1
$\begingroup$

In some comments on this forum I read that satellites have capacitance of some nF or pF, What is this about? How can I calculate it roughly?

$\endgroup$
3
  • $\begingroup$ Can you post a link to such a comment (or its question)? $\endgroup$ – DrSheldon Apr 8 at 4:39
  • 2
    $\begingroup$ Everything has capacitance. When you tap the screen of your phone, it detects your finger because you have capacitance. $\endgroup$ – SF. Apr 8 at 9:21
  • $\begingroup$ Consider it a metallic sphere of a given diameter. This is now a standard electrostatics calculation - put some charge on it, it has to be at some potential relative to ground then. Calculation of the capacitance follows directly from Q/V. $\endgroup$ – Jon Custer Apr 8 at 15:46
2
$\begingroup$

Why do satellites have capacitance?

Capacitance is some ratio of charge to potential difference, with a constant in front.

Whenever some charge is added to something, be it a person, a cow, or a bit of metal in space, it's electrostatic potential changes. Capacitance is simply the ratio that tells us; *for a given charge added, how does much does the electrostatic potential change?

Our spacecraft could be once capacitance relative to one thing, and at the same time a different capacitance to another.

For a parallel plate capacitor we have:

$$C = \frac{Q}{V} = \frac{\varepsilon A}{d}$$

with $A$ the area of each plate and $d$ the separation between them.

For an isolated conducting spherical cow shell we use the same definition. For two spherical shells

$$C = \frac{Q}{V} = \frac{4 \pi \varepsilon_0}{\frac{1}{R_1}-\frac{1}{R_2}}$$

and for a single isolated sphere of radius $R$ just set the larger radius $R_2$ to infinity and get

$$C = \frac{Q}{V} = \frac{4 \pi \varepsilon_0} R$$

With vacuum permitivity $\varepsilon = 8.854 \times 10^{-12} $ F/m a one meter diameter sphere will then have a capacitance of about 0.2 nF. If your spacecraft is not a spherical cow and instead has some pointy bits and corners then the charge will accumulate at the points and corners, the electric field will be higher and capacitance could be orders of magnitude higher.

However:

  1. For any realistic shape, you need a computer model, or you need to build a mock-up and test it.

  2. If in low-enough Earth obit, one needs to think about capacitance with respect to the very low density plasma of the ionosphere, rather than to infinity.

These are why a spacecraft don't actually measure their own charge. They instead measure their potential.

$\endgroup$
1
  • 2
    $\begingroup$ A spherical cow in a vacuum, as it canonically should be. +1. $\endgroup$ – Camille Goudeseune Apr 8 at 17:49

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