Kepler - constants and the mass of the Sun. Therefore we...

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We can derive his third law for the special case of a circular orbit. (A circle is an ellipse with zero ellipticity) The gravitational force of the Sun on a planet in a circular orbit of radius “r” is given by F = G . M S . m P /r 2 EQ. 1 This force provides the centripetal force to hold the planet in a circular orbit around the Sun. Thus we may write F = ma as follows: G . M S . m P / r 2 = m P . v 2 /r EQ. 2 But, recall that the velocity of the planet is given by the circumference of its orbit “2 π r” divided by the time it takes to travel once around its orbit, i.e. its period “T”. v = 2 π r/T EQ.3 Plugging this result into EQ. 2 we find, after canceling all the common factors on each side of the equation T 2 /r 3 = 4 π 2 /G . M S EQ.4 But the planetary data is all on the left side of the equation 4 and the right side contains only
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Unformatted text preview: constants and the mass of the Sun. Therefore we can write a similar equation for any planet that orbits the Sun. T 1 2 /r 1 3 = 4 2 /G . M S EQ.5 T 2 2 /r 2 3 = 4 2 /G . M S EQ.6 And therefore can write Keplers third law T 1 2 /r 1 3 = T 2 2 /r 2 3 EQ.7 This was a magnificent result for Newtons theory. Incidentally, scientists could then use these results to find the mass of the sun for the first time. Writing EQ.5 for the Earth-Sun system, we have T E 2 /r ES 3 = 4 2 /G . M S EQ.8 where T E = 3.16x10 7 sec r ES = 1.50x10 11 m G = 6.67x10-11 N m 2 /kg 2 Solving for the mass of the Sun we have M S = [4 2 /G] r ES 3 /T E 2 EQ.9 M S = 2.0x10 30 kg...
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Kepler - constants and the mass of the Sun. Therefore we...

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