Orbit problem

"Darrell" wrote in message ...
How do we know that the circle covers more distance? I was thinking of it
as a string tied at the ends to make a circle and no matter how I shape the
string, the distance is the same? I'm sure I'm missing sometime, what is
it?

The constraint is that the semimajor axis of the ellipse
is the same size as the radius of the circle. So the
ellipse will fit entirely inside the circle, the ends of
the major axis just touching the circle. It should be
clear that the ellipse must have a shorter perimeter than
the circle.

"Darrell" wrote in message ...
How do we know that the circle covers more distance? I was thinking of it
as a string tied at the ends to make a circle and no matter how I shape the
string, the distance is the same? I'm sure I'm missing sometime, what is
it?

The constraint is that the semimajor axis of the ellipse
is the same size as the radius of the circle. So the
ellipse will fit entirely inside the circle, the ends of
the major axis just touching the circle. It should be
clear that the ellipse must have a shorter perimeter than
the circle.

Two questions:
I'm just not getting it, so please bare with me.

How can the semimajor axis (which is the part I'm not understanding so well.
I'll look at a diagram, but I'm sure I'll need a formula to figure it out)
be the same as the radius of the circle?

I understand that the ellipse can not be wider than the circle, but why
wouldn't the ends of the ellipse extent beyond the edge of the circle?

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* @ @ *
* | *
* ~~~~ *
* V *
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"Greg Neill" wrote in message
.. .
"Darrell" wrote in message
...
How do we know that the circle covers more distance? I was thinking of
it
as a string tied at the ends to make a circle and no matter how I shape
the
string, the distance is the same? I'm sure I'm missing sometime, what
is
it?

The constraint is that the semimajor axis of the ellipse
is the same size as the radius of the circle. So the
ellipse will fit entirely inside the circle, the ends of
the major axis just touching the circle. It should be
clear that the ellipse must have a shorter perimeter than
the circle.

Two questions:
I'm just not getting it, so please bare with me.

How can the semimajor axis (which is the part I'm not understanding so well.
I'll look at a diagram, but I'm sure I'll need a formula to figure it out)
be the same as the radius of the circle?

I understand that the ellipse can not be wider than the circle, but why
wouldn't the ends of the ellipse extent beyond the edge of the circle?

****
* *
* *
* @ @ *
* | *
* ~~~~ *
* V *
****




"Greg Neill" wrote in message
.. .
"Darrell" wrote in message
...
How do we know that the circle covers more distance? I was thinking of
it
as a string tied at the ends to make a circle and no matter how I shape
the
string, the distance is the same? I'm sure I'm missing sometime, what
is
it?

The constraint is that the semimajor axis of the ellipse
is the same size as the radius of the circle. So the
ellipse will fit entirely inside the circle, the ends of
the major axis just touching the circle. It should be
clear that the ellipse must have a shorter perimeter than
the circle.

"Darrell" wrote in message
...
Two questions:
I'm just not getting it, so please bare with me.

How can the semimajor axis (which is the part I'm not understanding so
well.
I'll look at a diagram, but I'm sure I'll need a formula to figure it out)
be the same as the radius of the circle?
Hi Darrell,
Imagine you're looking straight down at the top of a mug. The top is
circular, and the diameter is easy to measure.

Now imagine you're looking at an angle at the top of the mug. The shape you
see is now an ellipse, which has one long axis, (left to right) and one
short axis (joining the front and back of the rim as you look at it). The
major axis is the long one, and the minor axis is the smaller one. Notice
that the major axis is the same as the diameter of the mug.

Take a piece of paper and draw a circle, now try and draw an ellipse shape
that fits exactly inside the circle.

Hopefully you can see that the circle gives the size of the major axis. Now,
just as the radius is half the diameter for a circle, the semi-major axis is
half the length of the major axis for an ellipse.

Does this help?

Owen

"Darrell" wrote in message
...
Two questions:
I'm just not getting it, so please bare with me.

How can the semimajor axis (which is the part I'm not understanding so
well.
I'll look at a diagram, but I'm sure I'll need a formula to figure it out)
be the same as the radius of the circle?
Hi Darrell,
Imagine you're looking straight down at the top of a mug. The top is
circular, and the diameter is easy to measure.

Now imagine you're looking at an angle at the top of the mug. The shape you
see is now an ellipse, which has one long axis, (left to right) and one
short axis (joining the front and back of the rim as you look at it). The
major axis is the long one, and the minor axis is the smaller one. Notice
that the major axis is the same as the diameter of the mug.

Take a piece of paper and draw a circle, now try and draw an ellipse shape
that fits exactly inside the circle.

