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Why can't a vehicle just fly into outer space at a constant velocity of, say, 50 mph instead of having to achieve "exit velocity"?

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closed as off-topic by Gert, user108787, JamalS, peterh says reinstate Monica, Jon Custer Dec 4 '16 at 4:18

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    $\begingroup$ I'm voting to close this question as off-topic because it shows no research effort. $\endgroup$ – Gert Dec 2 '16 at 22:43
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    $\begingroup$ Is the vehicle a rocket/powered or a projectile/unpowered? $\endgroup$ – Qmechanic Dec 2 '16 at 23:32
  • $\begingroup$ It took about 8 minutes for the Shuttle to get to lower earth orbit. Its top speed was on the order of 17,000 mph. The external tank held 143,000 gallons of liquid oxygen (1,359,000 pounds) and 383,000 gallons of liquid hydrogen (226,000 pounds). The fuel weighed almost 20 times more than the Shuttle. At launch, the Shuttle, external tank, solid rocket boosters and all the fuel combined had a total weight of 4.4 million pounds. The fuel weighed almost 20 times more than the Shuttle coolcosmos.ipac.caltech.edu/ask/… Go figure $\endgroup$ – user108787 Dec 2 '16 at 23:45
  • $\begingroup$ If you throw a stone upwards, would gravity pull it back down again? Or will it escape gravity's pull? Naturally, some speed will be just enough. (And I believe you meant "escape velocity", not "exit velocity") $\endgroup$ – Steeven Dec 3 '16 at 0:14
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    $\begingroup$ There is the space elevator, which is a rather exuberant idea for getting into space. A gondola rides this long column upwards to geosynchronous orbit. The velocity of the gondola is comparatively modest, say less than $1km/sec$ and it reaches geosynchronous orbit in a few days. $\endgroup$ – Lawrence B. Crowell Dec 3 '16 at 1:09
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The escape speed is computed by setting the kinetic energy $K$ and gravitational potential energy $U$, \begin{align} K &= \frac12 mv^2, & U &= - G\frac{Mm}{r}, \end{align} so that the total mechanical energy $E=K+U$ of your system adds up to zero. (We're assuming that the two masses are very different, $m\ll M$, which is true for planets.) This means that if the separation $r$ between the masses became very large, the kinetic energy will be zero --- but the two masses will be very far away from each other, and so will have "escaped."

Note that it's properly an escape speed rather than an escape velocity. The direction doesn't matter: if your little mass is moving faster than the escape speed, and there isn't some collision or drag or something to reduce its mechanical energy, it will eventually escape; if your little mass is moving slower than the escape speed, and there isn't some chemical reaction or gravitational encounter with a third body that increases its speed, it will never escape.

You raise an interesting point that there's not any obvious law of physics requiring you to reach this escape velocity entirely upon launch. If you could somehow maintain 50 km/h straight up, eventually you'd get to a high enough altitude $r$ that the escape speed is 50 km/h and you could shut off your car's engine and coast away into space from there. It's pretty far, though:

\begin{align} r_\text{drive away} &= \frac{2GM}{v^2} % \\&= % \frac{2 \cdot 7\times10^{-11}\rm\,J\,m\,kg^{-2} \cdot 6\times 10^{24}\,kg} % {\left(50\rm\,km/h \right)^2} % \times \left( \frac{ % 1\rm\,km\cdot3600\rm\,s % }{1000\rm\,m\cdot % 1\rm\,hour % }\right)^2 % \times\left(\frac{1\rm\,AU}{150\times10^9\rm\,m}{}\right) % \\& = 30\rm\,AU \end{align} What would actually happen is that once you got out of Earth's Hill sphere you'd have to change the heading on your vehicle to worry about other gravitational sources, like the Sun.

But you wouldn't want to escape this way, even if it were possible with an ordinary chemical rocket, because of the effect of the rocket equation. There aren't any filling stations once you leave Earth's surface, so all the fuel you use must be carried with it. In order to change a rocket's momentum you must also change the momentum of its fuel --- so the more fuel you have, the more fuel you need, exponentially.

Your idea of traveling slow and steady makes the rocket problem worse instead of better: not only do you have to change the momentum of the unburnt fuel, but you have to waste energy lifting it out of Earth's gravitational well before you use it. If instead you could use the fuel close to the ground, your rocket is more efficient, not less. The most energy efficient launch would be an explosion on the ground that instantly imparts the final speed to the rocket, but we squishy meatbags couldn't handle the acceleration. Gravity assists are even better.

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  • $\begingroup$ That's a nice and not obvious answer to a simple and interesting question. $\endgroup$ – user98038 Dec 3 '16 at 9:39
  • $\begingroup$ Thanks you for the responses. I deliberately did not specify a propulsion method. I understand the basics of exit velocity, but have always wondered whether the brains here that are far greater than mine might hypothesise rather than in, some cases, say I have made no effort to find out myself and vote to close the question. I wanted to keep it simple to give you the opportunity to invent not simply tel me how it's done now. $\endgroup$ – Fred812 Dec 3 '16 at 23:00
  • $\begingroup$ I didn't realise that comments are limited!! What I was wondering was whether it is impossible to leave the Earths pull other than by a rocket I suppose. I apologise if my questions are not technically correct, I'm not a physisist and rely on you folk knowing what I mean. $\endgroup$ – Fred812 Dec 3 '16 at 23:14

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