# Using parallax to find distance to the moon

I want to use the parallax method (from scratch) to find the distance between the earth and the moon. To get a significant parallax, I need to choose a large baseline. I live in India and my friend in the US and so the baseline will be about $$13,000$$ Km. That's quite a large so I suppose I would get a significant parallax.

What I don't think of, is taking of angle measurement? How can I measure the angle $$\alpha$$?

• Don’t forget that while ground distance is around 13,000 km, physical distance between you and your friend is less, as you need to account for Earth’s curvature, as your drawing makes clear. The Earth’s diameter is 12,756 km. Also, you can’t directly measure α, as this would imply you’re on the Moon, but you can measure the other two angles… May 8, 2021 at 5:59
• How can I, Can you show through figure and little calculation? And method through which I can measure those angle. May 8, 2021 at 6:41
• You basically need to compare where the moon is on the background field of fixed stars, for you and you friend (at the same time). Since the fixed stars are effectively infinitely far away, they appear at the same angles for both you and your friend. Take a picture each with the moon and some bright stars, find out the angular distance between the stars from a catalogue or software, and compare the moon's parallax to that in your pictures. May 8, 2021 at 6:51
• Isn't any other way, where I don't use the catalogs? May 8, 2021 at 7:46
• @YoungKindaichi: You don't need calatogues. You just need to compare the two photos with each other. (In fact, you don't even need a friend; you can take two pictures at different times.) May 8, 2021 at 20:42

I am sure that measuring the angle compared with distant stars is the best approach, but you can in fact work out the angle even without looking at distant stars, if you assume the Earth is spherical and you know its radius, and you know both the length and the direction of the great circle between you & your friend.

Here's a terrible drawing: So here, the radius of the Earth is $$R$$, and the great-circle distance between you and your friend is $$l$$. So the angle between you and your friend is $$\theta = l/R$$ (radians).

Now, you need to agree both to measure the Moon's angle above the horizon at a particularly good moment: that moment is when it crosses the plane defined by you, your friend, and the centre of the Earth. In other words, you need to measure its angle above the horizon as it crosses the great circle between you. This is annoying, because you have to know the direction of the great circle, but it's also nice because you don't need a clock.

So, when you measure its angle above the horizon you get two angles, $$\psi$$ and $$\phi$$. And now you have the quadrilateral I drew below the main diagram, and you know three of its angles and two of its sides. The total angle in a quadrilateral is $$2\pi$$, so this tells you the fourth angle, which is $$\alpha = \pi - (\theta + \psi + \phi)$$. And you can then go on to solve the quadrilateral for the distance to the Moon, $$d$$ (I am leaving this out because I can't remember how to do it!)

Note that this will only work if the Moon is above the horizon for both of you when it crosses the great circle: it may not be I think.

• Thanks a lot! but still, the main problem here is, How we determine the time when the moon passes through the great circle of me and my friend? May 8, 2021 at 13:27
• @YoungKindaichi: that's hard in practice, but so long as you know both where you and your friend are (and you need to know that to make any progress at all) then you can work out the great circle between you, and in particular you'll know which direction it points from where you are. Then you just need to wait until the Moon is directly over that compass point.
– user38308
May 8, 2021 at 14:42
• Still, You see, There is about $9$ Hour of difference between the two cities. So I don't think, there are many hours in which the moon is visible to both of us. May 8, 2021 at 14:53
• Yes, you're right: it needs to be above the horizon for both of you when it crosses the great circle and that may not be true. The time & start based one (other answer) doesn't have that limitation I think, so it's better probably. I've amended my answer to say that.
– user38308
May 8, 2021 at 15:03
• The moon being in the plane that contains the two observers and the center of the earth certainly simplifies things, but it does not seem necessary. Each observer can in advance figure out the straight line between them and the other observer and then measure the angle of the moon to that line. (Not sure what the required accuracy is, and whether it can be practically achieved.) May 9, 2021 at 17:48

You compare the position of the moon with that of the background of stars. If you both take a picture of the moon at the same time, and compare the angle that it makes with a distant star the moon will be in a different position, and from this you can calculate the angle. Here is your diagram showing, in slightly simplified way what I mean. In practice the stars won't be exactly on your baseline so you'll need to do some measuring of the position from the photos as in these simulated images. • And with the moon you want to make sure to really take photos at the same time. So make sure to synchronize your clocks reasonably well. It moves with about 30" per minute. Take that into account for your error analysis, given the accuracy of your concurrency of the photos to compare May 8, 2021 at 9:53
• Yes, that's alright with the method but there are two problems, first, there is about a $9$ Hour difference between the two cities so it's hard to find the moon with stars. And even if it does, there is too much light pollution and I don't have that sort of camera to capture such an image for measurement. May 8, 2021 at 13:30
• Yes. Nobody said it was easy. May 8, 2021 at 13:34
• I have the urge to put your example images in my stereopticon/Viewmaster... May 8, 2021 at 22:25
• imaginary.org/sites/default/files/… May 8, 2021 at 22:59

While the right answer is measuring the angles of the two telescopes with respect to each other this is not trivial.

The technique used here by Scott Manley side steps that by picking a time when two objects with a known angular distance between them will be in the same photo, and using that distance as a calibration to measure the moon parallax. He used planets which made the process easier but narrows the usable windows to achieve it. An earlier video did it with stars.

Second video also discusses a method using two photos taken from the same location at moonrise and 'moon noon' and determining the change in apparent size due to being closer due to Earth curvature.

To measure the distance $$D$$ of moon observe it from two different positions on the Earth (observatories in India and in the US) $$A$$ and $$B$$, separated by distance $$AB = b$$ at the same time. As the moon is very far away, $$\frac{b}{D}<<1$$ and therefore, $$θ$$ is very small. Then we approximately take $$AB$$ as an arc of length $$b$$ of a circle with center at $$S$$ and the distance $$D$$ as the radius $$AS = BS$$ so that $$AB = b = D θ$$. $$D=\frac{b}{θ}$$

Having determined $$D$$, we can apply a similar method to determine the angular diameter of the moon. If $$d$$ is the diameter of the moon and $$α$$ the angular diameter of the moon (the angle subtended by $$d$$ at the earth), we have: $$α=d/D$$. Using user38308's approach:

First, find the distance between you and your friend, knowing lat/long of your positions. You need to solve a spherical triangle with vertices at the north pole and the locations of you and your friend. Piece of cake. You have 2 sides And their included angle. The (included) angle at the pole is the difference in your longitudes. The sides are 90 degrees minus the latitudes of you and your friend. Using the law of cosines for spherical triangles find the arc distance between you and your friend. Don't stop there. Find the angles at you and your friend using the law of sines for spherical triangles. You'll need them later. Note that this calculation can/should be done months ahead of time. Note also that the sum of the angles in a spherical triangle is usually more than 180 degrees.

Good news is that we're done with spherical trig, and moving on to plane trig for the rest of the calculation. You and your friend are trying to find the altitude of the Moon when its center is on the plane having you, your friend and the center of the Earth. That will happen, for both of you, when its azimuth is equal to the angles of the spherical triangle (prior step) and, maybe, plus or minus 180 degrees.

Still with me?

Then we move on to the isosceles triangle with vertices at you, your friend and the center of the Earth. You know the angle at the center of the Earth from the spherical triangle analysis. You can find the unequal side knowing the other 2 sides (Earth radii) and that angle (memorized formula or law of cosines for plane triangles). That leaves you with a triangle having sides the chord between you and your friend, and the distances from you and your friend to the center of the moon.

Home stretch.

The Moon will pass through the plane of you, your friend and the center of the Earth when its azimuth is equal to the angles (or angles plus or minus 180 degrees) you calculated previously. You both should see that happen at about the same time, but a few minutes difference shouldn't matter. Measure the Moon's altitude at that time. After correcting for refraction and semi-diameter, you'll both find the Moon's altitude is less than you'd see if it was a very distant object. The angles of the triangle at you and your friend will be the observed altitudes minus half the angle of the isosceles triangle included between its radii. (Needs a picture - watch this space.) You now Know the parallax angle: 180 degrees minus the two angles at you and your friend. You can find the distances from you and your friend to the center of the Moon with the law of sines for plane triangles. You could carry on to find the distance between the Earth/Moon centers, but you wouldn't get a much more accurate answer.