Wednesday

December 7, 2016
Posted by **Jo** on Tuesday, January 27, 2009 at 11:38pm.

- space -
**drwls**, Tuesday, January 27, 2009 at 11:54pmAphelion it the point farthest away. Review what Kepler's second law says, and those facts will tell you if the statement is true or false.

- space -
**tchrwill**, Wednesday, January 28, 2009 at 1:04pmThe earth travels in an elliptical shaped orbit around the Sun.

An ellipse is the locus of a point, P, moving in such a way that the sum of its distances from two fixed points, F and F', called foci, is a constant. Draw yourself two perpendicular axes, X and Y with origin O. Locate two points, the foci, on the X axis, equidistant on either side of the origin O, say 1 inch, if you want to draw your own picture. Label the points F on the left and F' on the right. Each represents a focus of the ellipse. Now locate two points, on the X axis, 2 inches on either side of the origin. Label these A on the left and B on the right. AB is the major axis of our ellipse. With a compass, using AO as a radius and F as the center, swing an arc that intersects the +Y axis at C and the -Y axis at D. CD represents the minor axis of our ellipse. Let AO = a, CO = b, and FO = c, the typical terms used to represent the major and minor axes of an ellipse and the distance to the focus from the center O. Note that CF also equals "a" since we used radius AO to locate point B. (As you can see, from the pythagorean theorem, a^2 = b^2 + c^2.) Therefore if you are given the major and minor axes of an ellipse, you can determine the location of the focus points from c = sqrt(a^2 - b^2) where a = semi-major axis and b = semi-minor axis. The eccentricity of the ellipse is e = c/a = <1. The terms a, b, and c are also related by b^2 = a^2(1 -e^2) or e^2 = 1 - b^2/a^2.

That said, an elliptical orbit is the elliptical shaped path of a body, or satellite, around another body having a gravitational field. As Kepler's First Law states, the orbits of all the planets in our solar system, as well as the orbits of all Earth satellites, are ellipses. In the case of the planets, the Sun is the focus of the planetary orbits. In the case of Earth satellites, the Earth is the focus. Geometrically, an ellipse is the locus of a point moving in such a way that the sum of its distances from two fixed points, f and f', called foci, is a constant. Pictorially, draw yourself two perpendicular axes, X and Y, with origin O. Locate the two foci points, on the X axis, equidistant on either side of the origin O, say 1 inch for example, labeling the points f on the right and f' on the left. Each represents a focus of the ellipse. Now locate two points, on the X axis, 2 inches on either side of the origin O. Label these A on the left and B on the right. AB is the major axis of our ellipse. With a compass, using AO as a radius and f as the center, swing an arc that intersects the +Y axis at C and the -Y axis at D. CD represents the minor axis of our ellipse. Note that fB equals AO since we used radius AO to locate point B. (As you can see, from the pythagorean theorem, (fB)^2 = (fO)^2 + (OB)^2.) Therefore if you are given the major and minor axes of an ellipse, you can determine the location of the focus points from fO = sqrt[(fB)a^2 - (OB)^2] where AO = semi-major axis and OB = semi-minor axis. The eccentricity of the ellipse is e = fO/AO = <1. The terms a, b, and c are also related by b^2 = a^2(1 -e^2) or e^2 = 1 - b^2/a^2.

Now, unfortunately, we must change our terminology and parameter designations for use in the area of our orbital description. For our elliptical orbit, and referring to the picture created above, the basic terminology becomes AO = a = the semi-major axis, OB = b = the semi-minor axis, f and f' are the two foci, fO = c = the distance from the foci to the center of the ellipse, e = the eccentricity = c/a, p = a geometric constant of the ellipse called the parameter, sometimes referred to as the semi-latus rectum, and determined from p = a(1 - e^2), point B is now designated as p, the periapsis point, the closest the elliptical path comes to the focus f (not to be confused with p the parameter), point A is now designated as a, the apoapsis point, the farthest the elliptical path extends from the focus f (not to be confused with a the semi-major axis).

The polar equation of our elliptical orbit is defined as r = p/[1 + e(cos(v))] where r is the radius from the focus f to a point on the ellipse, p is our geometric constant, as defined above, e is the eccentricity, and v is the angle (true anomaly) between the radius r and the major axis of the ellipse, with the periapsis end being zero, and measured positively in the direction of flight, counterclockwise. If you know the semi-major and semi-minor axes and the eccentricity, the periapsis distance is calculated from r(p) = p/[1 + e(cos(0))] = p/(1 + e) = a(1 - e). Similarly, the apoapsis distance is calculated from r(a) = p/[1 + e(cos(180))] = p/(1 - e) = a(1 + e). It becomes clear that r(p) + r(a) = 2a and r(a) - r(p) = 2c. These expressions may be combined to yield e = [r(a) - r(p)]/[r(a) + r(p)].

The velocity of the orbiting body at the periapsis is defined by

.........................................______________

......................................../.........(2µ) r(a)

.............................Vp = \/ --------------------- ............in feet per second

............................................r(p) [r(a) + r(p)]

The velocity of the orbiting body at the apoapsis is defined by

.............................................______________

............................................/ (2µ) r(p)

.................................Va = \/ --------------------.............in feet per second

.............................................r(a) [r(a) + r(p)]

The velocity at any point on the orbit is given by

...................................._____________

.................................V = \ / 2µ/r - µ/a..................in feet per second

............................................................______

The orbital period is given by T = 2(Pi)../ a^3

.........................................................\/ ------................in seconds

..................................................................µ - space -
**Brandon**, Wednesday, January 28, 2009 at 1:46pmtchrwill: That was a little bit of overkill for a true/false question.