$Ar being a condensable gas at the temperatures and pressures that prevail on the lunar surface, condenses during lunar night. After lunar dawn, the $^{40}$Ar starts getting released to the lunar exosphere (blue shaded region in figure).

enter image description here

Variation of Argon-40 observed during one orbit of Chandrayaan-2 during dayside and nightside of the Moon. The observed partial pressure has to be refined for the background and other effects to infer the density of lunar exospheric argon. The observations when Chandrayaan-2 was on the nightside is indicated by the black solid rectangle at the top of the panel and the two vertical dashed lines. Being in a polar orbit, Chandrayaan-2 enters the dayside of the Moon crossing the north pole, traverses through the dayside and enters the nightside after crossing the southpole.

Source : Detection of Argon-40 in the lunar exosphere

We know that a satellite in a circular orbit, around a spherical celestial body of approximately uniform density, moves with constant speed (not constant velocity). So we must expect the time spent in the dayside to be equal to that of in the night side of the celestial body.

In the above figure, it can be seen that the duration of the satellite on the night side is comparatively lesser than the time spent in the dayside. I am not sure why this is so. I checked whether the time intervals on the $x$ axis or time axis are uniform or not. Yes! They are uniform. So why is there a huge difference between the duration of time spent in the dayside and night side, by the spacecraft?

Calculation of Orbital Time Period:

To check whether the graph represents the complete data of one full orbit, and for further analysis, I calculated the orbital time period (because no data was available on the Internet regarding this) of Chandrayaan-2 Orbiter, under the following assumptions:

  • The orbit is nearly circular

  • Moon is of uniform density and it is spherical

So, Time period $T$ is given by the following formula:

$$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$$

where $r$ is the radius of the satellite's orbit (sum of the radius of the Moon and altitude of orbit from the surface), $G$ is the Universal Gravitational Constant $(6.67408\times10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}})$, $M$ is the mass of the celestial body (and here it's our Moon).

Here,

$r=17.371\times10^5\ \mathrm m+10^5\ \mathrm m=1.8371\times10^6\ \mathrm m$

$M=7.34767309\times10^{22}\ \mathrm{kg}$

On substituting the values in the above equation for calculating the time period of the satellite, we get:

$$T=\frac{2\pi (1.8371\times10^6)^{3/2}}{\sqrt{6.67408\times10^{-11}\times7.34767309\times10^{22}}}\ \mathrm s$$

$$T= 7064.94\ \mathrm s$$

$$T= 1.96\ \mathrm h$$

$$T\approx 2\ \mathrm h$$

So, Chandrayaan-2 Orbiter completes a full orbit in about 2 hours. It spends half of this time i.e., 1 hour on the dayside and the other half on the nightside. From this, we can also infer that the graph represents data obtained in one full orbit, as the total time sums up to approximately 2 hours. There is also data consistency between the partial pressure of the gas at the start of the orbit and at the completion of the orbit. But from the same graph, we can see, the time spent on the night side is less than half an hour, instead of one full hour.

So, Why is the duration of time spent in the dayside greater than that of the night side of the Moon for Chandrayaan-2 Orbiter, as seen in the graph?

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  • $\begingroup$ There is not only the time spent in the dayside and that spent in the night side, there are also the times spent in dawn and dusk. If the period is about 2 hours, less than 1 hour may be spent on the day side or the night side. $\endgroup$
    – Uwe
    Nov 2, 2019 at 10:44
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    $\begingroup$ This is a really cool question, there's a lot going on here. $\endgroup$
    – uhoh
    Nov 2, 2019 at 11:04

3 Answers 3

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The degree of orbital shadowing experienced by an orbiting object with small orbital altitude is determined by its beta angle (normally used in reference to LEO objects but the concept applies to lunar orbiters as well).

The angle is taken between the satellite's orbital plane and the vector to the Sun. Depending on the value of the beta angle, a satellite can spend up to 100% of its time in sunlight.

Source: https://en.m.wikipedia.org/wiki/Beta_angle

enter image description here

Image from http://www.tak2000.com/data/planets/earth.htm

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  • $\begingroup$ Thank you for your answer. If the day and night time is defined with respect to the spacecraft being in sunshine or in the shadow of the Moon, then the rest of the explanation given in the website is bit contradictory and it forces us to assume it to be day and night time defined with respect to the lunar surface. For example, the partial pressure of the gas depends upon the surface temperature (whether it's day or night on the surface) and not with respect to the spacecraft being in the Moon's shadow or not. Kindly clarify this issue. $\endgroup$
    – Vishnu
    Nov 3, 2019 at 14:18
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There is no circular orbit that has a share of 50:50 between night and day. The possible times are a bit less than 50% to 0% night or, respectively, a bit more than 50% day to 100% day.

