Kepler's laws
Kepler's first law 




Kepler's first law.  
Now that we have polar basis vectors under our
wing (and the polar representations of velocity and angular momentum),
we are ready to proceed with the proof of Kepler's first law  that
the orbits of planets are ellipses with the Sun at one focus. To begin with, we will start off by applying Newton's law of motion and Newton's law of universal gravitation together to find that 

m a = (G m M/r^{2}) r  equation 1  
and, dividing both sides of the equation by m,  
a = (G M/r^{2}) r.  equation 2  
Recalling our work with polar basis vectors, we know dtheta/dtheta = r. Solving for r and applying the chain rule, we find that  
r = dt/dtheta dtheta/dt  equation 3  
r = 1/omega dtheta/dt.  equation 4  
Substituting this into our equation for a we find  
a = (G M/r^{2}) (1/omega) dtheta/dt  equation 5  
a = (G M)/(r^{2} omega) dtheta/dt.  equation 6  
If we multiply the right side of the equation by m/m (which is unity), we obtain  
a = (G m M)/(m r^{2} omega) dtheta/dt.  equation 7  
But l = m r^{2} omega (I told you this would come in handy as well), so we can rewrite this as  
a = (G m M/l) dtheta/dt.  equation 8  
We can multiply both sides by l/(G m M) and find  
l/(G m M) a = dtheta/dt.  equation 9  
But we know that a = dv/dt, and can substitute accordingly:  
l/(G m M) dv/dt = dtheta/dt.  equation 10  
This is a differential equation that we can now solve. Upon solving it, we find that  
l/(G m M) v = theta + C  equation 11  
where C is some constant vector. We'll solve this for v and find that  
v = (G m M/l) (theta + C).  equation 12  
This is a general solution to the differential equation. But we're not finished. This doesn't tell us much about the shape of a planet's orbit, although all the pieces are there. This is the general solution, and it could be an orbit of any of the possible shapes (though we can't be sure what they are yet) or any of the possible orientations. We're interested in knowing the shape, of course, so we want to restrict the possible orientations. To do that, we'll take a special case. It makes sense to have perihelion  that is, closest approach to the Sun  at time t = 0. We'll restrict the orientation so that, when perihelion occurs, the planet lies along the zero radian line from the Sun (or, in Cartesian terminology, along the positive xaxis)  that is, theta = 0. At this point, r, the position vector of the planet, will have only a component in the positive xaxis. We'll also assume that the planet orbits the Sun counterclockwise, through increasing measures of angles. If this is the case, then the velocity v at the instant of perihelion should be orthogonal to the position vector r, and it should have only a component in the positive yaxis. According to our expression for v, we have a scalar times the vector quantity theta + C. 

theta(theta = 0) = j;  equation 13  
that is, the unit transverse vector points "up" when the unit radial vector points "right." Since, at t = 0, theta points entirely in the ydirection, then our constant vector C must only have a component in the yaxis  this is the only way to get a resultant vector (v) that points entirely in the ydirection. So, we can rewrite C as a scalar times the unit basis vector in the ydirection:  
C = e j  equation 14  
where e is some scalar constant. (You will find that we have a very good reason for choosing the letter e in this case.) Substituting this into our equation for v, we get  
v = (G m M/l) (theta + e j).  equation 15  
This is the specific case when we want the orbit oriented so that
perihelion occurs at t = theta = 0. Now we are ready to finish up the problem. We can dot both sides of the equation with theta and get: 

v dot theta = (G m M/l) (theta + e j) dot theta  equation 16  
v dot theta = (G m M/l) [theta dot theta + e (j dot theta)].  equation 17  
A vector dotted with itself yields the square of that vector's magnitude, so theta dot theta = 1. Simplifying v dot theta, we find  
v dot theta = (dr/dt r + r omega theta) dot theta  equation 18  
v dot theta = dr/dt (r dot theta) + r omega (theta dot theta).  equation 19  
But the dot product of two orthogonal vectors is zero, so r dot theta = 0. We also already know that theta dot theta = 1. Therefore  
v dot theta = r omega.  equation 20  
The last part of our problem is finding an expression for j dot theta. We know that theta = sin theta i + cos theta j (by definition), so  
j dot theta = j dot (sin theta i + cos theta j)  equation 21  
j dot theta = cos theta.  equation 22  
We have the three pieces of the puzzle, so we'll put them together to find that  
r omega = (G m M/l) (1 + e cos theta).  equation 23  
We see r omega on the left side of this equation. We know that l = m r^{2} omega, and it would be nice to get rid of the mess that way. So we'll multiply both sides of the equation by m r to get  
m r^{2} omega = (G m^{2} M/l) [r (1 + e cos theta)].  equation 24  
Replacing the left side of the equation by l and moving the constants to the left side of our equation, we find that  
l = (G m^{2} M/l) [r (1 + e cos theta)].  equation 25  
We're almost finished now. Here it is clear that we have an explicit function in terms of r and theta  in other words, this should be the polar equation for our planet's orbit! All we need do now is solve for r:  
r = [l^{2}/(G m^{2} M)]/(1 + e cos theta).  equation 26  
The equation of a conic section with focusdirectrix distance p and eccentricity e is represented by the polar equation  
r = e p/(1 + e cos theta).  equation 27  
But this is exactly what we have, given that  
e p = l^{2}/(G m^{2} M).  equation 28  
The
focusdirectrix distance should be a constant, which p is: l, G, m,
and M are all individually constant; therefore the expression l^{2}/(G m^{2} M)
must also be constant. Therefore, Newton's laws of motion and
universal gravitation dictate that the orbits of planets follow conic
sections. This is Kepler's first law. ... Well, almost. Kepler's first law actually states that planets follow the paths of ellipses. An ellipse is only one type of conic section. So why is an ellipse allowed while the others are not? Because the others are allowed  we just don't call them planets. When Kepler said planet, he meant a body which returns to our skies over and over again. This means that the curve representing the orbit must be closed  that is, it must continue to retrace itself over and over. The only two conic sections which are closed are the circle and the ellipse (the circle is just a special case of the ellipse anyway). Thus, any body in the skies we see over and over must be orbiting the Sun in an ellipse. The other two conic sections  the parabola and hyperbola  are open curves and correspond to a position where the body has sufficient velocity to escape from the Sun's gravity well. The body would approach the Sun from an infinite distance, round the Sun rapidly, and then recede away into the infinite abyss, never to be seen again. So we have proved an extension of Kepler's first law: A body influenced by the Sun's gravity follows a path defined by a conic section with the Sun at one focus. 

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