In the case of bodies in circular or elliptical motion – such as in celestial mechanics – it often helps the solution to reframe the problem in polar coordinates (especially if an orbit is being considered confined to one plane only). The conversion from Cartesian (rectangular) to polar coordinates is very simple and is made using these transformations of the Cartesian coordinates, x and y (which readers can easily verify from Fig. 1 and basic trigonometry):
x = r cos(theta)
y= r sin(theta)
Thus, let’s take the Cartesian case shown in Fig. 1 for x^2 + y^2 = 25 (circle) and we’d like to transform it into polar coordinate form. We first need to obtain the correct equation in those coordinates (r, theta) then we can compare the figures. Substituting the polar transform equations into the original we obtain:
[r^2 (cos(theta)^2)] + [r^2(sin(theta)^2)] = 25
Or:
r^2 (cos^2(theta) + sin^2(theta)) = 25
or r^2 = 25
since: (cos^2(theta) + sin^2(theta)) = 1 by a trigonometric identity.
Now, specification of position (or velocity, acceleration) in any polar plane is helped immensely by the use of unit vectors u(r) and u(theta) which are defined as follows:
u(r) = icos(theta) + jsin(theta)
u(theta) = -isin(theta) + jcos(theta)
And these are seen to conform to the diagram in Fig. 3, where u(r) is directed along the radius vector outwards, while u(theta) is at right angles to the radius vector r and in the direction of increasing theta.
Of course, each of the unit vectors can also be differentiated, and I leave it to readers to show this and thence, show that such differentiation is equivalent to an operation whereby the vectors u(r) and u(theta) are rotated through 90 degrees counter-clockwise. (Hint: do the differentiation, then recast the result as the product of a 2 x 2 matrix in sin(theta, cos (theta) multiplied by i,j.)
What about the magnitude of the angular momentum vector h, which we saw in the earlier blog introducing energy constants? Well, in polar coordinates, r x r’ will be a vector perpendicular to the x-y plane and equal to: (0, 0, xy’ – yx’) = h.
In this guise, the vector r is just the length r along the unit vector u(r), so r = r u(r). Then from simple differentiation of a product: r’ = r’ u(r) + r q’ u(q) where I’ve taken the liberty of substituting the letter q for ‘theta’ to reduce clutter. (Blogger doesn’t tolerate Greek letters in the text, or can’t support them unless written out- so far as I’ve ascertained).
Then:
r x r’ = r u(r) x ( r’ u(r) + r q’u(q) ) = r u(r) x r’u(r) + r u(r) x r q’ u(q)
or, more simply:
r x r’ = r(rq’) sin 90 u(r) = r^2 q’ u(r)
For circular orbital motion in the polar plane then two equations need to be satisfied:
1) ½ (r’^2 + r^2 q’^2) – u/r = C
2) h = r^2 q’ u(r)
Again, recall u in (1) = G(m1 + m2)!
In general then, given an acceleration r” such that:
r" = -G(m1 + m2)*r/ r^3
One has circular motion provided the radius vector r = a (semi-major axis of orbit) at all times, so effectively r = constant. Thus, r’ = 0.
A final item of interest- especially as it connects to the Kepler second law for elliptical orbits, is the rate of change of angle theta (= q) related to h, and the period P.
From calculus, it is easily shown that:
(dq/dt) r^2 = 2 (dA/dt) = h
Where A denotes the area of the circular orbit, (A = pi x r^2)
Thus, dA = h/2 (dt) and A = pi a^2 = P(h/2) = h/2 (T – t_o)
Problem: Show that for the Earth, the period P ~ 366 days. Use r = a = 1.49 x 10^11 m. Earth’s mean motion per day n = (T – t_o) is ~ 0.985647 deg/day. However for purposes of the computation, radian measure must be employed, where 1 radian = 57.3 deg, so the mean motion per day will have to be reposed as radians/sec then re-converted into the appropriate units to get P in sec, then in days. (In the next blog this will be solved as well as the earlier energy constant problem to do with Jupiter).
x = r cos(theta)
y= r sin(theta)
Thus, let’s take the Cartesian case shown in Fig. 1 for x^2 + y^2 = 25 (circle) and we’d like to transform it into polar coordinate form. We first need to obtain the correct equation in those coordinates (r, theta) then we can compare the figures. Substituting the polar transform equations into the original we obtain:
[r^2 (cos(theta)^2)] + [r^2(sin(theta)^2)] = 25
Or:
r^2 (cos^2(theta) + sin^2(theta)) = 25
or r^2 = 25
since: (cos^2(theta) + sin^2(theta)) = 1 by a trigonometric identity.
Now, specification of position (or velocity, acceleration) in any polar plane is helped immensely by the use of unit vectors u(r) and u(theta) which are defined as follows:
u(r) = icos(theta) + jsin(theta)
u(theta) = -isin(theta) + jcos(theta)
And these are seen to conform to the diagram in Fig. 3, where u(r) is directed along the radius vector outwards, while u(theta) is at right angles to the radius vector r and in the direction of increasing theta.
Of course, each of the unit vectors can also be differentiated, and I leave it to readers to show this and thence, show that such differentiation is equivalent to an operation whereby the vectors u(r) and u(theta) are rotated through 90 degrees counter-clockwise. (Hint: do the differentiation, then recast the result as the product of a 2 x 2 matrix in sin(theta, cos (theta) multiplied by i,j.)
What about the magnitude of the angular momentum vector h, which we saw in the earlier blog introducing energy constants? Well, in polar coordinates, r x r’ will be a vector perpendicular to the x-y plane and equal to: (0, 0, xy’ – yx’) = h.
In this guise, the vector r is just the length r along the unit vector u(r), so r = r u(r). Then from simple differentiation of a product: r’ = r’ u(r) + r q’ u(q) where I’ve taken the liberty of substituting the letter q for ‘theta’ to reduce clutter. (Blogger doesn’t tolerate Greek letters in the text, or can’t support them unless written out- so far as I’ve ascertained).
Then:
r x r’ = r u(r) x ( r’ u(r) + r q’u(q) ) = r u(r) x r’u(r) + r u(r) x r q’ u(q)
or, more simply:
r x r’ = r(rq’) sin 90 u(r) = r^2 q’ u(r)
For circular orbital motion in the polar plane then two equations need to be satisfied:
1) ½ (r’^2 + r^2 q’^2) – u/r = C
2) h = r^2 q’ u(r)
Again, recall u in (1) = G(m1 + m2)!
In general then, given an acceleration r” such that:
r" = -G(m1 + m2)*r/ r^3
One has circular motion provided the radius vector r = a (semi-major axis of orbit) at all times, so effectively r = constant. Thus, r’ = 0.
A final item of interest- especially as it connects to the Kepler second law for elliptical orbits, is the rate of change of angle theta (= q) related to h, and the period P.
From calculus, it is easily shown that:
(dq/dt) r^2 = 2 (dA/dt) = h
Where A denotes the area of the circular orbit, (A = pi x r^2)
Thus, dA = h/2 (dt) and A = pi a^2 = P(h/2) = h/2 (T – t_o)
Problem: Show that for the Earth, the period P ~ 366 days. Use r = a = 1.49 x 10^11 m. Earth’s mean motion per day n = (T – t_o) is ~ 0.985647 deg/day. However for purposes of the computation, radian measure must be employed, where 1 radian = 57.3 deg, so the mean motion per day will have to be reposed as radians/sec then re-converted into the appropriate units to get P in sec, then in days. (In the next blog this will be solved as well as the earlier energy constant problem to do with Jupiter).
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