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Plotting E - e*sin[E] = M as E vs M, when given e as 5 different values [0 0.25 0.5 0.75 1]?

Asked by Daniel on 12 Sep 2013
Latest activity Answered by Youssef KHMOU on 30 Sep 2013

For a question in my homework , we are suppose to first find the fzeros of this function of E when given any values of e, and M. I already figured this out by the following:

function [ E,z ] = kepler( e,M )
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
f=@(E) E-e*sin(E)-M;
z=fzero(f,0)

But I cannot for the life or me figure out how I would plot this E vs M when only given a value of e. If anyone can help me out that would be greatly appreciated :)

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Daniel

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2 Answers

Answer by Walter Roberson on 12 Sep 2013

The usual way: loop over all the potential values of M for a fixed e, recording the outcome each time. Plot the potential values of M on one axis, and the recorded E values on another axis.

hint: ndgrid(), arrayfun(), and surf()

4 Comments

Matt Kindig on 12 Sep 2013

Another hint: to loop over values of M, use the 'for' statement. See the examples contained in:

doc for
Daniel on 12 Sep 2013

Also they have to be shown as five different curves on the same graph, but I know if you plotted each separately for e value given then E vs M and use "Hold on" command that would work.

Duy Nguyen on 29 Sep 2013

I have the same problems too, how can you graph E vs. M for e of values [0 0.25 0.5 0.75 1] with 5 curves in 1 figure?

Walter Roberson
Answer by Youssef KHMOU on 30 Sep 2013

Daniel, i find this issue not clear at all, besides you have to to give some Physical explanations about the constants, M as mean anomaly ,e as eccentricity :

function [ E] = kepler( e,M )
f=@(E) E-e*sin(E)-M;
E=fzero(f,0);
------------------------------
M=(0:0.01:6);
N=length(M);
e=[0 0.25 0.5 0.75 1];
n=length(e);
for x=1:N
     for y=1:n
 E(x,y)=kepler(e(y),M(x));
     end
end
figure, plot(M,E), grid on,

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Youssef  KHMOU

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