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- Thread starter Lonewolf
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Nereid

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Hmm, this may not be as easy as it seems at first.

Do you have some contraints? For example, "between two neighbouring planetary orbits"? or, "excluding gravitational slingshots"?

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Not sure if this helps, but oh well, I don't understand the question clearly enough to elaborate on mathematics. But I'm only 15, so I doubt I'd be any assistance there! I'll try though.

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E = mv²/2 - GMm/r

Now for a circular orbit, v= [squ](GM/r), so

E= GMm/2r - GMm/r = -GMm/2r

Now for an eliptical orbit, the body swings in and out from an average distance from the sun and this average distance is equal to the semimajor axis of the orbit (a) . as it does so, its orbital velocity increases and decreases also.

When the orbital distance is equal to 'a', the orbital velocity is equal to the velocity it would have in a circular orbit at that distance. Since the the energy of an orbit doesn't change, this means that any eliptical orbit has the same energy as a circular one of radius 'a', and we can rewrite our second equation as

E =-GMm/2a

As 'a' gets' larger E become less negative and thus is less.

If you are launching a craft from orbit one to orbit two, it is obvious that if 'a' is less than the average of the distance from the sun of the two planets, then the craft won't reach the second orbit. So 'a' must at least be that large. If the outer limb of the orbit swing out further than the second orbit, then 'a' has to become larger and the energy of the orbit increases.

Thus the energy difference between the first orbit and the transfer orbit increases (As does the energy needed to put the craft into the transfer orbit from oribt one.

This works even if the planet you are launching to orbits closer to the sun.

If you try and make the orbit so that it swings in closer to the sun than the target planet, then 'a' of this orbit gets smaller and the orbit has less energy. But such a tranfer orbit willhave less energy than the launch planet's orbit to start with, so decreasing its energy,

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Yup, that answers it. Thanks Janus. Thanks for having a go, Jeebus

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