| Topic: |
Science > Physics |
| User: |
"Peter" |
| Date: |
09 Apr 2007 09:28:56 AM |
| Object: |
Question about Kepler's second law |
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
.
|
|
| User: "Androcles" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 11:49:28 AM |
|
|
"Peter" <Poakfield@msn.com> wrote in message =
news:1176128936.116138.34440@y66g2000hsf.googlegroups.com...
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
Velocity has only one direction. Speed is the magnitude of velocity
and no special direction. If we ride bicycles alongside each other=20
and I suddenly reach out and push you sideways, that will change=20
your velocity and mine, but it may not change your speed.
.
|
|
|
|
| User: "" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 02:11:33 PM |
|
|
In article <1176128936.116138.34440@y66g2000hsf.googlegroups.com>, "Peter" <Poakfield@msn.com> writes:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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;
No, "radial" means that it has no component in the the direction
perpendicular to the radius joining the Sun and the planet. This is
not the same as the direction of motion of the planet, unless the
planet is in a circular orbit (in which case, indeed, its speed
doesn't change).
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
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 08:37:19 PM |
|
|
On 9 Apr 2007 07:28:56 -0700, "Peter" <Poakfield@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
With a circular orbit the velocity is constant. If it suffers an
impulse that changes the orbit to an ellipse you will have a component
rdot that steals energy from v. The total energy is:
W = .5m(v^2 + rdot^2)
The radial velocity rdot is an oscillatory function from +rdot to
-rdot. When rdot is 0, v is maximum. Otherwise v is reduced.
John Polasek
.
|
|
|
| User: "boson boss" |
|
| Title: Re: Question about Kepler's second law |
10 Apr 2007 04:47:44 AM |
|
|
One idea is that the planets are in a free fall - and they don't sense
much of anything just like zero gravity in falling elevator. Secondly
we use some math calculus to show that planets are just point like
bodies (with dimension yes, but calculate as point like bodies). Then
we notice that planets really have tidal waves and that gravity is
complex locally. Add to this GR for mercury and stuff like wobble of
ellipses, and finally realize that evolving solar system is beyond
comprehension all together.
.
|
|
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Question about Kepler's second law |
10 Apr 2007 09:20:31 AM |
|
|
On 9 Apr 2007 07:28:56 -0700, "Peter" <Poakfield@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
With a circular orbit the velocity is constant. If it suffers an
impulse that changes the orbit to an ellipse you will have a component
rdot that steals energy from v. The total energy is:
W = .5m(v^2 + rdot^2)
The radial velocity rdot is an oscillatory function from +rdot to
-rdot. When rdot is 0, v is maximum. Otherwise v is reduced.
John Polasek
.
|
|
|
|
| User: "Igor" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 12:46:36 PM |
|
|
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
Kepler II says that orbital angular momentum is conserved. This
corresponds to equal areas being swept out in equal times. Just
because angular momentum is constant, angular speed may not be.
Indeed, the only case where angular speed would be constant is for a
circular orbit. In general, angular speed varies as the inverse
square of the radial coordinate.
.
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 09:37:53 AM |
|
|
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 09:59:04 AM |
|
|
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet? Why?
Peter
.
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 10:39:43 AM |
|
|
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 12:45:35 PM |
|
|
On Apr 9, 11:39 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy- Hide quoted text -
- Show quoted text -
Thank you. But, isn't there a contradiction between "no torque acts on
the planet" and "there is a component of force in the direction of
motion [of the planet]?
Peter
.
|
|
|
| User: "PD" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 12:56:20 PM |
|
|
On Apr 9, 12:45 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 11:39 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy- Hide quoted text -
- Show quoted text -
Thank you. But, isn't there a contradiction between "no torque acts on
the planet" and "there is a component of force in the direction of
motion [of the planet]?
Peter
No. The angular momentum of the planet does not change, though its
speed does change.
If it helps, when an ice skater pulls in her arms in a spin, her
angular momentum does not change, either.
A torque would result in a change in angular momentum.
PD
.
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 12:56:40 PM |
|
|
On Apr 9, 1:45 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 11:39 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy- Hide quoted text -
- Show quoted text -
Thank you. But, isn't there a contradiction between "no torque acts on
the planet" and "there is a component of force in the direction of
motion [of the planet]?
No. Torque has a specific definition: T = r x F.
You might again be confusing this with circular motion. In
circular motion, "direction of motion" is perpendicular to
lever arm r. If there were any force in that direction, it would
be perpendicular to r and would give nonzero r x F.
However, in Keplerian motion, the direction of motion is
NOT perpendicular to r. "Component parallel to motion"
is not the same as "component perpendicular to r" and in
fact the the force is completely PARALLEL to r. Perhaps
you should draw a picture so you can understand how
something can be along the r direction but nevertheless
have a component along the direction of motion.
