Return-Path: Received: from mtain-ma05.r1000.mx.aol.com (mtain-ma05.r1000.mx.aol.com [172.29.96.13]) by air-dd08.mail.aol.com (v129.4) with ESMTP id MAILINDD084-86b04d8faf78214; Sun, 27 Mar 2011 17:43:20 -0400 Received: from post.thorcom.com (post.thorcom.com [195.171.43.25]) by mtain-ma05.r1000.mx.aol.com (Internet Inbound) with ESMTP id 2717C3800148F; Sun, 27 Mar 2011 17:43:15 -0400 (EDT) Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1Q3xj2-0007xD-95 for rs_out_1@blacksheep.org; Sun, 27 Mar 2011 22:42:20 +0100 Received: from [195.171.43.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1Q3xj1-0007x4-Ik for rsgb_lf_group@blacksheep.org; Sun, 27 Mar 2011 22:42:19 +0100 Received: from mail-iw0-f171.google.com ([209.85.214.171]) by relay1.thorcom.net with esmtp (Exim 4.63) (envelope-from ) id 1Q3xiz-00058c-LT for rsgb_lf_group@blacksheep.org; Sun, 27 Mar 2011 22:42:19 +0100 Received: by iwn8 with SMTP id 8so3349275iwn.16 for ; Sun, 27 Mar 2011 14:42:11 -0700 (PDT) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=gamma; h=domainkey-signature:mime-version:in-reply-to:references:date :message-id:subject:from:to:content-type:content-transfer-encoding; bh=4pEQT2Xd76kRO6ElX1VZj0/oin+YBWChDZaECxylX0I=; b=VGXuWt2lubTtLA7c0PhmyHfgHZabWp5plILPN77cTzKPqqsaIJUK4gCvXRQzfjwI43 XdG3gs97I2el02oZNRsaIEUDrFU1RPkZ3gPaFQ6+jLGM4iaKmdrHdeROxyPyc5eQqLH2 KcuBeo08Da0uNeXGRsgoWlCzfztJiwnqpSe1U= DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :content-type:content-transfer-encoding; b=Hke/l8QeeriGcRk0yGnMwuQpcb3Y8UXDKR95tcj+lUm5G4PFrkoVkN1BSJaGQilRux H4eL93ct1GBCb98lFgCjx01h57GNYM5DfrDev4o6zJKnGBXm3jL7osmm7je6KhSRYQPq lJShcx26KQDTR0hkihsX9gjiN38gIqI1DlXdU= MIME-Version: 1.0 Received: by 10.42.162.1 with SMTP id v1mr987504icx.56.1301262130974; Sun, 27 Mar 2011 14:42:10 -0700 (PDT) Received: by 10.231.199.132 with HTTP; Sun, 27 Mar 2011 14:42:10 -0700 (PDT) In-Reply-To: <4D8FA9B3.1020303@toya.net.pl> References: <4D8FA9B3.1020303@toya.net.pl> Date: Sun, 27 Mar 2011 22:42:10 +0100 Message-ID: From: Andy Talbot To: rsgb_lf_group@blacksheep.org DomainKey-Status: good (testing) X-Spam-Score: 0.0 (/) X-Spam-Report: autolearn=disabled,none Subject: Re: LF: Earth capacitance Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.0 required=5.0 tests=none autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false x-aol-global-disposition: G X-AOL-SCOLL-AUTHENTICATION: mail_rly_antispam_dkim-m008.1 ; domain : gmail.com DKIM : pass x-aol-sid: 3039ac1d600d4d8faf734202 X-AOL-IP: 195.171.43.25 X-AOL-SPF: domain : blacksheep.org SPF : none X-Mailer: Unknown (No Version) A slightly different way of doing it - from EM first principle, but yes spot on Very close to both my calc and the quoted figure seen in a paper. Worked out from the parallel plate capacitor equation : Area of Earth surface =3D 4.pi.6371000^2 m^2 Height of ionosphere d =3D 60000m C =3D Eo . A / d =3D 0.075F. Which is a large electrolytic as seen in a typical big amateur 12V linear= PSU And seemed awfully high when I first saw the figure. Especially when charged to quarter of a MV Andy 'JNT On 27 March 2011 22:18, Piotr Mlynarski wrote: > Andy Talbot pisze: >> >> Here's an interesting little thought excercise >> >> Consider the Earth as one plate of a capacitor, and the lower surface >> of the ionosphere as the other. =A0 Without actually doing the >> calculation, what do you 'feel' the capacitiance might be? >> >> Then do the calculation; the result is rather surprising >> >> > > Dear Andy, LF > > as i had completely no idea what might be the value of such defined > capacitance > i immediately thought that this "interesting little thought excercise" > =A0could be added to > my sunday evening cup of coffee ... :) > > Gauss theorem says that E field from charge Q on a sphere of radius R is > E =3D Q/(4*pi*eps*R*R) > having now two spheres: inner (earth) R1 and outer: ionosphere R2 > the potential V =3D integral from R1 to R2 over EdR which immediately gi= ves > V =3D (Q/4*pi*eps)*(1/R1 - 1/R2) so as C =3D Q/V one gets > C =A0=3D 4*pi*eps/(1/R1 - 1/R2) > assuming R1 =3D 6370km ; R2 =3D 6370km + 60km =3D 6430 km; eps =3D 8.854= *10^-12 F/m > C is about 0.076F or 76 miliFarad > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0pse, check my= "homework" =A0:) =A0 Piotr, sq7mpj > > qth: Lodz /jo91rs/ > > > >