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Relationship between characteristic energy and delta-v

  1. Feb 10, 2015 #1
    Firstly, apologies if this is in the wrong thread.

    I'm currently writing a presentation on the physics of getting a spacecraft from Earth to Mars in the near future. In my research I've come up against Porkchop plots which seem to plot contours of equal characteristic energy so you can find out the best dates to launch the spacecraft.

    What I'm struggling to understand is the relationship between characteristic energy and delta-V, I think that when the characteristic energy is 0 the delta -V is also 0 because you'll just stay in the orbit you were, but I'm not sure.
     
  2. jcsd
  3. Feb 11, 2015 #2

    D H

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    A characteristic energy of zero means the spacecraft is exactly on a parabolic escape orbit. It takes quite a bit of delta-V just to achieve that, over 3 km/s from low Earth orbit.
     
  4. Feb 15, 2015 #3
    Ok, so when I'm looking at the plots and it shows different curves of the characteristic energy is that the energy needed getting the craft from earth orbit into one that will coincide with Mars
     
  5. Feb 16, 2015 #4

    D H

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    That's correct.

    Porkchop plots are a mission planning tool. In many endeavors, mission planning has to make a number of simplifying assumptions to reduce the complexity of the search space to a level that can be studied. This is most certainly the case when planning interplanetary missions. The algorithms used to produce a porkchop plot make three key of simplifying assumptions: Patched conics, short times while escaping Earth's gravity well and entering that of the target planet, and impulsive changes in velocity.

    The patched conic approximation ignores the gravitational attraction of the Sun and other planets when the spacecraft is within the gravitational sphere of influence of a planet, and on the flip side ignores the gravitational attractions of the planets when the spacecraft is outside of the gravitational sphere of influence of any planet. With this assumption, gravitation is always a solvable two body problem. The assumption of short times inside a planet's gravitational sphere of influence means that the only issue of concern is getting from Earth's orbit about the Sun to the target planet's orbit about the Sun. The assumption of impulsive Δv means that solvers of Lambert's problem can be employed.

    Something that is never explicitly said (except perhaps in college lectures) is why the Δv's that result from solving Lambert's problem are expressed in terms of energy rather than in terms of Δv. There are a number of reasons for not using Δv. First and foremost, the simplifying assumptions ignore escaping Earth's own gravity well. This is not a good assumption from the perspective of a launch provider. Secondly, Δv is a bit misleading because Δv is not additive. Interplanetary vehicles are typically launched from the surface of the Earth into a temporary parking orbit and then given a short burn that sends them on the desired interplanetary trajectory. This means there's a huge Oberth effect involved. Thirdly, burns aren't instantaneous. Launch providers need to account for this. It turns out that energy is a better metric than is Δv.
     
  6. Feb 16, 2015 #5
    Thank you! It makes sense now :)
     
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