It isn't possible for a circle, but it is the case nearly
everywhere on an elliptical orbit.
- Randy
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 02:53:38 PM |
|
|
On Apr 9, 1:56 pm, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 1:45 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 11:39 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy- Hide quoted text -
- Show quoted text -
Thank you. But, isn't there a contradiction between "no torque acts on
the planet" and "there is a component of force in the direction of
motion [of the planet]?
No. Torque has a specific definition: T = r x F.
You might again be confusing this with circular motion. In
circular motion, "direction of motion" is perpendicular to
lever arm r. If there were any force in that direction, it would
be perpendicular to r and would give nonzero r x F.
However, in Keplerian motion, the direction of motion is
NOT perpendicular to r. "Component parallel to motion"
is not the same as "component perpendicular to r" and in
fact the the force is completely PARALLEL to r. Perhaps
you should draw a picture so you can understand how
something can be along the r direction but nevertheless
have a component along the direction of motion.
It isn't possible for a circle, but it is the case nearly
everywhere on an elliptical orbit.
- Randy- Hide quoted text -
- Show quoted text -
I agree there is a component of the force along the direction of
motion of the planet, but then, what is its lever arm?
Peter
.
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 07:36:02 PM |
|
|
On Apr 9, 3:53 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 1:56 pm, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 1:45 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 11:39 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy- Hide quoted text -
- Show quoted text -
Thank you. But, isn't there a contradiction between "no torque acts on
the planet" and "there is a component of force in the direction of
motion [of the planet]?
No. Torque has a specific definition: T = r x F.
You might again be confusing this with circular motion. In
circular motion, "direction of motion" is perpendicular to
lever arm r. If there were any force in that direction, it would
be perpendicular to r and would give nonzero r x F.
However, in Keplerian motion, the direction of motion is
NOT perpendicular to r. "Component parallel to motion"
is not the same as "component perpendicular to r" and in
fact the the force is completely PARALLEL to r. Perhaps
you should draw a picture so you can understand how
something can be along the r direction but nevertheless
have a component along the direction of motion.
It isn't possible for a circle, but it is the case nearly
everywhere on an elliptical orbit.
I agree there is a component of the force along the direction of
motion of the planet, but then, what is its lever arm?
The total force F is acting in a direction toward the sun.
You may think of this as having components along the
direction of the planet's motion, and perpendicular to that
motion: F = F_par + F_perp
The lever arm is the vector r from planet to sun. The
torque is equal to T = r x F, and r and F point in the
same direction.
If you want to think of it in components, as PD says
you would get two nonzero torque terms r x F_par
and r x F_perp, but they will cancel. For all of them
r is the vector from planet to sun.
- Randy
.
|
|
|
|
| User: "boson boss" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 03:38:22 PM |
|
|
On Apr 9, 9:53 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 1:56 pm, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 1:45 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 11:39 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet?
It means this statement of yours was incorrect:
"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;"
That would be true of a circular orbit, which is also a constant
speed orbit. It is not true of an elliptical orbit. The direction of
motion and the direction of force are not perpendicular, so there
is a component of force in the direction of motion.
Why?
Your statement about torque is correct: "the force exerted by the
Sun produces no torque on the planet." The reason for this is
that torque = r x F, and the vectors r and F are parallel.
Since there is no torque, angular momentum is constant. Kepler's
2nd law (equal area in equal time) can be proven to be equivalent
to "angular momentum is a constant".
- Randy- Hide quoted text -
- Show quoted text -
Thank you. But, isn't there a contradiction between "no torque acts on
the planet" and "there is a component of force in the direction of
motion [of the planet]?
No. Torque has a specific definition: T = r x F.
You might again be confusing this with circular motion. In
circular motion, "direction of motion" is perpendicular to
lever arm r. If there were any force in that direction, it would
be perpendicular to r and would give nonzero r x F.
However, in Keplerian motion, the direction of motion is
NOT perpendicular to r. "Component parallel to motion"
is not the same as "component perpendicular to r" and in
fact the the force is completely PARALLEL to r. Perhaps
you should draw a picture so you can understand how
something can be along the r direction but nevertheless
have a component along the direction of motion.
It isn't possible for a circle, but it is the case nearly
everywhere on an elliptical orbit.
- Randy- Hide quoted text -
- Show quoted text -
I agree there is a component of the force along the direction of
motion of the planet, but then, what is its lever arm?
Peter
What about tidal waves?
.
|
|
|
|
|
|
|
|
| User: "Dwib" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 12:50:33 PM |
|
|
On Apr 9, 9:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet? Why?
Peter
The sun applies a same force to every tiny piece of a planet causing
the planet to fall towards the sun.
Certainly, if the planet is symmetrical there will be no net torque.
For an assymetrical non-rotating planet, there will be a net torque
but any sort of planetary rotation will null out this torque over
time.
