Protons to accelerate a space probe

by josephcollins
Tags: accelerate, probe, protons, space
josephcollins is offline
Oct13-04, 01:51 PM
P: 60
Okay ppl, I have a question, I'd just like you to check that my working here is correct.

Q: A space probe of mass m=490kg is driven by an ion propulsion motor that involves the use of positive ions, each of mass m=2.18*10^-25 kg and having a charge of magnitude e=1.6*10^-19 C. These ions, having a negligible initial velocity, are accelerated by a potential difference of magnitude 1.2*10^3 V that exists between two electrodes which are a distance of 5cm apart. The ejection of these accelerated ions from the space probe generates a forward thrust that drives the probe.

I got the velocity with which the ions leave the probe, using the argument that kinetic energy and velocity are equal. So mv^2=QV and v turns out to be 419169.8 m/s

Then I had to calculate the Force exerted on the ions when the number of ions between the electrodes is 2*10^13. This I did by saying that F=EQ, and since E=V/separation, F=VQ/separation. I got an answer of 0.0768N.

Lastly I had to calculate the time taken to accelerate the probe from 0 to 100km/h. I did this by first getting the acceleration of the probe. It is equal to F/m where F=0.0768N and m=490kg. so a=1.567*10^-4 m/s^2. Then I used (v-u)/a gives the time and set v= 100000/3600 and used the accn just calculated to get a time of 49.23 hrs. Could someone just verify that I did this correctly in terms of the logic involved?

Thanks, Joe
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Soveraign is offline
Oct13-04, 05:02 PM
P: 54
I get the same final answers you do, 49.23 hours to get to 100km/h but [itex] \frac 1 2 m v^2 = QV [/itex] gives 4.197 x 10^4 m/s.

[tex] v = \sqrt{\frac{(1.6 \star 10^{-19})(1.2 \star 10^3)}{(.5)(2.18 \star 10^{-25})}} [/tex]

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