Some advocate sourcing all the world's energy with space-based solar power, on the order of 30 Terawatts. They suggest power satellite masses on the order of 4 kg / kW, or 0.004 kg per watt. That works out to 4e-3 kg/W * 3e13 W or 120e9 kg of power satellite mass delivered to Geostationary Orbit ( GEO, 42164 km radius, 3075 m/s ). This does not include construction equipment or longitudinal station-keeping fuel.

Modern comsats use noble-gas electric thrusters to climb to GEO. This is propellant-thrifty; however, this has driven up the price of Xenon and is shifting to Krypton. There is not enough available high-Z noble gas (xenon, krypton) to raise the orbits of 120 million of tonnes of power satellite. There is plenty of argon, but that must be chilled below 27 Kelvin to store densely for the long trip to GEO.

The transfer orbit, and GEO injection velocity

So, consider liquid hydrogen/liquid oxygen ( LH/LOX) chemical rockets, with typical 1:6 mass ratio (slightly fuel rich), producing a propellant plume of and H₂ and H. The mass ratio of water to total hydrogen ( H₂ plus H ), assuming complete combustion and no hydroxyl ( OH or HO ) or hydronium ( H₃O ).

Assuming a high expansion vacuum nozzle, the exhaust velocity of a liquid hydrogen engine might be as high as 4460 m/s.

Spacecraft typically launch to GEO from low earth orbit (LEO) via a geostationary transfer orbit (GTO). Presume that the GTO perigee starts from a 122 km equatorial altitude so that the GTO is an ellipse with perigee = 6500 km, apogee = 42164 km, semi-major axis 24332 km. The apogee velocity of the GTO ellipse is 1589 m/s. The GEO circularization delta V is (3075-1589) = 1486 m/s .

Presuming an exhaust velocity of 4460 m/s, and a delta V of 1486 m/s, the plume mass can be calculated using the Tsiolkovsky equation:

M_{plume} = M_{delivered} \times ( e^{ \Delta V / V_e } - 1 ) = 120e9 kg × ( exp( 1486 / 4460 ) - 1 ) = 120e9 kg × 0.395 = 47e9 kg.

The plume will be emitted in a retrograde direction from the vehicle as it accelerates from 1589 to 3075 m/s ; the plume velocity will range from (4460-1589) m/s to (4460-3075) m/s retrograde, or 2871 m/s to 1385 m/s.

What happens to the plume??

Even with a high expansion nozzle, the plume will be hot, with significant Maxwellian thermal velocity. Some species (molecules, atoms, ions) will travel more slowly than average, some faster. But let's ignore that to start with, and assume the gas molecules are all in retrograde elliptical orbits. What are the perigees of those orbits?

Well, to start with, we know a retrograde orbit with a velocity of 1589 m/s has a perigee of 122 km; slower species will descend to a lower altitude and fall into the atmosphere. Species with a retrograde velocity of 2871 m/s will have an perigee of 32600 km, and will remain above the atmosphere for a very long time. Species with a retrograde velocity of 1640 m/s will have a perigee above 600 km, and will remain in orbit for decades.

Lets consider the portion of the plume emitted between 1640 m/s and 2871 m/s to be in semi-permanent orbits, and the portion of the plume emitted between 1589 m/s and 1640 m/s to re-enter and stop being a problem. If the plume is 1640 m/s retrograde, the vehicle is travelling 4460-1640 = 2820 m/s prograde. So, the re-entering portion of the plume is the fraction emitted while the vehicle accelerates from 2820 to 3075 m/s, a delta V of 255 m/s. Let's use the Tsiolkovsky equation for that:

M_{entry} = M_{delivered} \times ( e^{ \Delta V / V_e } - 1 ) = 120e9 kg × ( exp( 255 / 4460 ) - 1 ) = 120e9 kg × 0.0588 = 7e9 kg.

So, the plume emitted into semipermanent orbits is (47-7)e9 = 40e9 kg.

Orbits with an apogee at GEO!