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Orbital Period

  1. Jun 24, 2017 #1
    Hi, everyone!
    I'm trying to find the period of a satellite orbiting the earth by the next formulas
    with: miu= 398600.5;
    a=36770.48 km;
    I. 2*pi*(a^3/miu)^1/2
    II. 84.489*(a/Re)^(3/2)min
    III. 0.00016587*a^(3/2)min
    And I got aprox. 19 hours from both formulas (I. and III.) which is correct.
    Can anyone please explain to me how do you get from the first formula to the second and third,what transformations should I make and also who is Re?

    Another problem that I have is that I'm trying to get from cartesian system to orbital and I have reached a point where I have to:
    Compute the time of periapse passage, T (note that EA must be in radians), with the formula

    T=t-(1/n)*(EA-ecc*sin(EA));
    where EA=eccentric anomaly=0.0335
    ecc=eccentricity=0.80324;
    n=sqrt(miu/a^3);
    I have taken t=270;
    With the following result:-73.7766 which I know is bad but I don't know where I've done wrong.Can anyone explain to me ,please?
    Also, Is this time the same as the period?

     

    Attached Files:

  2. jcsd
  3. Jun 24, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That is a weird and inconsistent way to handle units.

    Re should be the radius of Earth.

    You can just plug in numbers, then the formulas can be converted into each other. (I) is the one closest to the derivation via the forces on a satellite.
    As an easier example: You can rewrite "10*x" as "5*b*x" if you define b=2. If you have Re=6370, you can get it into formulas in the same way. Sometimes that makes the formula handier.


    For the second problem I think something is missing. Time relative to what? Where do the orbital parameters come from?
     
  4. Jun 25, 2017 #3
    ctk1.jpg ctk2.jpg ctk3.jpg
    First of all, thank you for your response, it has been very helpful.As for the second problem I've written the above matlab code. It uses the values of position vector and velocity vector to determine the orbital elements.As you can see the values for a and e and other elements are correct so I believe that t is the moment in time at which those values for the vectors.For my code I've use the above steps.
     
  5. Jun 25, 2017 #4

    gneill

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    Staff: Mentor

    Your images have too low resolution to read. Rather than screen or window dumps, a better tool is the Windows Snipping Tool which allows you to select and capture portions of the screen. Code can be cut and pasted as text (use code tags to surround it in order to preserve formatting). The best way to present equations is to use LaTeX syntax, which will automatically be rendered properly as equations. See the LaTeX guide: https://www.physicsforums.com/help/latexhelp/
     
  6. Jun 25, 2017 #5
    t=270;
    miu= 398600.5;
    Position values
    Rx=1.120260101155629e+03;
    Ry=-5.997082966407411e+03;
    Rz=-3.919879916453886e+03;
    Rxyz=[Rx Ry Rz];
    Velocity values
    Vx=5.0702;
    Vy=4.9659;
    Vz=-6.9805;
    Vxyz=[Vx Vy Vz];
    Angular momentum
    hez = cross(Rxyz,Vxyz);
    N = cross([0,0,1],hez);
    hx=hez(1,1);
    hy=hez(1,2);
    hz=hez(1,3);
    he=((hx^2)+(hy^2)+(hz^2))^(1/2)

    R=((Rx^2)+(Ry^2)+(Rz^2))^(1/2); radius
    Vel=((Vx^2)+(Vy^2)+(Vz^2))^(1/2);speed
    E=((Vel^2)/2)-(miu/R);%Energy
    ac= -(miu/(2*E)); semi-major axis
    Excentricity e
    e_vec = ((norm(Vxyz)^2 - miu/norm(Rxyz))*Rxyz - dot(Rxyz,Vxyz)*Vxyz)/miu;
    ecc = norm(e_vec);
    Inclination
    inc=acos(hz/he);
    ig=radtodeg(inc); in degree
    Omega
    OM=atan2(hx , -hy);
    if N(2) < 0 quadrant check;
    OM = 2*pi - OM;
    end
    OMg=radtodeg(OM)%OMEGA in degree;

    True anomaly
    tra=acos((ac*(1-ecc^2)-R)/(ecc*R));
    if dot(Rxyz,Vxyz) < 0 quadrant check;
    tra = 2*pi - tra;
    end
    trag=radtodeg(tra) ; in degree

    Argument of periapse w
    om= acos(dot(N,e_vec)/(norm(N)*ecc));
    if e_vec(3) < 0 quadrant check;
    om = 2*pi - om;
    end
    omg=radtodeg(om); in degree

    Eccentric anomaly
    EA=acos((ecc+cos(tra))/(1+ecc*cos(tra)));

    The time of periapse passage
    n=sqrt(miu/ac^3);
    T0=t-(1/n)*(EA-ecc*sin(EA))
    The values for the vector are from T=70172.6768628;%Period
    and t=270; Those values I've use when I've done the conversion from orbital to cartesian and the output was the values of the 2 vectors.
    Based on the following images. Sorry for the poor quality of the above images!
     

    Attached Files:

  7. Jun 25, 2017 #6
    Gneill, thank you for the advice, I didn't know about the LaTeX!
     
  8. Jun 25, 2017 #7

    gneill

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    Staff: Mentor

    You should state the units associated with your given values, position and velocity vector components included. Numbers alone are mostly meaningless.
     
  9. Jun 25, 2017 #8
    Sorry for that. For position and a we have km and for velocity km/s while T and t are in seconds.
     
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