Calculate Tangential & Normal Acceleration of Particle in Elliptical Orbit

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In an elliptical orbit with uniform speed, the tangential acceleration is zero because there is no change in tangential velocity. However, normal acceleration exists due to the continuous change in the direction of velocity, even though the speed remains constant. The discussion highlights that while tangential acceleration remains constant in magnitude, it does not imply that it is zero in all cases. For circular orbits, tangential acceleration is also zero since the radial distance does not change. Overall, both tangential and normal accelerations are important in understanding the motion of a particle in an elliptical orbit.
brasilr9
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A particle is moving in a elliptical orbit with uniform speed. How can I tell whether there are tangential and normal acceleration or not on the particle? (At A B and C )


thanks for help!
 

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You can show your work for a start.
 
I think I figure it out.
Since it's speed is constant, there's is no change in tangential velocity, hence tangential acceleration remain zero.
 
brasilr9 said:
Since it's speed is constant, there's is no change in tangential velocity, hence tangential acceleration remain zero.

That's correct. :smile: Now what about the normal acceleration?
 
siddharth said:
Is it? The direction of e_\phi continously changes with \phi. So, even if the speed is the same, the direction of velocity changes, doesn't it? So how can the tangential acceleration (ie, acceleration along e_\phi) be the same?

Ahh yes, I suppose constant magnitude would be an accurate term. Just re-reading through the question (and without looking at the picture obviously), I can't see the point. There is always going be tangental acceleration, and there also must always be normal acceleration, although this will change. :confused:
 
What I posted first wasn't exactly correct

What I mean is, if
\vec{r} = r \vec{e_r}

then according to the OP's question,
|\frac{d\vec{r}}{dt}| will be constant. So, for an ellipse, this doesn't mean that \frac{d^2\vec{r}}{dt^2} along e_\phi will be 0.

In fact, for a circular orbit, since \frac{dr}{dt} =0, the tangential acceleration will be 0.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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