You just totally missed the transit of Venus. And there’s more. It was awesome. No worries. The next one will be in 2117. You should be fine.
Originally, I was going to do a pre-transit post to talk about how awesome this event was going to be. However, that plan got smashed when Jennifer Ouellette (@JenLucPiquant) wrote a killer post about the history of the transit of Venus. I love talking about this historical aspect. There are so many cool stories associated with it, I just don’t know what else to say. Read Jennifer’s post.
Then what am I to do? How about I use a video of the recent transit of Venus to estimate the distance to the Sun. Instead of repeating what was done in the past — using multiple viewing locations — let me do it with one view. Now, let me give you a warning. I have no idea what I am doing. This might not even work and I am going to make some approximations. There, I said it.
Here is the video I will use.
There are many good ones out there, but this one is a time lapse and states both the frame rate (1 photo every 30 seconds) as well as the viewing location (Takapo Village, New Zealand). The viewing location will be useful if this method fails. One more thing. I have no idea about the angular width of this video. However, I do know the angular size of the Sun (as seen from Earth) is about 0.53° — yes, the distance from the Earth to the Sun does change, but I don’t mind. So, using the size of the Sun in the video I can get the angular speed of Venus with respect to the Sun. How? Using Tracker Video analysis of course.
After taking into account the relative motion of the Sun in the video, I get the following for the motion of Venus (relative to the Sun):
The units in this case are in radians and seconds. If I do a linear regression on both of these plots, I can find the angular velocity in both the x- and y-directions. This gives an angular speed in the x-direction of -8.15 x 10-8 radians/sec and -2.64 x 10-7 radians/sec in the y-direction. To get the magnitude of the angular velocity, I can just treat these two velocities as components (which they are).
This gives an angular velocity of 2.77 x 10-7 radians per second.
Now that I have this angular velocity of Venus with respect to the Sun, what am I going to do with it? Let me start with this simplified model of the orbits of Earth and Venus (not to scale).
For this model, I will assume that both the Earth and Venus are moving in circular orbits around the Sun. Since this orbital motion is due to the gravitational force between each planet and the Sun, I can get the following relationship between angular velocity (about the Sun) and distance from the Sun.
Just to be clear, rp is the orbital radius for a particular planet and ωp is the orbital angular velocity of that planet. The 1/r2 term is the gravitational force and ω2 times r is the acceleration of an object moving in a circle. The mass of the planet (mp) cancels. In short, the further a planet is from the Sun, the lower the angular speed. The angular velocity for the Earth and Venus are pretty easy to determine.
Of course, we don’t know the orbital distances for either planet but knowing the ratio of angular velocities will give the ratio of distances.
The key here is that both Earth and Venus have angular velocities that both depend on the same gravitational constant (G) and the same mass of the Sun.
Using the Video Data
Now I am going to make another assumption. I am not quite sure how valid this will be, but I will do it anyway. Let me assume that during this transit of Venus, the Earth is mostly stationary. Calm down … I know it isn’t true. But look at it this way. This transit lasts about 3 hours. The Earth only moves 0.0023 radians during this time.
If the Earth is stationary, then the motion of Venus in the video will be just like Venus is going around the Earth. Here, maybe this diagram will help.
Here I have the angular velocity of Venus with respect to the Earth (ωve) and the distance from Venus to the Earth (rve). With a stationary Earth, this would give me the linear velocity of Venus as:
Ah, but I don’t know the Venus-Earth distance. Yes, but I can say that at this transit position, the distance between Earth and Venus is related to the radius of the Earth and Venus orbits:
Things are still confusing, but I am getting somewhere. I still do not know either of the orbital distances. So now that I have an expression for the linear velocity of Venus, I can use that to get an expression for the angular velocity of Venus as it goes around the Sun. Yes, I know this seems like circular logic — but let’s just see if it works. Remember, vV is the linear velocity of venus, ωV is the angular velocity of Venus around the Sun and ωve is the angular velocity of Venus as seen from Earth.
Now I can bring in the relationship for the angular velocity of Venus with respect to the gravitational force from before. Also, I can use the ratio of angular speeds of the planets to the orbital distances (from above) to remove the radius of the Earth’s orbit.
