Question

1. Keplers harmonic law of planetary motion (his 3rd law) states that the square of the orbital period is proportional to the cube of the characteristic distance of the planet from the sun. Test this by fitting a power law to the data below. Give your answer to 4 decimal places. Data file
#period (days), distance (AU) # Data supporting Keplers third law # distance is measured relative to the earths mean distance, # which is about 9.26e7 miles and 1.4960e8 meters # Planets, in order: Mercury,Venus,Earth,Mars,Jupiter,Saturn # from Timoshenko and Young, page 92 87.96,0.3871 224.6,0.7233 356.2,1.000 686.9,1.52 4332,5.200 0759,9.540
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Answer #1

%%% Matlab code %%%

clc;
clear all;
close all;
d=[87.96 224.6 356.2 686.9 4332 10759];
t=[0.3871 0.7233 1 1.523 5.2 9.54];

% b)
% user defined function fitting
c= lsqcurvefit(@myfun1,[1 ,2], d, t);
y2=c(1)*d.^c(2);

figure;
plot(t,d,'*');
hold on
plot(y2,d);
xlabel('Time period');
ylabel('Distance from sun');
fprintf('fitting function is t=%f d^ %f ',c(1),c(2));

OUTPUT:

fitting function is t=0.019716 d^ 0.665874

12000 10000 c 8000 6000 4000 2000 0 1 234 5 678 9 10 Time period

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