Difference between revisions of "Soft Electric Landing on Luna"
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The energy stored is equal to th kinetic energy of the vehicle, plus the potential energy in the elevation of the overall orbit. | The energy stored is equal to th kinetic energy of the vehicle, plus the potential energy in the elevation of the overall orbit. | ||
With the same velocity of 1800 m/s | With the same velocity of 1800 m/s | ||
| − | Each kg caught would deliver e=1/2mv2 = 1800*1800/2 = 1 620 000 J or 1,6 MJ per kg | + | Each kg caught would deliver e=1/2mv2 = 1800*1800/2 = 1 620 000 J or 1,6 MJ per kg minus the system energy losses. |
| − | |||
| + | Launching to orbit would require the same energy, while launching to escape velocity would require 2380 km/s or 2,8 MJ plus the energy losses in the system. | ||
== See Also == | == See Also == | ||
Revision as of 20:21, 17 October 2022
Soft electric landing on Luna can be provided for an orbiting spacecraft in the following way.
- The spacecraft lines up with the runway on the equator at the same spot from which spacecraft are launched to L2.
- The incoming orbit is right down the middle of a rectangular cross section ditch.
- Two slots, one on each side of the ditch, lead to the linear electric motors that power the catcher that also races down this ditch.
- Structural members of the catcher disappear into these slots terminating in the moving portion of the linear electric motors.
- The spacecraft maneuvers so that its orbit brings it to the right place with the right velocity.
- The catcher adjusts its run to be there at the right time for a rendezvous.
- After catching the spacecraft, the catcher brakes electromagnetically, saving the energy as electricity that must be stored in an accululator, such as a capacitor bank, batteries or induction rings. This energy can later be used to launch payloads.
This arrangement may be easier than soft landing by tether because a tether cannot adjust the time that it arrives at a rendezvous with a spacecraft.
The facility described above is a high capacity version of a mass driver.
The same track should probably be usable for launching.
Sizing the landing system
This is landing at orbital speed, and would require a very long track to keep acceleration low. As V=a*t Orbital velocity= v = 1680 m/s if we add 10% to elevate the orbit a bit, with let's say 1800 m/s for an acceleration (a) of 2g, or 20 m/s2, the time required is 90 seconds. In that time d=a*t^2/2 = 81 000m or 81 km.
The energy stored is equal to th kinetic energy of the vehicle, plus the potential energy in the elevation of the overall orbit. With the same velocity of 1800 m/s Each kg caught would deliver e=1/2mv2 = 1800*1800/2 = 1 620 000 J or 1,6 MJ per kg minus the system energy losses.
Launching to orbit would require the same energy, while launching to escape velocity would require 2380 km/s or 2,8 MJ plus the energy losses in the system.