The two extreme cases are:

  • an orbit that is aligned with the terminator (the border between night and day on the surface) is in perpetual daylight.

  • an orbit that passes above the point where the Sun is in zenith spends almost half of the time on the night-side. The "almost" comes from the fact that due to the height of the orbit the satellite is still in sunlight after it passed over the terminator.

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    $\begingroup$ And then you get to elliptical orbits, which only intensify the extrema (towards more and more daytime). $\endgroup$
    – Draco18s no longer trusts SE
    Nov 2, 2019 at 15:32
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It all depends on how you define "dayside" and "nightside", and how you define "entering" or "exiting" either one of them for a satellite.

I suppose a big part of the confusion comes from this statement:

Being in a polar orbit, Chandrayaan-2 enters the dayside of the Moon crossing the north pole, traverses through the dayside and enters the nightside after crossing the southpole.

The above statement makes it sound like they are defining dayside and nightside as two perfect halves of the moon, and "entering" the dayside as flying 100km exactly above the dividing line between the two. If that was the case, then we would of course expect the satellite to spend exactly half of its orbit time "in the dayside" and half "in the nightside", or, more correctly put, "half the time over the dayside of the moon and half the time over the nightside of the moon".

But I am afraid that this is not what the sentence actually means. Notice that the statement is inconsistent with itself: it says that Chandrayaan-2 enters the nightside after crossing the south pole, not at the moment of crossing the south pole. Duh?

What is most probably happening is that this sentence was written in a completely haphazard way, without any attempt to use precise terms, without even much regard to reality. By nightside and dayside they are most probably referring to whether the satellite is in the sunlight or in the shadow of the moon, not whether it is day or night at the spot on the surface 100 km below the satellite.

So, as others have already pointed out, when flying at 100 km above the surface, the satellite sees the sun for a lot more time than the surface right under the satellite does.

The same can even hold true with airplanes flying at 10 km over earth: there are brief periods of time during the evening or the early morning when it is possible to see the sun from your window, but you look down and the surface is dark.

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    $\begingroup$ Also note: one of your initial assumptions is that "The orbit is nearly circular with a radius of 100 km". Of course, that's not true, and you know it, and later on in your math you use the correct radius, which is ~1900 km. $\endgroup$
    – Mike Nakis
    Nov 2, 2019 at 18:34
  • $\begingroup$ With respect to this statement "By nightside and dayside they are most probably referring to whether the satellite is in the sunlight or in the shadow of the moon, not whether it is day or night at the spot on the surface 100 km below the satellite." I understand if this is the way they define day and night, the length of the day is more than that of the night for the spacecraft. But, from the explanation in the webpage mentioned, the properties associated with day and night time relates with respect to moon and not with respect to the spacecraft. $\endgroup$
    – Vishnu
    Nov 3, 2019 at 14:09
  • $\begingroup$ For example, the partial pressure drops in night time of moon and I think it doesn't matter whether the spacecraft is in sunlight or in the shadow of the Moon. The partial pressure is a result of what happens at the surface (whether it's day or night on the Moon). Kindly explain this misalignment with the explanation in that website. $\endgroup$
    – Vishnu
    Nov 3, 2019 at 14:10
  • $\begingroup$ @Intellex okay, the discussion is now starting to become specific to Argon-40, about which I know close to nothing, (only that if you live in a one or two story house you should be airing it a couple of times per year,) but my guess is that when they speak of detection of Argon-40 in the lunar exosphere they really mean just that: they are only concerned with what's going on in the exosphere. Not what's happening on the surface. (Continued...) $\endgroup$
    – Mike Nakis
    Nov 3, 2019 at 14:23
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    $\begingroup$ I think, your last two comments contradict slightly with each other. If gas density is low, heat propagation is dead slow. I agree with this. But, as mentioned in the 3rd comment, I don't think heat will have effects on individual molecules rather than the bulk. So, the variation in the partial pressure must be related to the surface phenomenon. However, if the day and night is defined with respect to the spacecraft, your answer gives the best explanation. I also agree, the terms are not well defined in the website and cause confusion if we dig deep. $\endgroup$
    – Vishnu
    Nov 4, 2019 at 7:20

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