If I were orbiting the earth and could only see a “chord” of it when looking out the window, could I calculate the size of the earth? What measurements must be taken, and what is the math involved?
I understand how to calculate the size if you can see the whole globe (now 
), but what if you can see only a little bit of it?
If you could see just a segment of a circle, you can easily reconstruct the whole circle. This means you know the radius too.
But are you sure you know how to calculate the distance if you can see the whole circle? Its basic trigonometry, but you still have to know the distance between the observer and the earth. The only way I can think of, off the top of my head, is to observe from the moon and wait for a total lunar eclipse. Measurements of the moon’s shadow on the surface of the earth would reveal the distance between the moon and earth.
I forgot to mention that I do know how high I am above the surface of the earth. Sorry.
This was covered recently but I haven’t found that thread yet.
As Chas. E said you need to know the distance and there are a few ways to do that depending on what technology you prefer. It could be a simple pulse radar set or you could use a telescope and trig to measure a geographic feature of known size on the surface. Of course you’ll get mutiple answers depending on distance and orientation because the earth is far from a perfect sphere.
There are easier ways to measure the earth that don’t require space technology. Two observers a long, known distance apart can easily estimate the diameter of the earth with nothing more complex than a pole, plumb line, tape measure and a signal light.
Ah, I think this is a different problem entirely. I was assuming you were in an orbit high enough to see the whole earth, like halfway to the moon or something, and maybe you could only see a chord because the rest of the earth was in shadow.
But it appears you may be thinking about a low orbit, so low that you cannot see the earth as a disc, and the horizon is quite close. Different rules would apply, I can’t think of any obvious solution here.
Does that mean you don’t already know how high you are or that you don’t have any way of finding out? That information is critical to some methods. I think we need a few more conditions to answer your question to your satisfaction.
I know I am in an orbit 1000 miles above the surface of the earth, and can only see a little bit of it–the chord mentioned earlier.
It’s possible to determine the size without being in orbit at all, as was done a couple thousand years ago, with sticks, shadows, and pacing off distances.
Yeah, I know the size of the earth has been figured out various ways over time --I guess what I am looking for is a geometry/trig formula that can do the job in the situation as described. All ideas are welcome.
I TOLD you people you need an equation editor.
I have a solution, but putting it up on this board is going to be a pain in the neck without symbolic math.
The Radius of the earth, R, is given by your height above the surface, h, times the sine of theta, all divided by one minus the sine of theta. In other words, R equals h divided by cosecant theta minus 1).
Theta is one-half the angle subtended by the earth at your altitude. If your porthole is big enough to see the whole earth, then simply measure the angle it subtends with your pocket protractor, astrolabe, or sextant and take half of that.
But you said you could only see a chord. Presumably you know what proportion of a circle that chord is. Call the angle subtended by that chord “beta”. Assume that the angle that chord reresents on a circle is “alpha” (s if you could see the whole circle of the earth alpha would be 180 degrees or pi radians). Theta is given by (beta/2) divided by sin(alpha/2). Now just plug theta into the above formula.
This all assumes the earth is a perfect sphere an I don’t have to worry about refraction at the edges, yadda, yadda. Bear in mind that, if you are not an infinite distance away you CANNOT see the entire diameter of the earth, so when I talk about seeing angular subtenses they are to the observable horizon.
Have fun. Let me know how it works out.
Hey! How do they do that? I mean, I’ve wondered about this a lot but can’t figure it out.
Boris B;
Famous work by Eratosthenes.
Find a point where you know the sun is directly overhead at some exactly known time. (Eratosthenes used a well in Egypt.)
Go to a point a known distance directly north or south of this location (everyone always seems to leave that out in descriptions of this rick, but it’s essential.)
Measure the angle the sun makes with the vertical at that time. Let’s say it’s 10 degrees.
That ten degrees is also the angular distance between the location of the well and your present location.
You can now figure out the size of the earth. If that angle is ten degrees, then the distance between the place you are and the place with the well is 10/360 = 1/36 of the earth’s circumference, so just multiply the distance by 36. If you want the earth’s radius, divide the circumference by two times pi.
Tricky part: how do you know that you are measuring the angle at EXACTLY the same time that the sun is overhead at the well? You don’t have a telephone. You don’t have an accurate clock (See Dava Sobel’s book “Longitude”). Eratosthenes’ trick was to make the measurement at noon on the day in question, and to use as his referece city a place where the sun shone directly down the well at noon. You know it’s noon because the sun is as close to vertical as it’s going to get.
Trickier than it looks at first, but doable.
Cal–I very much appreciate the info. I am a little confused one place though. I can measure what I see out the window/porthole–say I can see a chord that from my position measures 45 degrees. How can I find what proportion of the circle that 45 degrees represents? I know I can assume I am directly above the “center” of the circle, but I don’t know how to continue. Sorry if this sounds confused–it is hard to explain without diagrams.
Imagine that the section of a circle you see has been expanded to a complete circle. Place your protractor in the center of tha circle and the “zero” of the protractor on one end of the window, then read off what angle/portion of the circle you can see at the other end of the window.
This is, as you say, hard without diagrams.
There are other ways, but describing them is even harder when I can’t wave my hands around in the air for explanation or dramatic emphasis.
You can e-mail me through my http://www.MedusaMystery.com site and I can send you a MathCAD file or a WORD document or something through snail mail, if you want.
Cal, you said h is the height above the surface. Isn’t it the distance to the center of the earth. I believe Newton made the same mistake in Principia. (Sitting back, puffing on pipe in my tweed jacket with elbow patches.)
Mipsman:
Nope. Drew out the figure myself. Defined “h” as the height above the surface, NOT the distance to the center of the earth.
(Choking on pipe smoke, taking off scratchy tweed jacket) Ok, let me work this out instead of relying on memory.
Cal, you probably don’t need the reinforcement but you are correct. But why not define H as the distance to the center of the earth and just use the sine theta. And then wouldn’t the distance to the center of the earth be determinable from your orbital characteristcs? (My orbit determination skills have 25 years of rust on them.)
Trig skills equally rusty, the use of sine theta could only be an approximation.
… Man, them ancient Greeks were smart. Think of all the cool stuff they figured out without Arabic numerals. Or cell phones.