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I am not versed in general relativity but I see a problem with your scheme. Let us take the ensemble of distant stars as our inertial frame. Whenever a body has acceleration $\textbf{a}$ w.r.t. inertial frame it experiences an inertial force equal to $\textbf{F}_{inertial}=-m\textbf{a}$, where $m$ is body's mass. Now suppose that the large mass $M$ in your ...


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I think your right. Due to equivalence principle, to the free-falling body it will seem like it has no acceleration, as opposed to a body standing on ground which is equivalent to the ground pushing the body so that it will accelerate upward, and hence experience the normal force from the ground. Because the acceleration due to gravity will depend only on ...


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Primary question: This is very similar to the old question of what would happen if you fell into a black hole. You are correct that you wouldn't feel the acceleration due to gravity per se, but you'd still need to worry about tidal forces. These have complicated geometric dependence - they're negligible near the center of a planar mass $M$, and for a ...


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The equation is built up from considering considering the mass $M$ moving with velocity $v$ splitting into $M+\delta M$ moving at ${\bf v}~+~\delta\bf v$ and a smaller piece with mass $\delta M$ moving with velocity ${\bf v}-\bf V$, for $\bf V$ the velocity of hot gases or plume from the rocket. The diagram below illustrates this Conservation of momentum ...


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The temperature in space is not as easy thing, because it is not clear, what is actually whose temperature we want to know. On the Earth, the temperature in the meteorology means the temperature of the air in shadow. But there is no air in the space. The cosmical microwave background has a temperature of around 2.7K, but it is -270C and not -246C. But in ...


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These schemes have been proposed and studied. A spacecraft with a magnetic field could steer charged particles away from it. The magnetic field would have to be much stronger than the Earth's magnetic field. The reason is pretty easy to see. The Lorentz for $\vec F~=~q\vec v\times\vec B$ for the charged particle velocity perpendicular to the magnetic field ...


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Although this is almost an engineering question, there are some simple physical constraints that mean the answer is an emphatic no. At staging, significant change in momentum / trajectory of the payload-bearing stages can only arise from high impulse transfer between the payload bearing and jettisoned stage; this statement is a reformulation of conservation ...


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If there is no gravity in your simulation, your thoughts are on the right lines. The amount of centripetal force (=thrust) required to maintain a space-craft of mass m in circular orbit of radius r at speed v is $F=mv^2/r$. Thrust has to be directed towards the centre of the circle and has to be maintained constantly - so unlike orbiting a planet using ...


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I'm just wondering how does the Probe/NASA determine the Position of the Probe? There are several things that can be directly measured from Earth, as discussed in detail at https://solarsystem.nasa.gov/basics/bsf13-1.php. I describe these briefly below. radial velocity toward/away from Earth: using the Doppler-shift of a well known downlink carrier ...



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