Universial Gravitation

1. Jan 27, 2007

fsm

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

2. Relevant equations
a=(GM/r^2)

3. 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.

2. Jan 27, 2007

arildno

HOW did you "sub in"?

3. Jan 27, 2007

fsm

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

4. Jan 27, 2007

arildno

And found what you were after how?

5. Jan 27, 2007

fsm

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

6. Jan 27, 2007

arildno

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

7. Jan 27, 2007

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.

8. Jan 27, 2007

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.

9. Jan 27, 2007

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.

10. Jan 27, 2007

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.

11. Jan 27, 2007

fsm

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

12. Jan 27, 2007

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.

13. Jan 27, 2007

fsm

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

Last edited: Jan 27, 2007
14. Jan 27, 2007

fsm

repost*******

15. Jan 27, 2007

arildno

i) d is in your case NEGATIVE!

16. Jan 27, 2007

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 :surprised . The answer is wrong.

17. Jan 27, 2007

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

18. Jan 27, 2007

arildno

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.

19. Jan 27, 2007

arildno

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

20. Jan 27, 2007

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.