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Orbit Semimajor Axis and Eccentricity after increasing V

  1. Jan 23, 2008 #1
    Question:

    A space shuttle is in an orbit about the Earth. At its apogee, it uses thrusters and increases its velocity by 400 m/sec. What is the new orbit semimajor axis, eccentricity and how much will the next perigee altitude be increased?

    Known:

    Original semimajor axis: 7000 km -> a
    Original eccentricity: 0.05 -> e
    Earth's Radius: 6378 km
    u= GxEarth's Mass=3.986x10[tex]^{5}[/tex]

    What I have done so far:

    I figured out the apogee and perigee of the orbit, as well as the velocity at the apogee before the firing of the thrusters.

    i) apogee: a(1+e) = 7350 km
    ii) perigee: a(1-e) = 6650 km
    iii) velocity at apogee:

    [tex]\sqrt{u*((2/r)-(1/a))}/[/tex] where r = apogee.

    I got v=7.17 km/s

    Now after the thrusters are fired, the new velocity is 7.57 km/s

    Using [tex]\epsilon[/tex] = V[tex]^{2}[/tex][tex]/2[/tex] - u[tex]/r[/tex] where r is the current position, aka the apogee and plugging [tex]\epsilon[/tex] into

    a = -u[tex]/2\epsilon[/tex]

    I found the new semimajor axis to be 7809 km. But then here is the problem. How do I find out the new eccentricity? Thanks!
     
  2. jcsd
  3. Jan 24, 2008 #2

    tony873004

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    There's 1 assumption you're making, and that's that your 400 m/s of delta V are in the prograde direction. You could fire your engines to accelerate away from the Earth instead, until your total velocity increased by 400 m/s. Or you could fire towards the Earth, or anywhere inbetween.

    But I imagine they're expecting you to make this assumption. This means that after a delta V of 400 m/s in a prograde burn, your new velocity will momentarily be all tangental. This means that you're either still at apogee, or at your new perigee. Just compute circular velocity for that distance and see if you're above it (which means you're at perigee), or below it (which means you're at apogee). Your now know your apogee or perigee and your semi-major axis, so eccentricity should be easy to find: e=(sma-perigee)/perigee, or e=(apogee-sma)/sma
     
    Last edited: Jan 24, 2008
  4. Jan 26, 2008 #3

    tony873004

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    I should have said e=(sma-perigee)/sma
     
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