• BenMur
In summary, the student is having trouble with an old OU science exercise that makes the assumption that the shadow cast by the Earth on to the surface of the moon during a solar eclipse is the same size as the Earth itself. He is trying to solve the problem by using photos of the moon to calculate the radius of the Earth, and is working to limits of uncertainty. He is trying to find a TV program or other source of information that explains how to correct the inital results.
BenMur
Hello.

I have been going through some old OU books on the sciences and started working through an excercise when I came to a stumbling block. I have come here in the hope that someone may be able to shed some light on my problem. This is all for personal study as I am not enrolled on any courses in the subject and just wish to expand my mind.

## Homework Statement

The exercise I am going through is part of an old OU science module that belonged to my mother. Unfortunantely she no longer has the videos which I need to explain to me what I need to do next to correct my results.

Basically the exercise makes the assumption that the shadow cast by the Earth on to the surface of the moon during a solar eclipse is the same size as the Earth itself. Now whilst I have an idea why this is not the case the assumption stands through most of the exercise in order to show a method of calculation to the student.

Through the use of the photos of the moon showing different phases of a partial eclipse I am asked to make measurements to estimate the radius of the Earth in the photograph.

Before I explain any more I would like to show everything I have calculated thus far in the exercise.

(Exact correct figures may not be used during the exercise as the learning experience is more to do with the approach than the results)

## The Attempt at a Solution

less than 7.2 cm (upper limit)
more than 6.4 cm (lower limit)

Radius of Earth's shadow in the photograph (RE) = 6.80cm +/- 0.40cm

Radius of the Moon in the photograph = (RM) 2.75cm +/- 0.1cm

Therefore the Radius of the Earth is 2.47 times bigger than the radius of the moon.

RE = 2.47 RM

Limits of Uncertainty

Max Ratio = 7.2/2.65 = 2.72
Min Ratio = 6.4/2.85 = 2.25

Therefore RE = (2.47 +/- 0.22) RM

Now RE = 2.47 RM (Text book uses a figure of 6200km for Earth's radius)

So 6200/2.47 = RM

RM = 2510

I then calculate the limits of uncertainty for distance using the earlier numbers and get:

RM = 2510 +/- 231 km

This is the extent of my working out

Having gotten this far through the exercise I read the following:

"One word of caution about this calculation of RM: throughout Section 3 [the above] you make the assumption that the shadow of the Earth at the Moon is the same size as the Earth itself. As you will see in the TV programme, this is an unjustified assumption that leads to a considerable error in the value obtained for RM. However, the programme also shows how to correct this error, so be prepared to adjust your value of RM after viewing."

Unfortunately I don't have this TV program and have been unable to obtain a copy or find anything similar avaliable. So I have come here in the hope that one of you brilliant minds can assist me in completeing this exercise by explaining to me what I am missing and how to correct my inital results.

Last edited:
It's possible that they are referring to the fact that the Sun is not at an infinite distance, so it's rays are not strictly parallel. During a lunar eclipse the Moon is an additional distance d ~ 384400 km from the Earth's distance to the Sun of about D ~ 1.496 x 1011 m. There will be a "projection effect". Draw a diagram and work out the magnification.

The sun is not a "point source". We see only the full shadow (umbra) really dark on the Moon, which is smaller than the size of Earth.

ehild

## 1. What is the ratio method for determining the radius of the Moon?

The ratio method for determining the radius of the Moon involves measuring the ratio between the length of the Moon's shadow and the length of the Earth's shadow during a lunar eclipse. This ratio is then used in a mathematical formula to calculate the radius of the Moon.

## 2. How accurate is the ratio method for determining the radius of the Moon?

The ratio method has been found to be accurate within a few kilometers when compared to other methods of measuring the radius of the Moon. However, the accuracy can vary depending on the precision of the measurements and the atmospheric conditions during the eclipse.

## 3. Can the ratio method be used to determine the radius of the Moon at any time?

No, the ratio method can only be used during a lunar eclipse when the Moon passes through the Earth's shadow. This typically occurs twice a year, so the method can only be used to determine the radius of the Moon a few times per year.

## 4. Are there any other methods for determining the radius of the Moon?

Yes, there are other methods for determining the radius of the Moon, such as radar and laser ranging techniques. These methods involve bouncing waves off the surface of the Moon and measuring the time it takes for the waves to return. However, the ratio method is considered to be the most accurate and widely used method.

## 5. Why is it important to accurately determine the radius of the Moon?

The radius of the Moon is an important measurement for understanding the Moon's composition, structure, and evolution. It also helps scientists to better understand the Earth-Moon system and its impact on other celestial bodies.

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