# In which case is the period larger? ( Above earth or below?)

1. Oct 28, 2016

### Vitani11

1. The problem statement, all variables and given/known data
A pendulum clock is adjusted so that it keeps excellent time on the ground. The clock is brought to a mine of depth h below the ground and then raised a height h above the ground to see the differences. In which case is the error larger?

2. Relevant equations
T = 2π √(L/g)

L = length of the pendulum
g = gravity on earths surface
T = period

g = GMm/Re2

G = gravitational constant
M = mass of earth
m = mass of pendulum (negligible)

3. The attempt at a solution
g1 = GMm/(Re - h)
g2 = GMm/(Re+h)

g1 = GMm/(Re - h) is for the pendulum underground. Plugging into the period equation gives T = 2π (Re - h) √(L/GMm)

g2 = GMm/(Re+h) is for the pendulum above ground. Plugging into the period equation gives T = 2π (Re + h) √(L/GMm)

It's obvious that the pendulum will tick slower below ground and faster above - but what about the error? I think that the errors would be the same. The question seems to imply that they are not though. I think there is a factor that I am not taking into account?

2. Oct 28, 2016

### drvrm

pl. show a work out
if it was obvious then the problem might not have been posed.
well i do not have the answer and can not work out as its for you....homework thing...

3. Oct 28, 2016

### Vitani11

I showed my work above. :P

4. Oct 28, 2016

### Vitani11

I'm not asking for the answer, I just want to know if I'm on the right track or if there is something I am missing here.

5. Oct 28, 2016

### drvrm

suppose your clock is slower then the time for a full period will be larger
and if your clock is faster then time for full period will be smaller , so ask again when the clock has more elapsed time for full period
above or below

6. Oct 28, 2016

### haruspex

Yes.
Both of those are wrong, even dimensionally wrong.
The first is, in a sense, doubly wrong. The gravitational attraction of a uniform sphere of radius R turns out to be the same as for a point mass, of the same mass, only at radii > R. Inside the sphere it is different. Do you have a formula for that?

7. Oct 28, 2016

### Vitani11

Okay - how does this look?

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8. Oct 28, 2016

### haruspex

You are still using the wrong formula for the gravitational attraction when inside the sphere.
Here's a clue. Inside a unform spherical shell, the shell exerts no gravitational field. So at distance r<Re, what part of the Earth is responsible for the gravitational field? What is the mass of that part? What is the resulting gravitational acceleration?

9. Oct 28, 2016

### Vitani11

This is the resulting gravitational acceleration.

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10. Oct 28, 2016

### Vitani11

I believe this is how it changes when you are underground. the g without the subscript is earths (9.8 m/s2)

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11. Oct 28, 2016

### Buffu

If $h$ is the distance from the surface to the particle, then yes.

Side note: Why don't you learn basic latex/mathJax to format maths then, you do need post a picture every time. It will also help others to understand what you mean.

12. Oct 28, 2016

### Vitani11

I didn't know that was a thing. Now that I'm in the physics major I'll likely be posting here often so I will look into that. Thank you for the resource

13. Oct 29, 2016

### Vitani11

I used percent error and found that it was greater for the pendulum above earth.

Last edited: Oct 29, 2016
14. Oct 29, 2016

### Vitani11

Method

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15. Oct 29, 2016

### PeroK

It's difficult to see what you've done there. To help you with latex, you are looking for something like:

$g_a(h) = g(1 \pm f_a(\frac{h}{R}))$

$g_b(h) = g(1 \pm f_b(\frac{h}{R}))$

Where $g_a(h)$ is the gravity a height $h$ above the Earth and $g_b(h)$ is the gravity a depth $h$ below the Earth's surface. $f_a$ and $f_b$ are functions you need to find and compare. $R$ is the radius of the Earth.

16. Oct 29, 2016

### Vitani11

Okay, but why is it plus or minus? I know below is the expression with a minus sign, and I've gotten those results but I figured that since for above ground the mass isn't changing, then the expression for how gravity changes would just be the typical ga = GM/(R+h)2 divided by g = GM/R2. Either way, I've tried it both ways and am getting that the error is larger above earth but with different percent errors for each function. I've been using h = 30m to compare.

17. Oct 29, 2016

### PeroK

That may be correct for $h = 30m$, but what about other values of $h$.

It was $\pm$ so I didn't give too much away! Did you get a formula for gravity a distance $h$ above ground, compared to surface gravity?

18. Oct 29, 2016

### Vitani11

Yes that is what I did. In my most previous post with pictures you can see the equations I used although I know it's hard to read. I'll try other values for h. Thanks for the help - I think I am in the home stretch, lol.

19. Oct 29, 2016

### PeroK

That's why I posted some latex, so you could hopefully just copy that to get you started.

20. Oct 29, 2016

### Vitani11

What is latex?