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Calc: Obi-Wan Kenobi's Force Push

Updated: Dec 17, 2022



In Star Wars: Revenge Of the Sith, Obi-Wan Kenobi duels the cyborg General Grievous, cutting off two of his arms before taking him by surprise with a force push.


Grievous travels upward at at an angle for 16 frames. To find the distance he travels, I tried to scale off Grievous's velocity, but it turns out that he accelerates midflight (implying that force pushes are long-duration applications of force, not simply impacts). Fine. Have it your way. I'll just guess the height.


If Grievous impacts at 12 m up, then we can calculate his gravitational potential energy from this and his mass, which Wookiepedia tells me is 159 kg. The gravitational acceleration on Utapau is "standard" which I guess means pretty close to Earth's. This gives me:

159 kg x 9.8 m/s^2 x 12 m = 19000 J


Don't forget Grievous also travels horizontally, so we have to figure that out as well. If Grievous travels about 4 m horizontally, and the horizontal force that Kenobi's force push exerts on him is 1kN if he accelerates at 6.4 m/s^2 (which is just a rough guess).

1000 N x 4 m = 4000 J


Now the total energy:

19000 + 4000 = 23000 J (23 kJ; Wall level)


As with all calcs on this site, this is pretty back-of-the-envelope. However, I'm pretty happy with it, and it makes sense.

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