10

First of all, it is a bit odd to talk about specific plans. If you think about emergency procedures for other natural phenomena, such as Hurricanes or Earth quakes, you need to keep in mind that a reversal is a long-term process which is supposed to take decades to centuries. This is what is called geological time scales. In a way, some scientist say, that ...


6

I'm going to attempt to answer your later questions : What kind of measurements are performed - in space - for better understanding the situation and corresponding processes? Is there any kind of monitoring - in space - on the current state of Earth's magnetic field, its development and space weather phenomena, which may influence the field? Yes, there'...


5

I've never heard of a geostationary satellite having magnetic torquers. Since the magnetic field is very weak on such high orbits, wheel desaturation is carried out with thrusters. Also, every geostationary satellite I'm aware of uses propulsion for station keeping, so even if any magnetic effect could cause orbit perturbations on them (as does solar ...


2

The short answer is yes, changes in the Earth's magnetic field would cause a lot of troubles. But here's a longer answer: Currently the Earth's magnetic field does a pretty good job of diverting particles from the Sun and protecting us from solar winds. The weakening of the Earth's magnetic field would change a lot of things: First of all solar winds ...


2

Geostationary is approximately 35,786 km (22,236 mi) above mean sea level. Earth's two main Van Allen Belts extend from an altitude of about 640 to 58,000 km (400 to 36,040 mi) above the surface. Changes to the Earth's magnetic field will change its Van Allen Belts, which may have an impact on geostationary satellites. Particularly if the Inner Belt ...


1

Supplementary information to @Mefitico's answer: Per Wikipedia the field strength on Earth's magnetic equator (about 11 degrees from the Earth's equator and plane of GEO orbits) is about 0.3 gauss times the ratio of the Earth's radius to the GEO radius $$\left(\frac{r}{R_{GEO}}\right)^3 \approx \left(\frac{6378}{42164}\right)^3 \approx 0.0034$$ That ...


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