Dwib
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 02:59:41 PM |
|
|
On Apr 9, 1:50 pm, "Dwib" <dwibd...@gmail.com> wrote:
On Apr 9, 9:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet? Why?
Peter
The sun applies a same force to every tiny piece of a planet causing
the planet to fall towards the sun.
Certainly, if the planet is symmetrical there will be no net torque.
For an assymetrical non-rotating planet, there will be a net torque
but any sort of planetary rotation will null out this torque over
time.
Dwib- Hide quoted text -
- Show quoted text -
Assuming the planet is a point mass with no spin, but in an elliptical
orbit, is there a tangential force on it?
Peter
.
|
|
|
| User: "" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 03:32:28 PM |
|
|
In article <1176148781.237043.31890@b75g2000hsg.googlegroups.com>, "Peter" <Poakfield@msn.com> writes:
On Apr 9, 1:50 pm, "Dwib" <dwibd...@gmail.com> wrote:
On Apr 9, 9:59 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet? Why?
Peter
The sun applies a same force to every tiny piece of a planet causing
the planet to fall towards the sun.
Certainly, if the planet is symmetrical there will be no net torque.
For an assymetrical non-rotating planet, there will be a net torque
but any sort of planetary rotation will null out this torque over
time.
Dwib- Hide quoted text -
- Show quoted text -
Assuming the planet is a point mass with no spin, but in an elliptical
orbit, is there a tangential force on it?
Depends what you define by "tangential". If it stands for
"perpendicular to the radius joining the planet with the Sun" then the
answer is "there is no tangential force". If, however, it stands for
"tangent to the direction of motion of the planet" then yes, in
general there is a tangential force.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
|
| User: "Dwib" |
|
| Title: Re: Question about Kepler's second law |
10 Apr 2007 11:58:09 AM |
|
|
On Apr 9, 2:59 pm, "Peter" <Poakfi...@msn.com> wrote:
Assuming the planet is a point mass with no spin, but in an elliptical
orbit, is there a tangential force on it?
The force is ALWAYS points toward the sun. That's the critical
fact.
Relative to the elliptical orbital path of the planet... "yes" there
can be a component of that force which is parallel to the orbital
path.
Dwib
.
|
|
|
|
|
|
| User: "Sam Wormley" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 10:16:29 AM |
|
|
Peter wrote:
On Apr 9, 10:37 am, "Randy Poe" <poespam-t...@yahoo.com> wrote:
On Apr 9, 10:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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
The planet's orbit is not circular but elliptical. During most of the
orbit, the direction of motion is not perpendicular to the direction
of the gravitational force.
- Randy
Thank you. Does this mean the force exerted by the Sun produces no
torque on the planet? Why?
Peter
Twist or Torque cannot be applied by gravity. r X F = 0
.
|
|
|
|
|
|
| User: "arvee" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 09:44:46 AM |
|
|
On Apr 9, 7:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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.
Correction: "...can change the sun-to-planet-direction *component* of
its velocity...". Velocity is a vector, while speed is the magnitude
of that vector. In a small interval of time the velocity component
perpendicular to the sun-planet direction does not change (here we are
neglecting curvature, etc.) but the component parallel to the sun-
planet direction does change. You can also get the explanation from
conservation of (total) energy = kinetic + potential. When the
potential energy goes up, the speed goes down, etc.
Here is a little puzzle for you to think about: if the sun's gravity
is always exerting an inward force on the planet, why is the planet
sometimes moving away from the sun after previously moving toward it?
Doesn't that mean the acceleration was outward, opposite to the pull
of gravity?
R.G. Vickson
Thanks.
Peter
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 03:07:23 PM |
|
|
On Apr 9, 10:44 am, "arvee" <C...@shaw.ca> wrote:
On Apr 9, 7:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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.
Correction: "...can change the sun-to-planet-direction *component* of
its velocity...". Velocity is a vector, while speed is the magnitude
of that vector. In a small interval of time the velocity component
perpendicular to the sun-planet direction does not change (here we are
neglecting curvature, etc.) but the component parallel to the sun-
planet direction does change. You can also get the explanation from
conservation of (total) energy = kinetic + potential. When the
potential energy goes up, the speed goes down, etc.
Here is a little puzzle for you to think about: if the sun's gravity
is always exerting an inward force on the planet, why is the planet
sometimes moving away from the sun after previously moving toward it?
Doesn't that mean the acceleration was outward, opposite to the pull
of gravity?
R.G. Vickson
Thanks.
Peter- Hide quoted text -
- Show quoted text -
When the planet is moving away from the sun, it is being decelerated,
not accelerated.
Peter
.
|
|
|
| User: "arvee" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 06:46:43 PM |
|
|
On Apr 9, 1:07 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 10:44 am, "arvee" <C...@shaw.ca> wrote:
On Apr 9, 7:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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.
