3
$\begingroup$

https://en.wikipedia.org/wiki/Ionosphere The ionosphere is a shell of electrons and electrically charged atoms and molecules that surrounds the Earth, stretching from a height of about 50 km (31 mi) to more than 1,000 km (620 mi)

On the other hand.

https://en.wikipedia.org/wiki/Ground_(electricity)

A "ground" is usually idealized as an infinite source or sink for charge, which can absorb an unlimited amount of current without changing its potential.

And We also Know that balloon named BU60-1 has attained highest altitude of 53.0 km (53 Km means it entered the ionosphere)

So, I am wondering if may be someday we should try this. Even is that possible to connect ground and ionosphere.

By the way, I love to think about something that is completely bizarre. Besides this I can understand problems in doing so like; Weight, material, amount of charges (if able to establish) and various others (honestly, at the moment I don't remember/know). Well, my choice of material for path is very non-conducting type such ordinary thread or anything else that is not much expensive. I have started wondering already.

If you don't want to be specific, that is great but please don't involve too much mathematics in explanation (if there is one!). Answer in such a way that everyone understand and make them love science and specially Physics.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ ask benjamin franklin $\endgroup$
    – user116941
    Jun 5, 2016 at 21:44
  • $\begingroup$ the principle seems correct, but I am not sure about the practicality or economics of it (just think that he amount of current will be highly variable depending on atmospheric conditions), perhaps a better question for engineering $\endgroup$
    – user83548
    Jun 5, 2016 at 21:49
  • $\begingroup$ The entire thing is a spherical capacitor of relatively low capacitance. If you put in optimistic numbers, you end up with something like 0.1Farad. You can buy capacitors that will fit into the palm of your hand that have several Farad capacitance. Let's assume that the potential difference between ground and the ionosphere is 300kV, then the entire energy in there would be on the order of $E=0.1F\times (300kV)^2/2=4.5GJ$. That's as much energy as you get from one large power plant in about 5 seconds. You can completely forget about it. $\endgroup$
    – CuriousOne
    Jun 5, 2016 at 22:35
  • $\begingroup$ So if I use $n_{e} \sim 10^{4} - 10^{6} \ cm^{-3}$, $h_{iono} \sim 50-1000 \ km$, I get a volume of ~$10^{18} \ m^{3}$ and total number of electrons to be ~$10^{28} - 10^{30}$. This corresponds to a total charge from electrons of $Q \sim 10^{9} - 10^{11} \ C$. If we assume $\Delta \Phi \sim 300 \ kV$, then we have $$C = \tfrac{Q}{\Delta \Phi} \sim 19 \ kF - 19 \ MF$$. Take this with a massive grain of salt because the ionosphere is not made of only electrons, so $Q$ is likely very different... $\endgroup$
    – honeste_vivere
    Jun 8, 2016 at 12:29

1 Answer 1

Reset to default
4
$\begingroup$

@CuriousOne is approximately right. Others have thought of it, the url below is a paper from one who concludes with slightly more careful calculations (but still relatively straightforward) that the most one could get in terms of a continuous power from it is 50 MW, and typically a lot less. 50 MW is 5% of a large power plant.

http://www.electrostatics.org/images/a2.pdf

The paper uses 200KV instead of 300, but both are in the right ballpark. The paper figures out the most power that can be extracted from an equivalent circuit diagram. It also shows how little can be extracted from local harvesting.

An interesting intro to the the whole topic of the global atmospheric electrical circuit can be found in the url below, the first lecture in a course on it. It shows nicely the fair weather global circuit. 200-300 KV is about average, and it shows other applicable parameters.

http://www.atmo.arizona.edu/students/courselinks/spring13/atmo589/ATMO489_online/lecture_1/lect1_global_elec_circuit.html

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.