What is K? K stands for Konstant. It is just that cubed root of the square of the angular velocity ratio. I was just too lazy to re-write that stuff that is just a constant. Now I just need to do one thing: Get a value for the mass of the Sun and the gravitational constant G. I don’t think I can get that here. I am pretty sure that the mass of Sun was calculated by using the orbital distance of the Earth (or other planets).
OK. I am going to cheat just to see if this method was even close to working. The gravitational constant is 6.67 x 10-11 N*m2/kg2 and the mass of the Sun is 1.99 x 1030 kg. Using this, I can solve for the distance from Venus to the Earth:
Plugging in my values, I get a distance from the Sun to Venus of 2.28 x 1011 meters. This is wrong — but not crazy wrong, just super wrong. The listed value for Sun-Venus distance is 1.08 x 1011 meters. Why am I off by a factor of 2? Maybe it is because I didn’t take into account the motion of the Earth. I guess I should add that in.
A quick note. Here, when I look at the velocity of Venus I am treating it as though it were moving in a straight line. This isn’t a terrible assumption since I am just looking at a 3-hour part of its 225-day motion in a circle. This makes it close to a straight line. In my velocity calculation for Venus, I assumed a stationary Earth. Since both the Earth and Venus are moving in the same direction (with Venus appearing to move faster), this becomes a relative velocity problem. I can write:
Where the “es” subscript means “velocity of the Earth with respect to the Sun”. The same convention applies for the other velocities. And like I said, since the planets are moving in the same direction, I can just add the magnitudes of these velocities without worrying about their vector nature. This means I have re-write my expression for the angular velocity of Venus (with respect to the Sun).
Now I can bring back my relationship between the orbital radius of Earth and Venus with that constant K:
Putting this back into the expression for angular velocity due to the gravitational force:
With the known values, I get 9.72 x 1010 meters. Wow. That is pretty close. Honestly, I can’t believe it is only a little bit off. Normally, I make some dumb algebra mistake or something. I will chalk this up as a partial victory. I still had to cheat on the value of G*M.
OK, I came up with another hare-brained idea. What if I assumed that Venus was the same size as the Earth? In that case, I could use the angular size of Venus as it goes across the Sun to get the distance to Venus. I guess that would be a pretty cheap solution also.
Then How Did They Do It?
There are some technical issues that I won’t go all the way over (because I am not sure). In short, the previous methods for measuring the distance to the Sun used multiple viewing positions. Just suppose that you were in Ecuador and I was in North Dakota. If we were looking at the transit of Venus, we would see Venus in slightly different locations. Here is another (not to scale) diagram.
The two red dots on Earth (the blue planet) are the two viewing locations. If you know the distance between those points and the angle between the different observed locations of Venus, you are all set. It is a simple problem to find the distance from Earth to Venus. Since (as I stated above) you can also find the relative distances of Earth and Venus, you can then find the distance to the Sun.
Here is the tricky part. How do you make your observations at the same time from different locations? That is the technical part scientists faced in 1761. One method uses the times of when Venus enters and leaves the Sun. But, like I said — it can get complicated.
But there is another way to get observations at the same time: flickr. Lots of people took some pictures of the transit of Venus and posted them to flickr. Almost all of them have the image time in the EXIF data (available in Flickr). Also, many of these photos include a location. So, after browsing around, I found two images that are almost at the same time with location data.
- Image 1 was from Chula Vista, CA (USA).
- Image 2 was in Brisbane, Australia.
If you look at both of those images, it is pretty cool to realize that the Sun has a different orientation. This is because the Earth is not flat. The people in Australia are sort of upside down compared to the California people. I can fix this orientation problem by lining up the sun spots (good thing there are some sun spots visible). Here is a sketch of what that looks like (without using the actual images).
Using the angular size of the Sun as a scale, I get the angular distance between these two locations as 0.00102 radians. Now I just need the distance between the two Earth-based locations. Using this calculator I get a distance of 1.161 x 107 meters.
With this super small angle, I approximately say
And putting in the values from above, I get an Earth-Venus distance of 1.139 x 1010 meters. This would put the Earth-Sun distance at 4.11 x 1010 meters – which is wrong. The accepted value is 1.49 x 1011 meters. I am not sure what went wrong.
I just realized something rather stupid. If I knew the mass of the Sun and the gravitational constant (G), I could easily find the distance to the sun using the angular velocity of the Earth. I will just chalk this up as a big mistake.Go Back to Top. Skip To: Start of Article.