Calculating Orbit Distance from Equator w/ Particle in Bowl

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In summary, the conversation discusses the problem of determining the distance from the center of a frictionless bowl to the orbit of a particle moving at a certain velocity. The equations used to solve the problem involve trigonometric functions and constants such as gravity. The final solution is a quadratic equation with two solutions, one of which is incorrect due to the nature of the problem. The conversation also touches on how to determine which solution is correct without a calculator or taking a derivative.
  • #1
Päällikkö
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I suppose I could've used the "Fast particle in a bowl" thread, but as this is a different problem, I decided not to.

If you set a particle to move in a frictionless bowl (radius R) at the velocity v, how far is the orbit from the equator?

I got the following equations:
[tex]tan\alpha = \frac{v^2}{gr}[/tex]
[tex]r = Rsin\alpha[/tex]
[tex]d = Rcos\alpha[/tex]
where g is the constant 9,81[tex]\frac{m}{s^2}[/tex] and r the radius of the orbit.

I ended up with [tex]d = \frac{\frac{v^2}{gR}-\sqrt{(\frac{v^2}{gR})^2+4}}{-2}R[/tex], which seems a little complicated but gives reasonable answers. I haven't yet figured out why it only works with - in front of the square root (I got both plus and minus when I solved the equations). Is my solution correct?
 
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  • #2
I got
[tex]
d = \frac {-v^2 + \sqrt {v^4 + 4g^2r^2}}{2g}
[/tex]
I think that's the same as what you got.
By putting v=0, I get d=R. Which is a reasonable answer.

I got a quadratic in d. That's where the minus sign came from.
 
  • #3
siddharth said:
I got
[tex]
d = \frac {-v^2 + \sqrt {v^4 + 4g^2r^2}}{2g}
[/tex]
I think that's the same as what you got.
Yep, it is the same.
By putting v=0, I get d=R. Which is a reasonable answer.

I got a quadratic in d. That's where the minus sign came from.

I got a quadratic too, from which I get two solutions (the plus/minus-sign). I'm uncertain why the other solution is wrong.
 
  • #4
I took the value of d to be from the center of the bowl to the projection of the particle on the vertical axis.

This value must be positive.If I take one of the solutions of the quadratic, then the value of d will always be negative.

Because this is not possible (ie, the value of d must always be positive as the particle moves only in the bowl), I rejected the value and took the other value.
 
  • #5
siddharth said:
This value must be positive.If I take one of the solutions of the quadratic, then the value of d will always be negative.
The outcome cannot be predicted before the final solution(s), or can it?

Is taking a derivative the only "proper" way to figure out which solution is correct?
 
  • #6
'd' must be positive because the way I measure it, a negative value of 'd' would mean that the particle is above the hemi-spherical bowl (Where there is no bowl!).Thus, in the context of the question, this answer can be neglected.
 
  • #7
siddharth said:
'd' must be positive because the way I measure it, a negative value of 'd' would mean that the particle is above the hemi-spherical bowl (Where there is no bowl!).Thus, in the context of the question, this answer can be neglected.
Yes, I know it. But without a calculator or taking derivative, is there a way to figure out which of the solutions is right?
 

What is the equation used to calculate the orbit distance from the equator with a particle in a bowl?

The equation used is distance = radius of bowl * sine of angle of inclination. This equation assumes a circular bowl and a particle orbiting at a constant speed.

How do you determine the radius of the bowl?

The radius of the bowl can be measured using a ruler or measuring tape. If the bowl is not perfectly circular, you can take multiple measurements and calculate the average radius.

What is the angle of inclination?

The angle of inclination is the angle between the plane of the equator and the plane of the particle's orbit. It can be measured using a protractor or calculated using trigonometric functions and the particle's orbital parameters.

Can this equation be used for any type of bowl and particle?

No, this equation is only accurate for a circular bowl and a particle orbiting at a constant speed. If either of these conditions is not met, the equation will not accurately calculate the orbit distance from the equator.

What are some limitations of using this equation?

Some limitations include the assumption of a perfectly circular bowl and a constant orbital speed. It also does not account for other factors that may affect the particle's orbit, such as gravitational pull from other objects or air resistance. Additionally, the equation does not take into account the curvature of the Earth, which may be significant for large distances from the equator.

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