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7

My guess is that the exhaust is not perfectly in line with the balloon's centre of mass. In other words, the exhaust is slightly off-centre. That will cause the thrust of the escaping gas to push the nozzle slightly to the side, causing the balloon to rotate sightly as it travels forward. If this deflection stays constant you will get a perfect circle. If ...

1

The Poynting vector $\vec{N}$ is the power per unit area of your beam. If the beam is perfectly absorbed, then the force is given by $$F = \frac{1}{c} \int \vec{N} \cdot d\vec{A}$$ So, providing you have the beam incident normally upon something, the force on it will just be the power of the laser divided by the speed of light. Of course, if the light is ...

0

Thank you both for working this problem. Rob Jeffries has it right, but I'll present a derivation that's purely SR, now that I better understand the problem. I take the relativistic Doppler effect as a given: $$\lambda'= \sqrt\frac{1+\beta}{1-\beta} \lambda$$ Since $t=\lambda/c$ , this is also how time transforms; i.e. $$t'= \sqrt\frac{1+\beta}{1-\beta} ... 0 The thrust for a rocket (can be demonstrated with global conservation of momentum on control volume set around the rocket) is, in general:$$\mathbb{T}=\dot{m}u_e + A_e\left( p_e - p_a \right) $$where \dot{m} is the mass flow rate \left(\frac{M_p}{t_b} \right) (being M_p the mass of propellant and t_b the burning time) and u_e the gases exit ... 1 So within 2 minutes of posting this I think I have the answer, and it turned out to be incredibly simple. I guess I've just been out of it for some time :) The exit area is constant so: The density multiplied by velocity gives us the mass flow rate. Substituting this result into the equation for dynamic pressure shows us the the dynamic pressure is equal to ... 0 To answer your question: yes, it's perfectly possible to achieve a sizable \Delta V with an Electric Propulsion device as long as you take enough time and have enough propellant. EP is in general low thrust but high I_{sp}. I assume they have enough time and/or planned a low-thrust transfer rather than a more traditional method so we wont consider that. ... 1 In an inertial frame, the sail acceleration is A. [...] P Watts of photon power ... let's say "photon power, or wattage" of the source: P_{\text{source}} ... is beamed towards the sail It seems therefore that you're thinking of a beam source which remains a member of an inertial frame, emitting a constant power beam (i.e. constant as judged ... 2 I think I understand your problem, but I think there is at least one mistake. If I work in the frame of reference of the spacecraft, its mass is constant. The Poynting vector intercepted from the photon beam and hence force exerted is diminished by a factor \gamma^2 (1-v/c)^2 (see below). Thus I think that$$ a = \frac{2P \gamma^2 (1 - v/c)^2}{mc} = ...

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