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Orbit Transfer to Lower Apoapsis of a Mars Orbit

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider an orbit transfer to lower the apoapsis of a Mars mapping orbit. The initial orbit has a periapsis radius of 3800 km and an eccentricity of 0.5. The final apoapsis radius is to be 1/2 of the initial one. Note that only one burn at periapsis is needed to perform this change and that the periapsis radius of the final orbit is the same as the initial orbit. Find the [tex]\Delta V[/tex] required.


    2. Relevant equations
    General Velocity Equation: V=[tex]\sqrt{{2\mu/R}-{\mu/a}}[/tex], where [tex]\mu[/tex]=GM, and a is the semi-major axis

    Relation of radii to semi-major axis: [tex]r_p + r_a[/tex] = 2a, where [tex]r_p[/tex] is the radius at periapsis, and [tex]r_a[/tex] is the radius at apoapsis


    3. The attempt at a solution
    First I calculated the radius at initial apoapsis using the formula [tex]r_a = r_p((1+\epsilon)/(1-\epsilon))[/tex]

    I got 11400 km, which means the final apoapsis is 5700 km.

    Using the relation for radii to semi-major axis I calculated the initial semi-major axis and got 7600 km. Doing the same for the final semi-major axis, I got 4750 km.

    Then, I used the general velocity relation to find the initial and final velocities based on the respective a value, the value for [tex]\mu[/tex] (mass of Mars times the Gravitational Constant, given in my textbook to be 4.293 E4 km^3/s^2), and the radius of Mars (3397 km).

    I got an initial velocity of 4.43 km/s and a final velocity of 4.03 km/s, resulting in a [tex]\Delta V[/tex] of -0.4 km/s. I assume the negative value points to the fact that the delta V needs to be applied in the reverse direction to slow the vehicle down, resulting in a smaller apoapsis.

    I just did this on a wing so I don't know if it is correct. Am I missing something here or does this seem correct?
     
  2. jcsd
  3. Sep 13, 2007 #2
  4. Sep 13, 2007 #3

    D H

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    This R is not the planetary radius. Think about it. Suppose you wanted to transfer from a circular orbit of radius a to a circular orbit of the same radius (i.e., no change in orbit). The delta-V for this maneuver is zero. You do not get zero if you use the planetary radius.
     
  5. Sep 13, 2007 #4

    learningphysics

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    yeah, R is just the distance from Mars
     
  6. Sep 13, 2007 #5

    D H

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    The distance from the center of Mars, to be precise. The distance from Mars does involve the radius. It is the vehicle's altitude.
     
  7. Sep 13, 2007 #6
    I understand why the delta V would be zero if you did not want to change altitude, but why would you not get zero if you use the planetary radius? [tex]\mu[/tex] and the apoapsis would not change if you were at the same distance.

    I am still confused as to what R I am supposed to use then. Is it the radius of Mars plus the distance to apoapsis? Or is it R the distance of Mars from the Sun?
     
  8. Sep 13, 2007 #7

    learningphysics

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    Yeah, I agree... don't see why delta V wouldn't be zero but...

    I think R is the distance of the orbiting body from the center of mass of the body being orbitted, as DH mentioned. ie... distance of the orbiting body from the center of mass of Mars.

    In this case wouldn't it just be the periapsis radius?
     
    Last edited: Sep 13, 2007
  9. Sep 13, 2007 #8
    I used R= Radius of Mars + Distance to Periapsis. Is this correct? I figure if the delta V burn is going to be performed at the periapsis where the velocity vector is tangental to the radial component, the R at that point is the Radius to the surface of mars (3397km) + the distance to the periapsis (3800km). Does this seem correct?
     
  10. Sep 13, 2007 #9

    learningphysics

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    I thought the periapsis and apoapsis radius was measured from the center of mass of Mars... I'm not sure about this.
     
  11. Sep 13, 2007 #10
    You are correct. I just read that the periapsis is "the least distance of the elliptical orbit of an astronomical object from its center of attraction, which is generally the center of mass of the system."

    So, I just use R=3800 km, and everything else looks alright?
     
  12. Sep 13, 2007 #11

    D H

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    That's correct.

    ColdFusion, if you are working with altitudes rather than distance from the center of the planet, everything you did in computing the eccentricity and semi-major axis is wrong. While apoapsis and periapsis sometimes are referenced with respect to planet surface, the semi-major axis never is. The equations you used for eccentricity implicitly assume planet-centered apoapsis/perapsis.

    If these are indeed distances from the center of the planet, the only thing you did wrong was to add the planet radius to them.
     
  13. Sep 13, 2007 #12

    learningphysics

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    I think so.. initial a = 7600km... final a = 4750km... R = 3800km... looks right.
     
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