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Let's assume there is no change in weight due to fuel consumption. That's a rather unrealistic assumption. If you get rid of that assumption, landing requires considerably more energy than does ascent. Some realistic assumptions: The lander Starts docked to some orbiting spacecraft, Separates from the orbiting spacecraft, Performs a de-orbit maneuver ...


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It is a bit more complicated. You should be able to land without the use of energy but you have to pass your momentum to something - and this passing on of momentum will be the same in both cases. Let us consider a "spherically-symmetric-chicken-in-a-vacuum" model to see this explicitly. Momentum conservation and propelled mass The spacecraft of mass ...


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Now it occurred to me that the escape velocity/breaking may be symmetrical - that means you need the same amount of energy to counter the gravity on the way up as you need on the way down - but what also matters is how long you stay "hoovering" in the gravitation filed. This is what consumes fuel no matter what way you go and in reality both descend and ...


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Landing is gravity-assisted, so requires less energy. A spacecraft on the moon has used up fuel resulting in its mass being less. If it is significantly less, then it takes less energy to lift a less massive craft back into orbit, than landing a heavier craft. Escape velocity applies only to non-powered projectiles, such as shooting a canon ball. There is ...


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I do not see any other possibility than doing the integral along the path. Do your telemetry data include the horizontal position? because if does you can calculate an approximation to the integral. Of course, there is no guarantee that the error will be larger than the effect you want to measure. You should have to do some pre-tests (using situations that ...


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The exobase is defined as the effective end of the atmosphere, and is a gray area between 500 km and 1,000 km. Presumably, once a craft's orbit is outside of the exobase, drag is negligible and stationkeeping basically not a necessity (as in won't return to earth for a century or more, until we can repark it or utilize it). From my time in Kerbal, I would ...


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Let's take the things step by step, from linear momentum conservation. Of course, the rigorous treatment is by Tsiolkovsky's law. Here I am saying something less rigorous but intuitive. To increase the velocity from zero to $\Delta V$ you consume a mass of fuel $\Delta m$. When I say from zero, I mean that I consider a discrete series of quick fuel ...


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There is an indigenous group of people in South America, the Aimaras. When they mention something that happened in the past, they do not signal to their back, but to their front. To their perception, the past is in front of them, because they can see it. And the future is behind just for the opposite reason. At least in western countries, our perceptions is ...


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Why is mine wrong? What is the intuition? To see what happens, use finite differences to write $$(M - \Delta m) \Delta v = v_{rel}\Delta m$$ which leads to $$M \Delta v - \Delta m \Delta v = v_{rel} \Delta m$$ Dividing through by $\Delta t$ yields $$M \frac{\Delta v}{\Delta t} - \Delta m\frac{ \Delta v}{\Delta t} = v_{rel} \frac{\Delta m}{\Delta ...


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Newton's 2nd Law states that the 'the change in momentum (p) of a body is proportional to the force (F) exerted on the body'. Mathematically, we write: $$F_{ext}=\frac{dp}{dt}$$ Where a body of mass $m$ and velocity $v$ has momentum $p=mv$. For a body of constant mass $m$, this becomes: $$F_{ext}=\frac{dp}{dt}=m\frac{dv}{dt}=ma$$ Where the mass $m$ ...


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Newton's second law relates force to acceleration: $$ F = Ma \tag{1} $$ but force can also be expressed as the rate of change of momentum: $$ F = \frac{dp}{dt} \tag{2} $$ In the case of a rocket it's the rate of change of momentum of the exhaust gas has that produces the force. Momentum is given by $p = mv$ (note $M$ is the mass of the rocket and $m$ is ...


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The downstream side of the nozzle is much more important to maintaining the efficiency of the nozzle by controlling expansion wave. It also influences the uniformity of the flow exiting the nozzle. Symmetry would be perfectly fine, but you'd end up making the converging section bigger than it needs to be. Here's a bit more about the design of the nozzle ...


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A good estimation would be an Hohmann transfer from LEO to MEO. It might be slightly more efficient if the burn would be performed in the upper atmosphere, due to the Oberth effect. When you calculate the total required $\Delta$v for the two burns of the Hohmann transfer, you get about 2.3 km/s, so the total $\Delta$v to get from Earth to MEO would be about ...


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Specific impulse is usually defined as $I_{sp} = \frac{F_T}{\dot m ~ g_0} $ That's true only if you use standard metric units. With force expressed in pounds-force and mass flow rate expressed in pounds, one simply divides the force (in pounds-force) by the mass flow rate (in pounds/second) and voila! you have specific impulse in lbf·s/lb. For example, ...



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