Calculating Altitude Using Gravimetry

[fsm]

Homework Statement

A sensitive gravimeter at a mountain observatory finds that the acceleration due to gravity is 5.00×10−3 {\rm m/s^2} less than that at sea level.
What is the observatory's altitude?

Homework Equations

a=(GM/r^2)

The Attempt at a Solution

Gravity should be 9.8-.005=9.795 and then subbing into the above formula I get 1582m. Which it wrong but is says I'm close.

 

[arildno]

HOW did you "sub in"?

 

[fsm]

G and M are constants. a=9.795. Subbed in that way.

 

[arildno]

And found what you were after how?

 

[fsm]

Unless I'm missing something see #2 minus the Earth's radius

 

[arildno]

Eeh?
I have no idea what you are vaguely trying to say.
Post your MATHEMATICAL solution attempt, in detail.

 

[fsm]

Sorry I was only going off the vagueness of the questions and thought I was obvious what I did. I did use 9.83-.005=9.825 as g. So, 9.825=((6.67*10^-11)*(5.98*10^24))/(6.37*10^6+x)^2. For x got 1580 which is wrong but close. I did try 9.8 but this answer is completely wrong.

 

[arildno]

My questions were precise.
You, however, have gotten a wrong answer and are STILL refusing to tell exactly HOW you got your answer!

It is obvious to me how this problem should be done PROPERLY; what is NOT obvious is which wrong way you have calculated the quantity.

 

[fsm]

Well you are right that I have the wrong answer and I NEVER said that it was not obvious for you to solve the problem. I just don't how I you want me to answer your questions when the post right above your's is how I attempted the solution.

 

[arildno]

There are loads of ways through which you may have calculated wrongly.

What you HAVE posted in #7 includes a minor flaw, 9.83-0.05=9.78, unless I'm much mistaken.
Possibly, that is enough to rectify your answer, but I do not know if you've made more mistakes as well.

 

[fsm]

ummm doesn't 5*10^-3=0.005?

 

[arildno]

Hmm...yes. I read your decimal number wrongly, sorry about that. Such mistakes are easy to make, you have probably done one yourself somewhere (getting an incorrect answer).

It is a lot better to first transform your equation in an algebraic manner, and only in the final step plug in the values:

Let the acceleration discrepancy be d.

Then, we have the equation:
$$(g+d)=\frac{GM}{(R+x)^{2}}\to(R+x)=\sqrt{\frac{GM}{g+d}}\to{x}=\sqrt{\frac{GM}{(g+d)}}-R$$
Much more simple!

Now, plug in numbers to your heart's content.

 

[fsm]

I must be doing something wrong. I am still getting the wrong answer. $$x=\sqrt{\frac{GM}{(g+d)}}-R$$
plugging values in:
$$1660(rounded)=\sqrt{\frac{6.67*10^-11*(5.98*10^(24))}{(9.83+(5*10^(-3))}}-(6.37*10^(6))$$

 

[fsm]

repost*******

 

[arildno]

i) d is in your case NEGATIVE!
ii) On your calculator, remember the parenthesis about your whole radicand, and, for your radicand fraction, remember a parenthesis about your denominator as well!

 

[fsm]

You are correct. Mistakes are easy to make . Fixing that I STILL get the same answer 1581m. Very close to my answer of 1582m in Post#1 . The answer is wrong.

 

[arildno]

In normal calculator language, you should write:

sqrt((6.67*10^(-11))*(5.98*10^24))/(9.83-5*10^(-3)))-6.37*10^6

 

[arildno]

You are correct. Mistakes are easy to make . Fixing that I STILL get the same answer 1581m. Very close to my answer of 1582m in Post#1 . The answer is wrong.

Well, I doubt that it matters, but shouldn't you use 9.81 rather than 9.83?

Anyhow, whatever you get out, when doing the calculator stuff properly, IS correct, WHATEVER the answer in the book says. Books are WELL-KNOWN in giving wrong answers.

 

[arildno]

No. It is correct, since the algebraic answer is the correct one.

 

[fsm]

When I use 9.81 or 9.8 the answer is wrong. When I use 9.83 I'm told the answer is "close". This is not from a book. This is a class website where the class does homework on-line. There are 3 possible responses for an answer-right, wrong, or close.

 

[arildno]

Then, most likely, you are using decimal approximations of the other quantities too inexact to yield "right". Look again at your resources:
Are G, M or the radius of the Earth given by more decimal accuracies than you've used?

The other alternative is that the computer has been fed incorrect values from the programmer/professor.

 

[fsm]

G, M, and R are straight out of my textbook. I imagine if the answer was wrong this issue would have been brought to the professor's attention but that would be a gross assumption.

 

[arildno]

Okay, that leaves one silly issue (apart from the professor having made an undiscovered mistake):
That of "significant digits".

Have you rounded off your answer according the rules of significant digits you've learned properly?

 

[fsm]

Sig Fig's is not an issue because that is not considered a wrong answer if you type the actual answer. It rounds your answer to the proper sig fig's.

 

[arildno]

Well, as Sherlock Holmes used to say:
"When you have eliminated all possibilities, except one, what remains is the truth, however improbable it seems".

That is to say:
Your professor has made a mistake.

i) It is ABSOLUTELY TRUE that the algebraic equation is correct.
ii) You have sworn, on your mother's life, that you have typed the numbers correctly into your calculator.
iii) Calculators do not make mistakes
iv) Sig Fig issues are not relevant

Still getting a wrong answer means that what your answer is measured up to is itself wrong.

 

[fsm]

I know you've asked why I didn't use 9.8 but could 9.8 or 9.83 both be wrong?

 

[arildno]

No, your professor has made a mistake.

Now, we are delving into decimal irrelevancies totally contrary to the spirit of physics&maths. Ask your prof. what he was after, and why your answer is given the value "close", rather than "right".

I can't help you any further; in my mind, the problem was solved when you got the algebraic, EXACT answer.

 

[fsm]

Just making sure. Thank you for your patience and help!

 

[arildno]

Just a final word:

You knew, from the start, what was physically correct, and transformed your equations correctly thereafter.

THAT is what is important (because it shows you understand the problem), not that your and your professor's answer differ in the decimals.

 

[arildno]

There is one final test you can make:

i) On the measured discrepancy's place, plug the number 5.05*10^(-3) into it instead, and compare it with the answer gotten by plugging in the number 4.95*10^(-3) there.

ii) What is the uncertainty interval you're getting?