Resistance of a wire around the Earth

[godiswatching_]

Homework Statement

Homework statement in the image.

Relevant Equations

$$R=\rho\frac{l}{A}$$
$$l=2\pi r$$

Hey! I had a question about this problem.

I did (1) Using
$$R_{0}=\rho\frac{l}{A}$$

For (2) I assume the question means that the radius increases by a meter.
So I used $$\bigtriangleup L = 2\pi (r_{E}+1) - l$$
and then I used that L to find the new R. Then I said $$\bigtriangleup R = R-R_{0}$$

Does that seem right? This seems too simple to be right.

My final answer was:
$$\bigtriangleup R = 3.4 \cdot 10^{-4} \Omega$$

 

[BvU]

'Too simple' is not an argument ...
What could possible be wrong ?

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[godiswatching_]

'Too simple' is not an argument ...
What could possible be wrong ?

$\$

I don’t see anything wrong with my steps. I think they are all logically sound. It’s more so that I’m paranoid and don’t want to lose points on homework for no reason really.

 

[BvU]

PF isn't in the business of stamp-approving homework. It wouldn't help anyone. You do your calculation, get a result and check it. Twice if you want, three times if you are paranoid

Have some faith in your work ...

$\$

 

[vela]

I find it pretty hard to believe that adding a mere 6.28 m of wire is enough to lift it by 1 meter off the ground all the way around the Earth, but it does indeed.

 

[Mark44]

I find it pretty hard to believe that adding a mere 6.28 m of wire is enough to lift it by 1 meter off the ground all the way around the Earth, but it does indeed.

It seemed unbelievable when I first heard this one, but the math bears it out.
$C = 2\pi r \Rightarrow \Delta C = 2\pi \Delta r$ -- this is an equality due to the linearity of this function.
If $\Delta r = 1 \text{ meter}$, then $\Delta C = 2\pi \cdot 1 \approx 6.28 \text{ meters}$