# Pulse echo technique to find the distance of the Moon

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1. Jun 4, 2017

### Javeria

1. The problem statement, all variables and given/known data

Over 40 years ago, the Apollo astronauts placed reflectors on the surface of the Moon. These are still used by a number of observatories on Earth to monitor the distance to the Moon by reflecting pulses of laser light from them and detecting the reflected signal. Scientists have determined that the Moon is at a distance of 363104 km at its closest and 405696 km at its furthest. It has also been determined that the Moon is getting about 3.8 cm further away from the Earth each year.

(a) Describe how the reflected pulses can be used to determine the distance to the Moon. (2)
Ans a- (already done) d = (s * t) / 2, where s is the speed of light and t is time for return trip.

(b) An observatory sends out pulses of laser light of duration 2.0 × 10^-10 s when it is determining the distance to the Moon.

(i) Calculate the pulse length. (2)
Ans b(i)- (already calculated) 0.06m or 6cm

(ii) Discuss whether the levels of precision quoted for the distance to the Moon and its rate of increasing distance from the Earth are justified. (2)
Ans b(ii)- (cannot understand)

3. The attempt at a solution
markscheme(ms):
Distance is to the nearest km but pulse length is to the nearest mm, so acceptable (accept pulse length to nearest cm) (1)
Not acceptable because 6.0 cm pulse is longer than 3.8 cm, Or The distance is calculated from a difference over 40 years, so it is over a metre, so it is acceptable compared to 6.0 cm (1)

examiners report(er):
Few students scored on this question. Many focused on significant figures, missing the point that using a pulse a few centimetres long justifies quoting an answer to the nearest kilometre despite there being 6 significant figures in the kilometre value. They did not realise that the uncertainty should be less than the quoted value, or link the length of the pulse with uncertainty at all.

I looked into these after failing to understand the question. The answer i wrote was completely wrong.
My answer: The distances are in km while the answer in cm. Then i calculated the % increase of distance to the moon using the rate of increasing distance from the Earth for some reason.
405696 * (100.000038/100) = 405696.15
(0.154/405696) * 100 = 3.8 * 10^-5 increase so not justified
Completely wrong. After reading the ms and er i still dont understand this question.

2. Jun 4, 2017

### jbriggs444

Let's start simple.

Suppose that we transmit the pulse toward the moon, obtaining a perfectly accurate time stamp for the time of transmission of the pulse midpoint. Suppose that we get one photon back and obtain a perfectly accurate time stamp for the time of reception of this photon.

How much could our resulting calculated distance vary from the true distance due to the finite pulse width?

[Note that the one photon supposition is reasonably correct. Wiki says... "Out of 1017 photons aimed at the reflector, only one is received back on Earth every few seconds, even under good conditions."]

3. Jun 4, 2017

### Javeria

Are you asking how the pulse length and distance are related?
The pulse length has to be double the distance being calculated
Is this what you were asking? Im so sorry, im really confused right now.
How can the pulse width affect the distance? Isnt length the only thing that matters here?

4. Jun 4, 2017

### jbriggs444

How are pulse length and measurement error related? I wrote "width" but meant length -- the distance between leading and trailing edges. Sorry, did not mean to confuse you with that.

The pulse length is most certainly not double the distance being calculated. The distance to the moon is much greater than 3 cm.

5. Jun 4, 2017

### Javeria

Ohh its okaay
Yeps yeps got that too
The calculated and true value wont be as different?

6. Jun 4, 2017

### jbriggs444

How much different can they be?

7. Jun 4, 2017

### Javeria

Less than the pulse length or around the length of the pulse

8. Jun 4, 2017

### jbriggs444

How about a calculation and a numerical result?

9. Jun 4, 2017

### Javeria

You won't just give an answer, will you?
Supposing 6cm difference then
Uncertainty = (6*10^-5/405696)*100
= (+-)1.48*10^-8%
Im not sure if im doing it right

10. Jun 4, 2017

### jbriggs444

Simply giving the answer is against the rules here.

This is a 6cm difference between the maximum possible measured round-trip distance and the lowest possible.

That is not the same thing as the maximum possible error in the one-way distance. I am not trying to get you to calculate relative error. I am trying to ask for the maximum absolute error in the calculated one-way distance measurement.

11. Jun 4, 2017

### Javeria

This is a 6cm difference between the maximum possible measured round-trip distance and the lowest possible.

That is not the same thing as the maximum possible error in the one-way distance. I am not trying to get you to calculate relative error. I am trying to ask for the maximum absolute error in the calculated one-way distance measurement.[/QUOTE]

So, 3cm?
'This is a 6cm difference between the maximum possible measured round-trip distance and the lowest possible'.
How was i supposed to know this? Its not mentioned anywhere

12. Jun 4, 2017

### Javeria

So, 3 cm?
How was i supposed to know its been the max and lowest? How did you know? It doesnt say this anywhere in the question.

13. Jun 4, 2017

### jbriggs444

We send a pulse. We start a timer at the halfway point between when the leading edge of the pulse has been transmitted and when the trailing edge of the pulse has been transmitted. We stop the timer when a photon is detected.

The detected photon could have been from the leading edge of the pulse. In which case we started the timer 3 cm too late to get a perfect measurement. When we compute the delta time, we'll be 3 cm short.

The detected photon could have been from the midpoint of the pulse. In which case we started the timer exactly right.

The detected photon could have been from the trailing edge of the pulse. In which case we started the timer 3 cm too early to get a perfect measurement. When we compute the delta time, we'll be 3 cm long.

If we were reporting the round trip distance this would give an error range of ±3 cm around the correct result. But we are not reporting the round trip distance. We are reporting the one way distance.

Note: I am using "cm" as a unit of time meaning "the time it takes light to go 1 cm".

14. Jun 4, 2017

### Javeria

Oh now i get it, thank youu
So for one way (+-) 1.5 cm?

15. Jun 4, 2017

### jbriggs444

Yes.

Now ±1.5 cm is plenty good enough to justify quoting distance figures to the nearest 1 km. So there is no indication [from this evidence] that the figures for the current distance to the moon, e.g.
are reported with too much accuracy. If anything, their accuracy [based on this error analysis alone] is understated.

Taken at face value, an error range of ±1.5 cm would mean that a figure of 3.8 cm per year is questionable. If one were to make two measurements, twelve months apart, subtract them and report the delta as the "annual rate at which the moon is receding", what would be the maximum possible error (in cm/year)?

But, if one were to make two measurements, 40 years apart, subtract them and report the $\frac{delta}{40}$ as the "annual rate at which the moon is receding", what then would be the maximum possible error (in cm/year)?

Now, for extra credit, go back to the ±1.5 cm error we computed based on pulse length. Can you think of any way to improve upon that figure? What if we ran multiple experiments, for instance?

Last edited: Jun 5, 2017
16. Jun 5, 2017

### Javeria

I gett it noow, thank youuu soo muuch.
Im not sure how to calculate the 1 year and 40 years max possible error but i get that for a year its gonna be less than the 40 year difference, right?
Yess and taking the average of the results or using a shorter length of pulse so the maximum possible errors are smaller

17. Jun 5, 2017

### jbriggs444

Well, let's go over it.
You have a starting distance that you've measured to within ±1.5 cm. You have an ending distance that you've measured to ±1.5 cm.

If you measured the starting distance and came out 1.5 cm high and then measured the ending distance and came out 1.5 cm low, you could calculate that the moon is approaching the earth at 3.0 cm per [whatever time interval you chose]. Similarly if the starting distance came out low the end ending distance came out high, you'd calculate that the moon is receding at 3.0 cm per time interval.

Are you with me so far?
Yes, the one I was aiming for was taking the average of the results.

18. Jun 5, 2017

### Javeria

Yesss

19. Jun 5, 2017

### jbriggs444

OK. So ±3 cm over a one year interval is ± 3 cm per year in terms of a movement rate. But ±3 cm over a 40 year span is only 3/40 ~= .1 cm per year.

That matches the precision of the reported value (3.8 cm per year).

20. Jun 5, 2017

### Javeria

I finally got it, thank youuu