How much bigger than earth's moon would my habitable moon's primary appear in the nightsky? And is there a way to calculate how bright it would be in lux when it is illuminated in full?

Here is the information for the two scenarios:

The gas giant has a mass of 3 times that of jupiter and about 1.04 times the radius in both cases and it has 4 times the moon's albedo

case 1: The moon is at about 544439 kms from the gas giant

case 2: The moon is at about 1002864 kms from the gas giant

The planet orbits an F-star 1.3 times the mass of the sun at about 2.5-3 billion years of age, the distance is about 2 AUs.


1 Answer 1


Albedo & Reflectance

Scientists can calculate the amount of solar output Earth receives at a given distance from the sun. Earth reflects some of this light into space, reducing the total radiation absorbed. This effect is described by the term albedo, which is a measure of the average amount of light reflected by an object.

Albedo is measured on a scale from zero to one. An object with an albedo of one would reflect all light reaching it, while at zero albedo, all light would be absorbed. Earth's albedo is about 0.39, but changes over time such as cloud cover, ice caps or other surface features change this value. (Source)

Since you know the albedo, you know how much of the star's light will be reflected. The Moon's albedo is about 0.12, so your gas giant's albedo is 0.48, meaning it will reflect 48% of the light that falls on it. How much that is in lux requires knowing how far the planet is from its star and some specifics about the star, which you haven't provided.

Note: Lux is the amount of illumination when one lumen is evenly distributed over one square meter. In other words, it's a measurement that makes sense to an observer (because distance must be involved)... and the only observers you've mentioned in your question are on the moons. That's kinda an ugly computation since part of it (to be accurate) would need to accommodate the penumbra of both moons. Granted, you're probably trying to determine if the reflected light of the primary can sustain life... but calculating lumens would be a whole lot simpler. Nevertheless, it would help if you explained why you need that measurement. There might be a simpler solution.

Angular Size

The angular size of an object is determined uniquely by its actual size and its distance from the observer. For an object of fixed size, the larger the distance, the smaller the angular size. For objects at a fixed distance, the larger the actual size of an object, the larger its angular size. For objects with small angular sizes, such as typical astronomical objects, the precise relationship between angular size, actual size and distance is well approximated by the equation:

angular size = (actual size ÷ distance)

However, when using this equation you must be very careful about the units in which quantities are measured. If the actual size and the distance are measured in the same units (metres or kilometres, or anything else as long as it is used for both quantities), the angular size that you calculate will be in measured units called radians. A radian is equal to a little more than 57° so, in order to obtain angular sizes in degrees, the following approximation can be used (as long as the angular size is not too great):

angular size = 57 × (actual size ÷ distance) (Source)

Your primary is 1.04X the radius of Jupiter (r = 69,911 km) so the "size" of your primary is 145,414.88 km.

  • Moon #1 (distance 544,439 km), angular size of primary = 0.267 radians or 15.3° of the sky.

  • Moon #2 (distance 1,002,864 km), angular size of primary = 0.145 radians or 8.3° of the sky.

Note that I used the first equation and a handy online radians-to-degrees converter for better accuracy. For comparison, Earth's moon has an angular size of about 0.5°. So you have some hefty views.

  • $\begingroup$ Join JBH on Codidact Oh right, apologies. I added the required information, 2 AUs from an F-Star about 1.3 times the mass of the sun 2.5 billion years to 3 billion years since formation. About the reason why I asked for the lux it is because most of the comparative scales I found work in lux and not in lumen. The main reason why I'd like to know is because I wanna get an idea of the gas giant may impact my world's daynight cycle, so really I don't need extreme accuracy an approximation would be okay even if it were just a comparison with earth's full moon and quarter moon. $\endgroup$ Aug 17, 2021 at 9:32
  • 1
    $\begingroup$ @JuimyTheHyena In that case, a simple estimate may do. Not only would your gas giant be 4X brighter when looking at any given point on the surface of the planet, it's filling (in the case of the 1st moon) 31X the amount of sky (assuming a "full moon" kind of condition). Result: 124X the illumination of a full moon on Earth. Earth's moon's lux (full moon) is generally about 0.075 lux, but it can reach 0.32 lux at perigee w/a full moon. Let's stick with the average. 0.075 * 124 = 9.3 lux. It's a reasonable rough estimate. Moon #2's would be 5 lux. Sunlight on Earth (equator/noon) is 120,000 lux. $\endgroup$
    – JBH
    Aug 18, 2021 at 19:44
  • 1
    $\begingroup$ Thank you, I'd consider this settled then $\endgroup$ Aug 18, 2021 at 22:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .