Measuring the distance to Mars using simple equipment (UPDATE: first estimates of distance!!)

[Gargamel]

Eagerly awaiting the results……

[JoeSchmuckatelli]

1 hour ago, Gargamel said:

Eagerly awaiting the results……

Also a nod for TOTM? 

[Starshot]

22 hours ago, cubinator said:

A more perfect opportunity is unlikely to come by chance,

What are the conditions for these kinds of experiment, just occultation?

[cubinator]

1 hour ago, Gargamel said:

Eagerly awaiting the results……

You will have to wait patiently, as I need to write a report on a far less exciting experiment for a week first.

[K^2]

4 hours ago, Starshot said:

What are the conditions for these kinds of experiment, just occultation?

Preferably, when Mars is close, because that both increases the parallax, and gives you a full moon to work with, making the occultation a bit easier to anticipate. And since both the Mars and the Moon have a significant inclination, this doesn't align nearly as often as you might otherwise think.

[kerbiloid]

The Dam Busters should help.

Two guys with directional flashlights are shining at the Mars surface from two coasts of the ocean.
The third one is watching the light spots in the telescope and commands when two spots become one.

Now they can just fix the flashlight angles and calculate the result.

G: Geometry.

[sevenperforce]

12 hours ago, kerbiloid said:

Two guys with directional flashlights are shining at the Mars surface from two coasts of the ocean.
The third one is watching the light spots in the telescope

Directional flashlights?

[cubinator]

I will be able to give a try at the value of the distance later on, but for now I can say that Mars is definitely farther away than the Moon.

[JoeSchmuckatelli]

48 minutes ago, cubinator said:

Mars is definitely farther away than the Moon

That's a relief!

[cubinator]

I've found something interesting. The first calculation I wanted to do was the angular velocity of Mars against the Moon. To do this, I measure the amount of time between Mars' disappearance and reappearance from my own location - simple enough.

I decided to use the first pair of times in the sequence, since Mars' angular size would mess up the calculation otherwise. These would be the first and third times recorded. I got a difference of 3977 seconds. Then I decided to check my value against the other pair, the second and fourth times, expecting that they should match if the measurements are good. I got a value of 3968.8 seconds! That's 8.2 seconds shorter, which is a bit alarming for an error considering that's in the range of what I'm expecting to see for the parallax difference!

I found that the amount of time it took Mars to rise was 8.2 seconds shorter than the time it took to set as well. The rise and set times should be the same length. I think there are a few things that could be going on here.

The first option is that I just don't know what is happening to any precision smaller than 8.2 seconds, and I wasn't able to measure the timing well. Errors like this might throw off my calculation quite a bit. Because the time difference is the same amount in both measures I think that multiple data points are affected and not just one.

Another possibility is that the Moon's non-roundness has struck again. Mars has some horizontal motion as it passes the Moon, which could cause surface variations to affect the exact timing of the occultation. A difference of a few seconds could occur if, say, Mars began to touch the Moon at the top of a mountain and then "slid" into a valley.

 

Now, the 8.2 second difference is 0.2% of Mars' entire trip behind the Moon. This means that my estimate of Mars' angular velocity will be off by 0.2% - but Mars is very slow, so this may not completely wash out the parallax. There are a lot of seconds in the Atlantic Ocean, though, even at the Moon's blistering speed. We will see once I do the vector math on the observers' positions.

[JoeSchmuckatelli]

30 minutes ago, cubinator said:

We will see once I do the vector math

Following with interest! 

[cubinator]

I've been writing a MATLAB script to deal with these angles and rotations. Because the inertial positions of each observer are different at each observation, I need to calculate the new positions based on the location on Earth and the time elapsed. 

The motion of the observers is based on the rotation of Earth, which effectively changes the longitude of the observers as they move in a circle around the polar axis. The axis is at an angle to the ecliptic by 23.5 degrees, the axial tilt angle we're all familiar with. It is ALSO at an angle to the Sun-Earth-Moon line by a "season angle" which effectively quantifies where the Sun shines on the Earth and depends on the day of the year. To get the most accurate motion of the observers, both of these angles need to be accounted for to find the orientation of Earth at the times of observation.

I could try to get a rough velocity vector for each observer and use that since it's not a huge amount of time so the motion of each observer will mostly be in the same direction, but I would need the position of the Moon in the sky from at least one location, and I didn't bother to gather that myself and I don't want to use data from an astronomy program except to check my work. Instead, I am using a sunrise/sunset table, which has been widely available for centuries...and MATLAB. I'll still write an explanation of all the calculations, it's all stuff that can be done by hand.

[cubinator]

This graph shows the positions of the observers in inertial (Earth-centric, but non-rotating) space at each observation time. The markers are small, so you may have to zoom in, but triangles represent the California observation, circles the Minnesota one, and squares represent Derbyshire. The sphere represents the Earth, but its axial tilt is not represented visually.

The X axis points away from the Sun and towards the Moon, which is assumed to be directly in the X axis at its previously measured distance at 0 UTC. The Y axis is the direction of motion of the Moon. The XY plane is the ecliptic plane, and Z is normal to the ecliptic. 

I think I may need to adjust the "time of day" angle of the Earth because California looks to be a little too close to sunset from this graph based on what actually happened, but the positions look correct relative to each other and the direction of Earth's axial tilt. 

Once I verify that the positions are accurate, I will be able to draw vectors between the observation positions and the observed direction of Mars through the Moon, and by finding where they cross I will locate Mars in 3D space.

[cubinator]

I think I will also need to come up with a more accurate estimate of the Moon's position and velocity...the expected duration of transit is different for all three locations and depends on the Moon's exact location. So I might have to end up calculating the Moon's inclination too...

[Gargamel]

Do you have temperature data for each location?  I would think different atmospheric conditions in each location might account for some of the discrepancies in times.      (I have no idea of it Would be significant, that was just my first thought when seeing your time numbers) 

[cubinator]

16 minutes ago, Gargamel said:

Do you have temperature data for each location?  I would think different atmospheric conditions in each location might account for some of the discrepancies in times.      (I have no idea of it Would be significant, that was just my first thought when seeing your time numbers) 

I would be surprised if it had any effect, the Moon and Mars are both out of atmosphere so I'd think it'd affect them both equally. But science is full of surprises.

The duration of transit is mostly dependent on how the Moon moves through space and how 'deeply' Mars goes behind the Moon from each vantage point. The timing is affected by the Earth's rotation and the motion of Moon and Mars.

[Gargamel]

15 minutes ago, cubinator said:

I would be surprised if it had any effect, the Moon and Mars are both out of atmosphere so I'd think it'd affect them both equally. But science is full of surprises

Like I said, I have no idea of it would have any effect, but to me, large temperature differences might have enough changes in refraction of the atmosphere to affect the timings by that tiny % you listed.   

[DeadJohn]

3 hours ago, Gargamel said:

Like I said, I have no idea of it would have any effect, but to me, large temperature differences might have enough changes in refraction of the atmosphere to affect the timings by that tiny % you listed.   

I don't think atmospheric effects are significant here. Light slows a tiny amount in air, but it overall spends such a small amount of time passing through our atmosphere that it's not measurable with conventional amateur timing methods.

Speed of light in vacuum is 299792 km/s.  Air at sea level has a refractive index of 1.0003. That reduces the speed of light to 299792/1.0003 = 299702 km/s at sea level.

For discussion's sake, temporarily ignore the thinning effect of air at altitude, and pretend that the atmosphere remains at sea-level-like density to 100 km altitude. How long would it take light to travel 100 km?

• 0.000333565 seconds in vacuum.

• 0.000333665 seconds in atmosphere.

That's only a 100 nanosecond difference and I don't think we can tell the difference with the techniques used for this experiment. And that 100 is a worst-case difference between vacuum and air, while you're talking about the smaller difference between different air densities, plus the real atmosphere begins thinning much earlier than 100km.

I'm an amateur astronomer who likes learning new things, not a professional scientist, so corrections are welcome.

 

[Gargamel]

20 minutes ago, DeadJohn said:

I don't think atmospheric effects are significant here. Light slows a tiny amount in air, but it overall spends such a small amount of time passing through our atmosphere that it's not measurable with conventional amateur timing methods.

Speed of light in vacuum is 299792 km/s.  Air at sea level has a refractive index of 1.0003. That reduces the speed of light to 299792/1.0003 = 299702 km/s at sea level.

For discussion's sake, temporarily ignore the thinning effect of air at altitude, and pretend that the atmosphere remains at sea-level-like density to 100 km altitude. How long would it take light to travel 100 km?

• 0.000333565 seconds in vacuum.

• 0.000333665 seconds in atmosphere.

That's only a 100 nanosecond difference and I don't think we can tell the difference with the techniques used for this experiment. And that 100 is a worst-case difference between vacuum and air, while you're talking about the smaller difference between different air densities, plus the real atmosphere begins thinning much earlier than 100km.

I'm an amateur astronomer who likes learning new things, not a professional scientist, so corrections are welcome.

 

Excellent!  Thanks!   I wasn’t sure, so I decided to test Godwin’s Law: “The best way to get a correct answer in the internet is not to ask a question, but to offer the wrong answer.”

[DeadJohn]

25 minutes ago, Gargamel said:

Excellent!  Thanks!   I wasn’t sure, so I decided to test Godwin’s Law: “The best way to get a correct answer in the internet is not to ask a question, but to offer the wrong answer.”

I don't recall that quote before, but I thought Godwin postulated that all big online discussions end up with ad hominem attacks and mention of Hitler/[the word for an evil WW2 German political party beginning with "N" is censored].

Google shows your quote as Cunningham's Law. Wrong again? 

 

Edited December 27, 2022 by DeadJohn
trying to keep context within forum rules

[Gargamel]

2 hours ago, DeadJohn said:

I don't recall that quote before, but I thought Godwin postulated that all big online discussions end up with ad hominem attacks and mention of Hitler/[the word for an evil WW2 German political party beginning with "N" is censored].

Google shows your quote as Cunningham's Law. Wrong again? 

 

Gotcha!  
 

That was the joke. 

[DeadJohn]

41 minutes ago, Gargamel said:

Gotcha!  
 

That was the joke. 

"Sarcasm is like a little child and easily lost!" -Someone's Law

[kerbiloid]

Cool things shrink, large things expand.

We should distinguish the winter and summer Moon and Mars diameters and introduce a seasonal planet size coefficient.

[PakledHostage]

Atmospheric refraction does significantly affect celestial navigation, particularly when the object being sighted is close to the horizon. It sounds like K2's observations, at the very least, were close to the horizon? I don't know how much that would affect these results though, because both objects are outside the atmosphere (as opposed to measuring angles of objects from the horizon)? But maybe atmospheric refraction does need to be considered in some way? 

(I tried something similar to this real world experiment in KSP some years ago, by observing a transit of Eve across Kerbol from near Kerbin's north and south poles, but I couldn't get the numbers to work out. My explanation to myself was that the error was because of how the telescope mod "magnified" the view, by moving the point of view closer rather than actually magnifying the view. That explanation was easier on my ego than that my math skills suck... :-))

[cubinator]

Assuming the Moon's orbital period is 2360592 seconds, and that while Mars is rising and setting its own motion through the sky is negligible, 

the Moon's angular velocity is w = 2*pi/T, and Mars' angular size is w*time to rise or set.

The rise and set times observed in Minnesota were different for an as-of-yet unknown cause, but they can be used nonetheless to estimate the angular size of Mars:

Rise time estimate	Set time estimate
21.4665 arcseconds	25.9684 arcseconds