{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Kepler

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 π 2 /G . M S EQ.5 T 2 2 /r 2 3 = 4 π 2 /G . M S EQ.6 And therefore can write Kepler’s third law T 1 2 /r 1 3 = T 2 2 /r 2 3 EQ.7 This was a magnificent result for Newton’s 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 π 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 π 2 /G] r ES 3 /T E 2 EQ.9 M S = 2.0x10 30 kg...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online