3 ms·

> Ions entering are then propelled from the positive to the negative screen and then out beyond, while electrons are reflected. There are thus two exhausts, but

by tbabb 8y ago

> Ions entering are then propelled from the positive to the negative screen and then out beyond, while electrons are reflected. There are thus two exhausts, but because the protons are much more massive than the electrons, the thrust of the ion current is more than 42 times greater than the opposing electron thrust, providing net thrust. Does this follow? I thought the force of an E-field on a particle is proportional to charge and field strength-- so the electrons are lighter, but will be accelerated with the same force to a much higher velocity. So wouldn't the reactive force be the same for both?

andbberger 8y ago
You're right, the author's analysis is incorrect. But I don't think this discredits the concept completely.
nickparker 8y ago
The particles are passing through a fixed distance (between the screens), wherein they experience a fixed force due to the electric field. An electron will shoot through that distance in no time flat and accrue very little momentum. A proton will plod through and transfer far more. For a 250V potential difference with 20m between screens, you can check it yourself here: https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F+mass+of+proton+*+250+V+%2F+20+m+)+*+sqrt(2+*+20+m+%2F+(1+electron+charge+%2F+mass+of+proton+*+250+V+%2F+20+m+))+*+mass+of+proton https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F... https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F+mass+of+electron+*+250+V+%2F+20+m+)+*+sqrt(2+*+20+m+%2F+(1+electron+charge+%2F+mass+of+electron+*+250+V+%2F+20+m+))+*+mass+of+electron https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F... The ratio of those two values is 42.8. This also makes it easy to see why heavier ions full of neutrons are _awesome_. They give a ratio of 5343.
- ananonymoususer 8y ago
  Thank you. I was going to post the same concerns about the result being electrons accelerated much more quickly out the front, thus no net propulsion. I am glad I read the comments first.
T-A 8y ago
Put an electron halfway between the screens. It is subject to a force F directed toward the positive screen, so it accelerates in that direction. Once it's through the screen, the force changes sign, but quickly drops to zero (for infinitely large screens, it is exactly zero everywhere outside the space bounded by the two screens). The work done by the electric field is W = F * d / q where d = distance from the electron's start position to the positive screen q = the electron's charge All this work is now kinetic energy carried by the electron, so we also have W = (m_e/2) * v_e^2 where m_e = mass of the electron v_e = final speed of the electron Therefore, v_e = sqrt( 2 * F * d / (m_e * q) ) If we repeat the experiment with a proton, it accelerates in the opposite direction, toward the negative screen, and its mass m_p is larger, but other than that it all goes as with the electron; the absolute values of q and F (which only depends on electric field strength and q) are the same. So, v_p = sqrt( 2 * F * d / (m_p * q) ) Divide the two equations to get v_e / v_p = sqrt( m_p / m_e ) Momentum is conserved, so we also have M * v + m_e * v_e - m_p * v_p = 0 where M = mass of the screens (and rest of the spacecraft) v = speed gained by the screens (and rest of the spacecraft) and the minus sign accounts for the opposite direction of travel of electron and proton. Use the expression for v_e / v_p to eliminate v_p, solve for v, and find v = (sqrt(m_p / m_e) - 1) * (m_e / M) * v_e In numbers, m_p ~ 1837 * m_e, so v ~ 41.86 * (m_e / M) * v_e and Zubrin's argument is essentially correct.
grogers 8y ago
Even if there wasn't a difference between electron and proton mass, it would still provide net force if the craft is already moving relative to the background plasma (I.e. swimming upstream). Incoming electrons from the front get reflected at their original velocity, but protons get accelerated out the back at arbitrarily high speed (limited by the electric field). The massive difference in electron and proton means it can provide force even if the plasma is coming from behind, which is much more useful.
deleted 8y ago
[deleted]

by tbabb 8y ago

andbberger 8y ago

You're right, the author's analysis is incorrect. But I don't think this discredits the concept completely.

nickparker 8y ago

The particles are passing through a fixed distance (between the screens), wherein they experience a fixed force due to the electric field. An electron will shoot through that distance in no time flat and accrue very little momentum. A proton will plod through and transfer far more. For a 250V potential difference with 20m between screens, you can check it yourself here: https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F+mass+of+proton+*+250+V+%2F+20+m+)+*+sqrt(2+*+20+m+%2F+(1+electron+charge+%2F+mass+of+proton+*+250+V+%2F+20+m+))+*+mass+of+proton https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F... https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F+mass+of+electron+*+250+V+%2F+20+m+)+*+sqrt(2+*+20+m+%2F+(1+electron+charge+%2F+mass+of+electron+*+250+V+%2F+20+m+))+*+mass+of+electron https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F... The ratio of those two values is 42.8. This also makes it easy to see why heavier ions full of neutrons are _awesome_. They give a ratio of 5343.

ananonymoususer 8y ago
Thank you. I was going to post the same concerns about the result being electrons accelerated much more quickly out the front, thus no net propulsion. I am glad I read the comments first.

T-A 8y ago

Put an electron halfway between the screens. It is subject to a force F directed toward the positive screen, so it accelerates in that direction. Once it's through the screen, the force changes sign, but quickly drops to zero (for infinitely large screens, it is exactly zero everywhere outside the space bounded by the two screens). The work done by the electric field is W = F * d / q where d = distance from the electron's start position to the positive screen q = the electron's charge All this work is now kinetic energy carried by the electron, so we also have W = (m_e/2) * v_e^2 where m_e = mass of the electron v_e = final speed of the electron Therefore, v_e = sqrt( 2 * F * d / (m_e * q) ) If we repeat the experiment with a proton, it accelerates in the opposite direction, toward the negative screen, and its mass m_p is larger, but other than that it all goes as with the electron; the absolute values of q and F (which only depends on electric field strength and q) are the same. So, v_p = sqrt( 2 * F * d / (m_p * q) ) Divide the two equations to get v_e / v_p = sqrt( m_p / m_e ) Momentum is conserved, so we also have M * v + m_e * v_e - m_p * v_p = 0 where M = mass of the screens (and rest of the spacecraft) v = speed gained by the screens (and rest of the spacecraft) and the minus sign accounts for the opposite direction of travel of electron and proton. Use the expression for v_e / v_p to eliminate v_p, solve for v, and find v = (sqrt(m_p / m_e) - 1) * (m_e / M) * v_e In numbers, m_p ~ 1837 * m_e, so v ~ 41.86 * (m_e / M) * v_e and Zubrin's argument is essentially correct.

grogers 8y ago

Even if there wasn't a difference between electron and proton mass, it would still provide net force if the craft is already moving relative to the background plasma (I.e. swimming upstream). Incoming electrons from the front get reflected at their original velocity, but protons get accelerated out the back at arbitrarily high speed (limited by the electric field). The massive difference in electron and proton means it can provide force even if the plasma is coming from behind, which is much more useful.

deleted 8y ago

[deleted]