Re: Question about Kepler's second law

On Apr 10, 11:23 am, "Peter" <Poakfi...@xxxxxxx> wrote:
On Apr 10, 9:19 am, "PD" <TheDraperFam...@xxxxxxxxx> wrote:



On Apr 10, 8:05 am, "Peter" <Poakfi...@xxxxxxx> wrote:

On Apr 9, 5:44 pm, "PD" <TheDraperFam...@xxxxxxxxx> wrote:

On Apr 9, 3:21 pm, "Peter" <Poakfi...@xxxxxxx> wrote:

On Apr 9, 1:42 pm, "PD" <TheDraperFam...@xxxxxxxxx> wrote:

On Apr 9, 9:28 am, "Peter" <Poakfi...@xxxxxxx> wrote:

Could someone help me, please? I must be missing something. According
toKepler'ssecondlaw, the force exerted by the Sun on a planet is a
radial force that has no component in the direction of motion of the
planet; however, the speed of the planet is not uniform: it changes
all the time. How can that be? I understand that only a force in the
direction of travel of a planet can change the magnitude of its
velocity. Thanks.
Peter

First things first.
Get a piece of corrugated cardboard, two pushpins, a length of string,
a ruler, and a pencil.
Place two pushpins some distance away on the cardboard.
Tie the string to make a loop that is longer than the distance between
the two pins.
Wrap the string around the pins and stretch the loop out with the tip
of the pencil, and then draw with the pencil in such a way that the
loop of string remains taught. You will draw an ellipse by
construction.
Take the pencil and draw a Sun at the location of one of the pushpins.
Remove both pushpins and the length of string.
Now draw a little circle at four or five places around the ellipse.
This will be one of the planets, viewed at different times during that
planet's year.
Take a ruler and draw a line between the Sun and one of the locations
of the planet. Draw a little arrow from the planet along this line
toward the sun.
Repeat the previous step for the other planet locations. Make the
arrow shorter if the planet is further away from the Sun than the
first location, longer if it is closer to the Sun.
These arrows are the force vectors of gravity between the planet and
the Sun, acting on the planet, in each of those locations.
Choose an orientation (clockwise or counterclockwise) for the planet's
movement around the ellipse.
You will notice that when the planet is approaching the sun (coming
toward perigee), then the force has a component along the planet's
motion, which is why it speeds up as it gets closer to the sun.
You will notice that when the planet is receding from the sun (headed
toward apogee), then the force has a component against the planet's
motion, which is why it slows down as it gets further away from the
sun.

Kepler'slaws are all completely consistent with this.

PD

I agree withKepler'slaws. I also agree there is a component of the
force along the planet's motion, but what is its lever arm?

Well, there are two components, recall: one that points along the
planet's motion and one that is perpendicular to the planet's motion.
Neither of these point toward the sun. One component changes the
magnitude of the planet's velocity, the other changes the planet's
direction. If you were to calculate the torque from both of these
*components*, you would find that they cancel. For the total force,
the one that points directly toward the sun, the lever arm is
obviously zero.

PD- Hide quoted text -

- Show quoted text -

I am not too sure, but I made a graphic, and I don't see the torques
cancel, because the force perpendicular to the planet's motion
necessarily points toward the sun.

Then you have made a mistake, and you should conduct the drawing
exercise I recommended to you.
Recall that for an ellipse, there is a major axis (the longest
distance between two points in the ellipse) and a minor axis
(perpendicular to the major axis and through the center of the
ellipse). Take for example, the point where the planet crosses the
minor axis. The direction perpendicular to the planet's motion points
to the center of the ellipse, but that is not where the sun is. The
sun is at one of the foci of the ellipse (where one of the pushpins
was) and that's where gravity points toward.

There are only two points on the entire ellipse where the force of
gravity is perpendicular to the planet's motion, and that's where it
crosses the major axis at perihelion and aphelion.

If you'd like, compare your graphic to the ones at the following
links:http://csep10.phys.utk.edu/astr161/lect/history/kepler.htmlhttp://www...

PD

Thank you for your help. I'll tell you what I did. You know that any
vector can be decomposed into its constituent parts. I drew
a line from the Sun to the planet, when the planet is at the top of
the minor axis. Then I drew a horizontal line going through
the planet, and then a vertical line from the Sun to the horizontal
line. The horizontal and vertical lines are the components of
the vector line from the Sun to the planet. Clearly, the horizontal
component of the force can be decomposed into a force
perpendicular to the line from the Sun to the planet; this force
multiplied by the distance from the Sun to the planet,
would be the torque. Where am I wrong?


Yes indeed, now I understand where you're going wrong. When you are
doing a force diagram (sometimes called a free-body diagram) where
torques are analyzed, the *point of application* of the force is very
important. Your method of breaking the force of gravity into
components by drawing a triangle and using the legs of the triangle
for the force components, will lead to an error here. Why is this
important? Well, obviously, if you "slide" a force from where it is
actually applied to an object to someplace else, then you're obviously
changing the lever arm without knowing it.

The force acting on the planet is a vector that is drawn as
originating *at the planet*, and the components of the force have to
be drawn also as originating at the planet. In the case as you've
drawn it, you *should* have a horizontal arrow starting at the planet
(and this is the component parallel to the planet's motion) and a
vertical arrow starting *at the planet* and pointing toward the center
of the ellipse (and this is the component perpendicular to the
planet's motion). The relative lengths of the arrows should be the
same as what you get from using your triangle method. But notice the
key difference is that the perpendicular component should *not* point
toward the sun but toward the center of the ellipse.

Once you have your corrected drawing, now you can see there are two
torques from the two components, one clockwise around the sun and one
counterclockwise, and they in fact cancel.

Does this help?

PD

.

Relevant Pages

  • Re: Question about Keplers second law
    ... planet; however, the speed of the planet is not uniform: ... Take the pencil and draw a Sun at the location of one of the pushpins. ... Now draw a little circle at four or five places around the ellipse. ... motion, which is why it speeds up as it gets closer to the sun. ...
    (sci.physics)
  • Re: Question about Keplers second law
    ... planet; however, the speed of the planet is not uniform: ... You will draw an ellipse by ... Take the pencil and draw a Sun at the location of one of the pushpins. ... motion, which is why it speeds up as it gets closer to the sun. ...
    (sci.physics)
  • Re: Question about Keplers second law
    ... planet; however, the speed of the planet is not uniform: ... You will draw an ellipse by ... Take the pencil and draw a Sun at the location of one of the pushpins. ... motion, which is why it speeds up as it gets closer to the sun. ...
    (sci.physics)
  • Re: Question about Keplers second law
    ... planet; however, the speed of the planet is not uniform: ... Take the pencil and draw a Sun at the location of one of the pushpins. ... Now draw a little circle at four or five places around the ellipse. ... motion, which is why it speeds up as it gets closer to the sun. ...
    (sci.physics)
  • Re: Question about Keplers second law
    ... planet; however, the speed of the planet is not uniform: ... Take the pencil and draw a Sun at the location of one of the pushpins. ... Now draw a little circle at four or five places around the ellipse. ... motion, which is why it speeds up as it gets closer to the sun. ...
    (sci.physics)