# How to calculate the orbital velocity, the high-orbit (quasigeostationary) holding force and the shadow of the disk on Earth? [closed]

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The following data are given:

• The circular orbit of the disk D (quasigeostationary orbit) parallel to the geostationary orbit. • Located 4,505 km to the north of the equatorial plane, above the 45 degree parallel. The radius of orbit with respect to the gravitational center of the Earth 41,922 km (calculated by the uhoh we thank) • Disc diameter: 10 km. • The disc consists of a support frame and solar panels. The specific weight approx. 4.5 kg per square meter. • The time when the disc makes a complete, retrograde motion, around the earth: 25 hours.

Required: 1 Orbital speed of the disk so as to achieve a complete circular motion in 25 hours.

2 The force required to maintain the disk in orbit and to compensate gravitational attraction. 2.1 How can this force be assured? 2,2 with thermochemical propellants (rockets with liquid fuel that could be started periodically? 2.3 electromagnetic, ionic, photonic propellants with cold gases. 2.4 And so on

3 What is the diameter of the disk shadow on the Earth's surface?

Figures 1: 2 and 3 graphically reflect the problem.

Any critical comments or comments are welcome.

• I'm voting to close this question as off-topic because questions about propulsion of artificial satellites is not related to Astronomy. It is however on-topic in Space Exploration SE where your previous question on this topic received three answers. – uhoh Oct 14 '19 at 8:40

## 1 Answer

Let's just pick up the second point. The disk is 42000 km (more or less) from the centre of the Earth, do the gravitational acceleration is $$9.81\times \left(\frac{6371}{41922}\right)^2 = 0.22 ms^{-2}$$ The component of that along the Earth's axis (Southwards) is $$0.22\times\frac{4505}{41922} = 0.024 ms^{-2}$$

The mass of the disk is $$5000^2\times\pi\times 4.5 = 3.53\times 10^8 kg$$ so the total force which must be counteracted by thrust is $$3.53\times 10^8\times 0.024 = 8.5\times 10^6 N$$

This is roughly the thrust of 4 of SpaceX's new Raptor engines burning continuously, each consuming about 1 ton per second of propellants. The best ion engines I can see quickly on wikipedia produce about $$2\times 10^{-5} N/W$$ (thrust/power consumption), so you'd need over $$10^{12} W$$ of power to achieve this thrust with ion engines. The disk collects at most about 200GW of incoming sunlight, so the numbers don't add up that way. You'd also need quite a lot of Xenon and the ion engines would add a lot of mass.

Moving on to the shadow. Seen from Earth's surface the disk subtends an angle of about $$\frac{10}{41922} = 2.5\times 10^{-4}$$ radians. The Sun subtends an angle of about $$0.01$$ radians (half a degree). So the disk would not come close to blocking the Sun at all. That is there would be no umbra at all. In the penumbra the amount of sunlight reaching the Earth would be reduced by about 0.1%

• It seems like a huge force is needed. But where is the quasigeostational orbital velocity which, calculated to surround the Earth in 24 hours, is 3,057 km / s and which counteracts much of the gravitational force? Then if the specific weight of the solar panel disk is reduced to 1 kg / sqm the required force is reduced by 4.5 times. Intuitively I feel that the force of holding the disk in a quasigeostationary orbit must be much smaller. And which should be applied from time to time only for corrections. – Ion Corbu Oct 14 '19 at 1:33
• I'm not talking about the force needed to launch and place the disc in orbit. I'm just talking about keeping the disc in that orbit. – Ion Corbu Oct 14 '19 at 1:33
• I edited the post for clarity – Ion Corbu Oct 14 '19 at 2:00
• The difference between the $0.22 ms^{-2}$ total gravitational acceleration at that distance and the $0.024 ms^{-2}$ component of that along the Earth's axis is the effect of the quasigeostationery orbital velocity, Without that, you would need 10 times as much thrust. – Steve Linton Oct 14 '19 at 6:18