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Elon Musk discussed in a question/answer session after the recent Falcon 9 launch that interplanetary missions are likely to use a landing barge, due to high horizontal launch speeds required for such launches. However, for an interplanetary mission, it seems that firing straight up might be preferred, or at least near to it, thus increasing the speed of the rocket, but still essentially being vertical. High horizontal speed isn't necessarily needed for interplanetary missions, as they don't need to orbit Earth, however, a high horizontal speed isn't precluded.

What is typically done for interplanetary missions as far as the launch profile goes, and will the Falcon 9 change this launch profile to make it easier to recover the rocket?

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marked as duplicate by PearsonArtPhoto Apr 12 '16 at 18:56

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Most interplanetary flights are put into a parking orbit first, not "straight up" direct ascents.

In any case, assuming a low energy Hohmann transfer is required, you'd want to depart Earth's orbit at a tangent. This would mean starting with a high horizontal velocity, compared to Earth's surface. So there wouldn't be anything to gain by leaving vertically.

For an alternative way of looking at this: a near-vertical launch doesn't require less velocity/energy to escape Earth. In fact it may need more, as you would lose the usual benefit of launching East. (Several hundred metres per second.)

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  • $\begingroup$ It would be somewhat reduced I believe by less of a gravity drag. The tangential launch could easily be managed by launching at local noon or midnight, depending on the direction. The gained speed would still be included, one just has to ensure that one calculates the launch correctly. $\endgroup$ – PearsonArtPhoto Apr 12 '16 at 15:55
  • $\begingroup$ Also - with the proposed vertical launch the upper stage would still be expending propellant at a relatively high altitude. (Unless it was a relatively fast-burning upper stage.) There would be less benefit from the Oberth effect. $\endgroup$ – Andy Apr 12 '16 at 16:28
  • $\begingroup$ Oberth might be the answer. Hmmm.... $\endgroup$ – PearsonArtPhoto Apr 12 '16 at 16:40

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