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

[Vitani11]

Homework Statement

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?

Homework Equations

T = 2π √(L/g)

L = length of the pendulum
g = gravity on Earth's surface
T = period

g = GMm/R_e²

G = gravitational constant
R_e = Earth's radius
M = mass of earth
m = mass of pendulum (negligible)

The Attempt at a Solution

g₁ = GMm/(R_e - h)
g₂ = GMm/(R_e+h)

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

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

 

[drvrm]

It's obvious that the pendulum will tick slower below ground and faster above -

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...

 

[Vitani11]

I showed my work above. :P

 

[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.

 

[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

 

[haruspex]

g = GMm/Re²

Yes.

g1 = GMm/(Re - h)
g2 = GMm/(Re+h)

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?

 

[Vitani11]

Okay - how does this look?

 

[haruspex]

Okay - how does this look?

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<R_e, what part of the Earth is responsible for the gravitational field? What is the mass of that part? What is the resulting gravitational acceleration?

 

[Vitani11]

This is the resulting gravitational acceleration.

 

[Vitani11]

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

 

[Buffu]

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

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.

 

[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

 

[Vitani11]

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

 

[Vitani11]

Method

 

[PeroK]

Method

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.

 

[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 g_a = 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.

 

[PeroK]

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 g_a = 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.

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?

 

[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.

 

[PeroK]

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.

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

 

[Vitani11]

What is latex?

 

[PeroK]

What is latex?

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

That's your equation for gravity below ground, which is correct, using latex.

The above ground is harder. Do you know about binomial expansions for negative powers?

 

[haruspex]

What is latex?

You can Google that. E.g. http://www.colorado.edu/physics/phys4610/phys4610_sp15/PHYS4610_sp15/Home_files/LaTeXSymbols.pdf. To type it in on these forums you can wrap it in two pairs of hash symbols, #. E.g. If you write ??x^2??, but change the four ? to # then click Preview it will come out as $x^2$.
If you want to see how a previous post wrote the latex you can right click on a line of algebra and select View as...

But you can do quite well without latex. Just use the x₂ and x² buttons above the typing area.

Posting algebra as images is a bad idea. Most people won't bother struggling to read it, and it makes it hard to comment on specific items in it.

Back to the problem. You still seem to be making errors in the algebraic forms of the two cases (but it is hard to be sure from your posts). If R is the Earth's radius, and g is the gravitational acceleration at the surface, what is the gravity at (a) height h above ground and (b) depth h below ground? Express the answers just using R, g and h.