Talk:Magnetoplasmadynamic thruster

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I really strongly feel that this technology will advance the future of deep space exploration.

[edit] Advantages

What does this mean: In theory, MPD thrusters could produce extremely high specific impulses (I_sp) of up to and beyond 11,000 s (110 km/s exhaust velocity)...? 11,000 seconds? Should it be m/s? But that's 11 km/s. Ehn 19:37, 16 October 2005 (UTC)

See specific impulse - for some reasons the specific impulse is traditionally given as (real specific impulse, i. e. the exhaust velocity)/(gravitational acceleration at earth's surface). So you've to multiply by a factor of about 9.8 to get the exhaust velocity in m/s from I_sp in s. 193.171.121.30 10:42, 17 October 2005 (UTC)

MPD thrusters should make a good second stage engine when launching satellites into Low Earth Orbit (LEO) from a Space Elevator. The electric motors that lift 20 metric ton climbers will need about 1.8 MW of electrical power, this is similar to a MPD thruster. For a 400 km LEO orbit climb the Space Elevator to 23 804 km release the satellite and circularise the orbit with a delta-v of -2.132 km/s. A MPD thruster supplying 200 Newtons will take about 7.5 days to circularise the orbit.

Nice. Now your satellite is additionally carrying at least 1-ton nuclear reactor to power your MPD? And why do you want LEO satellites anyway - they are dangerous for the Space elevator.

Using Newton's Laws of Motion Force F is mass m time acceleration a, F = m a and final velocity v equals initial velocity u plus constant acceleration a times time t, v = u + a t then t = delta-v * m / F = -2.132 km/s * 20 000 kg / 200 N = 213 200 seconds (about 2.5 days)

In a highly eccentric Hohmann transfer orbit the space craft can only use its engine to slow down at the bottom near the periapsis, say a third of the time. (Change of inclination burns are performed near the apoapsis.) If the climber climbs at 200 km/h the journey time becomes T = (23 804 km / 200 km/h) / 24 h + 2.5 days * 3 = 12.5 days

With high-Isp, low thrust engines you don't fly Hohmann transfer orbits. You calculate a curve on which you can thrust continuously.

Andrew Swallow 19:49, 13 November 2006 (UTC)

Nuclear reactors are not needed for LEO orbits. Space Elevator climbers will probably be powered by large photovoltaic panels. After releasing the payload only a quarter of the electrical power is needed so three quarters of the panels can go with the satellite and MPD thruster.

Earth monitoring satellites need to be in low orbits to get a detailed view of the Earth. The Space Elevator operators will just have to allow for this.

Objects leaving the Space Elevator below GEO automatically fall into a Hohmann orbit rather than a circular orbit. The simplest option is pick the drop height that produces a Hohmann orbit whose periapsis is the same as the required final orbit. Under these conditions a continuous burn has a large Delta-V taking longer than assumed unless heavy chemical thrusters are used.

Andrew Swallow 08:31, 16 February 2007 (UTC)