47

Delta-V to LEO is about 10 km/s. From there to C3 (Earth escape) is another 3.2 km/s. It's just another 30% delta-V. The problem is the Tyranny of the Rocket Equation. More delta-V means more fuel. More fuel means more mass. More mass means more fuel. How much more? Fuel costs scale according to $e^{\frac{\Delta V}{v_e}}$, that is e to the power of the ratio ...


34

It's not hard, it's just expensive. We know exactly how to do it. Compare this to building computer processors with 1nm transistors, or making reliable self-driving cars. Those are both things that we currently don't know how to do, and we don't even know exactly how to get better at doing them. Even going past low Earth orbit to another planet, like Mars, ...


29

The velocity of a rocket can exceed its exhaust velocity. It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. ...The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum. Source https://courses.lumenlearning.com/suny-...


27

An intuitive way to think about it: You have a big rocket, composed of two parts of equal mass: the payload part and the fuel part. You launch the exhaust (fuel) backwards at 1 km/s (for simplicity: all at once so you don't accelerate any of it forward first), and so clearly the payload part is now moving at 1 km/s forward. Open the fairing and reveal the ...


14

Wikimedia has the following graph for the heliocentric velocities of both Pioneer probes: (SVG) As far as I can tell it's accurate, since it clearly shows the velocity change of Pioneer 10 during its Jupiter flyby on new year 1974, and Pioneer 11's two flybys of Jupiter on new year 1975 and Saturn in 1979. And New Horizons: (SVG) (This answer also has one.) ...


9

For a rocket that is not subject to external forces, conservation of momentum dictates that $$m(t)\,\dot v(t) + v_e(t)\,\dot m(t) = 0$$ where $m(t)$ is the mass of the rocket, including propellant, at time $t$, $v(t)$ is the rocket's velocity at time $t$, relative to some inertial observer, $\dot v(t)$ is the rocket's acceleration at time $t$, $v_e(t)$ is ...


8

Return trips are harder The main "problem" with crewed trips is that we generally want those people to return back. This means that we don't just need to accelerate the manned part to the required velocity, but we also need to accelerate a sufficient amount of fuel and engines for the return trip, which is a significant increase - as the other ...


5

Consider a rocket floating motionless in space. You'd like to start moving, so you throw some mass out the back in the form of exhaust, which accelerates your craft in the opposite direction. The speed of the mass exiting your ship doesn't matter at all - so long as it's moving, your ship will have equal momentum in the opposite direction. Now consider the ...


5

In addition to Mark Foskey's answer relating to the implausible strength required for this there are a number of other complications. A traditional space elevator is placed in a circular orbit, the moon is not in a circular orbit so the system will need to change length by 42 800 km twice each month, which comes out to more than 100 kmh. Not something you do ...


2

There is a misconception here: [...] the space elevator idea faces the challenge that, up to geostationary orbit height, the entire structure has to be supported from beneath [...] This is wrong. One can not simply support anything up to geostationary orbit, it's way too far out. Rock at the bottom of such a structure would behave more like a liquid, ...


2

Consider, for comparison, the space elevator concept. It would extend from the surface of the Earth to a point past geostationary orbit, and weighted in such a way that geostationary orbit is where the center of mass is. This is actually conceptually very similar to your idea. For instance, it also is meant to benefit from descending loads balancing ...


1

Whatever is used in space has to be sent up from Earth either way. Missions that can do it with a direct launch should. For missions with greater mass requirements than direct launch can do an orbital depot may be an answer but it adds the complexity and costs of making, supplying and using the depot to that of future missions. Sending up eg the full fuel ...


1

There is no lower limit; however, as you lower the exhaust velocity, more mass must be exhausted. Rocket propulsion works by the conservation of momentum. The change in momentum of the exhaust (its mass times its velocity) is equal but opposite in sign of the momentum of the rocket. I've illustrated this below. Notice how the mass of the exhaust (drawn ...


1

The answer is entirely gravity loss. When a rocket burns straight up it constantly is incurring gravity losses equivalent to local gravitational acceleration (9.8 m/s on the Earth's surface). A rocket that is burning straight up at the surface of the Earth at 9.8 m/s is exactly counteracting the gravitational force and hovering and going nowhere. The ...


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