Correction: "...can change the sun-to-planet-direction *component* of
its velocity...". Velocity is a vector, while speed is the magnitude
of that vector. In a small interval of time the velocity component
perpendicular to the sun-planet direction does not change (here we are
neglecting curvature, etc.) but the component parallel to the sun-
planet direction does change. You can also get the explanation from
conservation of (total) energy = kinetic + potential. When the
potential energy goes up, the speed goes down, etc.
Here is a little puzzle for you to think about: if the sun's gravity
is always exerting an inward force on the planet, why is the planet
sometimes moving away from the sun after previously moving toward it?
Doesn't that mean the acceleration was outward, opposite to the pull
of gravity?
R.G. Vickson
Thanks.
Peter- Hide quoted text -
- Show quoted text -
When the planet is moving away from the sun, it is being decelerated,
not accelerated.
Of course. But, the radial velocity changed from inward to outward, so
that's an outward acceleration, for a while at least. (I know, I
know... it's a trick question. But it's along the lines--no pun
intended-- of the OP's original query.)
R.G. Vickson
Peter
.
|
|
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 10:20:16 AM |
|
|
On Apr 9, 10:44 am, "arvee" <C...@shaw.ca> wrote:
On Apr 9, 7:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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.
Correction: "...can change the sun-to-planet-direction *component* of
its velocity...". Velocity is a vector, while speed is the magnitude
of that vector. In a small interval of time the velocity component
perpendicular to the sun-planet direction does not change (here we are
neglecting curvature, etc.) but the component parallel to the sun-
planet direction does change. You can also get the explanation from
conservation of (total) energy = kinetic + potential. When the
potential energy goes up, the speed goes down, etc.
Here is a little puzzle for you to think about: if the sun's gravity
is always exerting an inward force on the planet, why is the planet
sometimes moving away from the sun after previously moving toward it?
Doesn't that mean the acceleration was outward, opposite to the pull
of gravity?
R.G. Vickson
Thank you. I am not questioning the validity of Kepler's second law, but I do not understand the mechanics. Isn't it true that only a force in the direction of motion of the planet can change its speed? However, the books say there is no torque acting on the planet, which I understand it means there is no component of the force in the direction of its motion. I am thinking about your puzzle.
Peter
- Show quoted text -
.
|
|
|
|
|
| User: "PD" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 12:42:58 PM |
|
|
On Apr 9, 9:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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's laws are all completely consistent with this.
PD
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 03:21:04 PM |
|
|
On Apr 9, 1:42 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 9:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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's laws are all completely consistent with this.
PD
I agree with Kepler's laws. I also agree there is a component of the
force along the planet's motion, but what is its lever arm?
Peter
.
|
|
|
| User: "PD" |
|
| Title: Re: Question about Kepler's second law |
09 Apr 2007 04:44:40 PM |
|
|
On Apr 9, 3:21 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 1:42 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 9:28 am, "Peter" <Poakfi...@msn.com> wrote:
Could someone help me, please? I must be missing something. According
to Kepler's second law, 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's laws are all completely consistent with this.
PD
I agree with Kepler's laws. 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
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
10 Apr 2007 08:05:41 AM |
|
|
On Apr 9, 5:44 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 3:21 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 1:42 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 9:28 am, "Peter" <Poakfi...@msn.com> 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.
Peter
.
|
|
|
| User: "PD" |
|
| Title: Re: Question about Kepler's second law |
10 Apr 2007 08:19:03 AM |
|
|
On Apr 10, 8:05 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 5:44 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 3:21 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 1:42 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 9:28 am, "Peter" <Poakfi...@msn.com> 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.html
http://www.windows.ucar.edu/tour/link=/physical_science/physics/mechanics/orbit/ellipse.html
http://www.sciencebyjones.com/ellipse%20lab%202004.htm
http://www.go.ednet.ns.ca/~larry/orbits/ellipse.html
PD
Peter- Hide quoted text -
- Show quoted text -
.
|
|
|
| User: "Peter" |
|
| Title: Re: Question about Kepler's second law |
10 Apr 2007 11:23:01 AM |
|
|
On Apr 10, 9:19 am, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 10, 8:05 am, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 5:44 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 3:21 pm, "Peter" <Poakfi...@msn.com> wrote:
On Apr 9, 1:42 pm, "PD" <TheDraperFam...@gmail.com> wrote:
On Apr 9, 9:28 am, "Peter" <Poakfi...@msn.com> 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.windows.ucar.edu/tour/link=/physical_science/physics/mecha...http://www.sciencebyjones.com/ellipse%20lab%202004.htmhttp://www.go.ednet.ns.ca/~larry/orbits/ellipse.html
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?
Peter
Peter- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -
.
|
|
|
|
|
|
|
|
|
|
Related Articles |
|