Hopefully you can see that the circle gives the size of the major axis. Now,
just as the radius is half the diameter for a circle, the semi-major axis is
half the length of the major axis for an ellipse.

Does this help?

Owen

"Darrell" wrote in message ...
Two questions:
I'm just not getting it, so please bare with me.

How can the semimajor axis (which is the part I'm not understanding so well.
I'll look at a diagram, but I'm sure I'll need a formula to figure it out)
be the same as the radius of the circle?

The semimajor axis can be any length at all. But in this particular
case it is stipulated to be 2AU, the same as the radius of the
circular orbit.


I understand that the ellipse can not be wider than the circle, but why
wouldn't the ends of the ellipse extent beyond the edge of the circle?

See above; it's just a fact for this particular problem.

"Darrell" wrote in message ...
Two questions:
I'm just not getting it, so please bare with me.

How can the semimajor axis (which is the part I'm not understanding so well.
I'll look at a diagram, but I'm sure I'll need a formula to figure it out)
be the same as the radius of the circle?

The semimajor axis can be any length at all. But in this particular
case it is stipulated to be 2AU, the same as the radius of the
circular orbit.


I understand that the ellipse can not be wider than the circle, but why
wouldn't the ends of the ellipse extent beyond the edge of the circle?

See above; it's just a fact for this particular problem.

"OG" wrote in message...
...

Following Kepler, since a^2 = k r^3,
where a = period
and r = semi major axis
( and k is there because I can't write a 'proportional to' squiggle) ;-)

it is clear that the orbital period of identical for both bodies.

The distance travelled in the elliptical eccentric orbit is less than the
distance in the circular orbit so it is clear that the average speed (over
the full orbit) is less for the elliptical case.

Here is a page with animations

http://home.cvc.org/science/kepler.htm

Thanks, Owen!... Now another question is raisined in my
brain:...

You'll notice that the animation for the first of Kepler's laws
is a true ellipse. Even though the speed of the orbiting
object seems to increase as it approaches the Sun, and
then decrease as it leaves the Sun, the object does not
appear to do this at the other end of the major axis. And
intuition followed by math would lead us to expect that the
speed would increase and then decrease at the other end
of the ellipse as well...

Now, the animation in the second of Kepler's laws seems
to me to be more egg-shaped rather than elliptical. And
if a planetary orbit is egg-shaped, then it's speed would
be minimum at its farthest distance from the Sun (aphelion).

To be clearer, velocity will be maximum at both ends of the
major axis of an ellipse, and minimum at both ends of the
minor axis, correct? Therefore, the planets would go the
fastest speeds at both perihelion *and* aphelion?

I realize that the planetary orbits are not that far from being
circles, and therefore both of the foci of all the planets'
orbits are close together and inside the Sun (except for
Jupiter). But i'm having a problem with the egg-shape...

Are planetary orbits true ellipses? or are they egg-shaped
as depicted in the animation for Kepler's Second Law?

--
happy days and...
starry starry nights!

Paine Ellsworth

"OG" wrote in message...
...

Following Kepler, since a^2 = k r^3,
where a = period
and r = semi major axis
( and k is there because I can't write a 'proportional to' squiggle) ;-)

it is clear that the orbital period of identical for both bodies.

The distance travelled in the elliptical eccentric orbit is less than the
distance in the circular orbit so it is clear that the average speed (over
the full orbit) is less for the elliptical case.

Here is a page with animations

http://home.cvc.org/science/kepler.htm

Thanks, Owen!... Now another question is raisined in my
brain:...

You'll notice that the animation for the first of Kepler's laws
is a true ellipse. Even though the speed of the orbiting
object seems to increase as it approaches the Sun, and
then decrease as it leaves the Sun, the object does not
appear to do this at the other end of the major axis. And
intuition followed by math would lead us to expect that the
speed would increase and then decrease at the other end
of the ellipse as well...

Now, the animation in the second of Kepler's laws seems
to me to be more egg-shaped rather than elliptical. And
if a planetary orbit is egg-shaped, then it's speed would
be minimum at its farthest distance from the Sun (aphelion).

To be clearer, velocity will be maximum at both ends of the
major axis of an ellipse, and minimum at both ends of the
minor axis, correct? Therefore, the planets would go the
fastest speeds at both perihelion *and* aphelion?

I realize that the planetary orbits are not that far from being
circles, and therefore both of the foci of all the planets'
orbits are close together and inside the Sun (except for
Jupiter). But i'm having a problem with the egg-shape...

Are planetary orbits true ellipses? or are they egg-shaped
as depicted in the animation for Kepler's Second Law?

--
happy days and...
starry starry nights!

Paine Ellsworth