accereration 101

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The orbit of a body has speed but not velocity according to the above post?
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A moving body ALWAYS has velocity. Speed is one of the two pieces of velocity. Either demands the existance of the other.

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So, then it is a constant change in direction that accounts for the fact that the orbit is as Kepler says?
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Not by itself, no. But when combined with the speed piece, yes. BTW, the speed part is only constant for circular orbits, so in general both peices will be changing.

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And the acceleration is used by the body and therefore somehow must be absorbed or radiated away, right?
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I'm tempted to say yes, but I'm not sure I understand your meaning here.

Consider this. Suppose that at a particular time in the Earth's orbit it is moving directly toward the center of the Milky Way.

This means that if Sol suddenly disappeared the Earth would move in a straight line <b>toward</b> the center of the galaxy, with a speed of 30 km/sec.

But if Sol stays put the Earth remains in orbit, and six months later the Earth is moving directly away from the center of the Milky Way. If Sol disappears now, the Earth would move in a straight line <b>away from</b> the center of the galaxy, at a speed of 30 km/sec. Sol's constant acceleration of the Earth is responsible for this total reversal of velocity.

A drawing often helps. Draw two small circles about 15 cm apart. Label one MW and the other Sol. Draw one line through the centers of both. Draw a second line through the center of Sol, perpendicular to the first line. Draw a third circle centered on Sol with a radius of about 5 cm.

This larger circle should cross the other two lines a total of four times. At each crossing draw a small circle. Label the one closest to MW as E1, then lable the rest (in a clockwise sequence) as E2, E3 and E4. These represent the Earth at 3 month intervals as it orbits Sol, assuming a clockwise orbit for this drawing.

Draw a line from E4 toward MW, parallel to the line between MW and Sol. Make it about 6 or 7 cm long, and put an arrow head on the end farthest from E4. Label it Ve @ E4.

Draw a similar line from E1 (it should be at a right angle to the line between MW and Sol, and should not cross Ve @ E4). Label it Ve @ E1.

Continue this process for E2 and E3. None of these last four lines should cross. If any do, draw them in the other direction from their starting point.

The magnitude of the Earth's velocity (speed) at each point in its orbit is the same (30 km/sec). But the direction at each point in its orbit is different (90, 180, 270 and 0 degrees, using the line between MW and Sol as the angular reference).

The velocity at each point of the orbit is equal in speed but opposite in direction to the velocity "on the other side" of the orbit.

Draw this picture, study it, then ask some more questions.

[NOTE to MODERATOR] the ability to insert a drawing frame and add simple objects like lines and circles and labels would sure be nice.

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As for mass of a photon; it may be something more than zero as you seem to say so f=>0ma for one photon? Is this better or worse?(I'm quite a bit less neat than you are so if I have >0 where I should have <0 or other omissions and errors please forgive and if you wish correct them. thanks)
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I'm not sure what you mean by "f=>0ma".

The mass of all photons (or any quantity of them) is zero. Even if they really exist.

Regards,
LB

For my part there is no need for a drawing here-I get it. Just to be clear on the meaning of the terms being kicked around; acceleration is v/t according to prior posts and also your posts,right or not? I wonder if an orbiting body such as Earth has velocity perturbed by the central mass or sol as your model shows? I'll get to the other issues you commented on at another time.

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For my part there is no need for a drawing here-I get it. Just to be clear on the meaning of the terms being kicked around; acceleration is v/t according to prior posts and also your posts, right or not?
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Right. It would be better to think of it as change-in-velocity / change-in-time (a = dv/dt). The smaller the time interval involved the closer you get to instantaneous acceleration. The larger the time interval involved, the closer you get to average acceleration.
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I wonder if an orbiting body such as Earth has velocity perturbed by the central mass or [of?] sol as your model shows?
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I don't understand your question.

All reasonably close masses will accelerate, or perturb, each other by a noticeable amount. Sol accelerates Earth, so Sol perturbs Earth.

Regards,
LB

Larry is correct. a=v/t is a ballpark equation. It will only give you the average acceleration over the time period (t). As such, its use is limited. In the physics world we need to be more precise, therefore we specify which acceleration term we're addressing (instantaneous, average, centripetal, etc.).

Maybe central mass = center of mass of the earth-sun system?

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[Jim]acceleration is v/t according to prior posts and also your posts,right or not? I wonder if an orbiting body such as Earth has velocity perturbed by the central mass or sol as your model shows? I'll get to the other issues you commented on at another time.

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I'm sorry to say that Jim and JUU but your posts show you're lacking basic knowledge of the principles of kinematics and mechanics. This isn't intended as a personal attack but to serve as a suggestion to you; if you are interested in learning these concepts you must get formal training in a classroom even as a non-matriculated student. I commend Larry for trying to answer your questions but it's hard to convey the meaning of these concepts in message boards.

Anyway, regarding the question above, earth has a velocity and sol perturbs its motion in such a a way as for the changes in the velocity to be directed towards the center of forces, resulting in a centripetal acceleration, which in turn is the cause of the elliptical orbit. In the absence of sol, earth would move on a straight line until perturbed by another planet or star causing it to either orbit it or move in some other trajectory, a hyperbola or parabola, defined by the conic sections and depending on earth's total mechanical energy.

The formula a =v/t is valid only in the presence of a constant acceleration, such as in a free fall situation near the surface of earth. This is true becaue a = dv/dt and if a = constant,then by integration we get v = a x t + v0.

I knew if I posted enough times I would get someone telling me I needed to go study the basics. My question at the end of my last post was directed at Jim, trying to get clarification on his 'central mass' statement. I work in the space program, and have multiple degrees. My statements about accelerations and forces have been correct.