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Help with basic orbital mechanics

  1. Oct 1, 2008 #1
    My first post on this site, been reading for a while but joined up today so please bear with me :)

    I've got an assignment for my University orbital mechanics course. I couldn't find the place in the 'Homework' section, so i thought here would be ok.

    I've got a few pretty simple questions, however I can't find any info in my lecture notes, or off the internet that answers them enough.

    When leaving a low earth orbit (LEO) to another planet, or moon, or anywhere on a hohmann transfer, do you have to leave from an equitorial orbit, where the inclination is 0? If so, why?
    Why can't you just be on an orbit of 28 degrees latitude, and then fire the rockets to get into a Hohmann transfer?

    If doing an inclination change for a Low earth orbit (LEO), the formula is delta V = 2vsin∆i/2. For example, if the initial circular orbit velocity was 10km/s and the deltav was 0.5km/s , would the final velocity be 10.5km/s? Would both orbits still be the same radius? Or would the spacecraft have to reduce 0.5km/s after it does the plane change to get back to 10km/s to stay in the same radius LEO?

    A Hohmann transfer is the most fuel efficient (lowest deltav) method only if the r2/r1 < 11.8 (from my lecture notes). If I have radioactive material in a LEO around Earth, and I want to send it into the sun, the ratio of r2/r1 is WAY bigger than 11.8. Does this mean, that the best deltav (lowest deltav) is not a Hohman transfer? If not, what is the best deltav?

    How do you calculate the deltav for going from LEO to a polar orbit around Mars.
    I have so far gathered the following, but am a bit unsure about some parts:
    Leave LEO and go into a Hohmann transfer to Mars orbit
    When in mars orbit, change inclination to 90 degrees

    Firstly calculate the velocity required to get into an elliptical orbit with Mars at the apohelion, and Earth at the perihelion. Then minus this velocity from the velocity the Earth is travelling relative to the Sun.
    My lecture notes go on to find another deltav, for the hyperbolic escape from LEO. So you find the perigee velocity (or apogee, i forget which is which), using the velocity found just before, and minus that from the initial speed you have in the LEO, and that is the deltav. Why do you have to find two deltav ? Which one is the actual deltav required to send the spacecraft to Mars?

    Finally, and hopefully my last question:
    How do you escape from the solar system. I know you need a hyperbolic escape (because parabolic has zero energy at radius infinity). But do you only need hyperbolic escape from Earth, or do you take into account the Sun also. I mean, do you have to do two Hyperbolic escapes? One for Earth, then another for the Sun also?

    I really really really appreciate any feedback, comments, help, detailed help :)

    Thanks in advance for anything and sorry if this is a repeat of a topic made earlier today, or yesturday. I did a quick search, but i admit i didnt spend long enough searching. I actually don't know what to search for, and I'm sick of searching after spending the last two days on the computer searching.

    Last edited: Oct 1, 2008
  2. jcsd
  3. Oct 1, 2008 #2
    I've solved the two parts in the "Spoiler"

    Still need help with the other 3. I dont really need a huge explanation, just a quick one to say why etc... This is introductory orbital mech :)

  4. Oct 2, 2008 #3


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    You could post this in "Advanced Physics" in the homework section.

    Your inclination with respect to the ecliptic is not 0 in an equitorial orbit. Most planets orbit the sun near the ecliptic plane without regard to what direction Earth's poles are pointing.

    Look up the actual definition of Hohmann transfer to make sure the answer isn't obvious in the exact wording. I would guess that in a Hohmann, you basically want to "stall" at your destination, which you won't be doing on an inclined trajectory, as you'll have some z-component velocity in addition to the overwhelming x-y component.

    I've never heard that before. What is the more efficient route if r2/r1>=11.8? What if you sent it out to 11.8 AU, then did a Hohmann into the Sun from there, where your delta V requirements are less? This is an interesting question.

    You need to find your LEO escape velocity with excess. If you blast away from Earth with escape velocity, you're on a parabolic orbit that (in the simplest case, ignoring all other bodies) would climb to apogee but take an infinite amount of time to do so before falling back to Earth. If you blast away from Earth with a velocity greater than this, then you're on a hyperbolic orbit. You will never return, not even at time = infinity. Add this extra velocity to the escape velocity formula, and square it. Escape velocity with excess is sqrt((2GM/r) + Vexcess^2). Plug in your first delta v, for getting from Earth's distance from the Sun to Mars' distance from the Sun - Earth's current velocity, as the excess velocity. This gives you your second delta V. So for example, (I'm just making up these figures). You might find that escape velocity from LEO is 11 km/s, and the Hohmann transfer velocity to Mars is 33 km/s, and Earth is travelling around the Sun at 30 km/s, so after escaping Earth, you need an additional 3 km/s of velocity at infinity. This is your excess velocity in the above formula.

    This is like #2. You need to find your solar escape velocity at Earth's distance, then subtract Earth's velocity from that. That tells you your delta V to escape the Sun. This delta V becomes your excess velocity in the "escape plus excess" formula, which will tell you how fast you need to be going in LEO to escape Earth with excess velocity that equals the Sun's escape velocity at 1 AU.
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