Can you find orbital velocity from circle equation y^2+x^2=r^2?

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Discussion Overview

The discussion revolves around the calculation of orbital velocity using the equation of a circle, specifically focusing on the relationship between gravitational acceleration and centripetal acceleration. Participants explore the implications of falling under gravity while maintaining a circular path, and they attempt to derive the necessary horizontal distance to prevent falling.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to derive the horizontal distance an object must travel to maintain altitude while falling under gravity, using the circle equation r² = x² + y².
  • Another participant questions the assumption that an object would fall 9.8 meters in one second, prompting a reevaluation of the calculations related to average velocity and acceleration.
  • There is a discussion about the correct interpretation of the equations of motion, with one participant expressing confusion over the relationship between distance, time, and acceleration.
  • A later reply suggests that the average velocity during the first second of free fall should be considered, leading to a revised understanding of the distance fallen.
  • Participants express varying levels of mathematical background, with some indicating they are new to calculus and advanced algebra, which affects their understanding of the problem.
  • One participant acknowledges a mistake in their earlier reasoning after graphing the relationship between velocity and time, leading to a clearer understanding of the average velocity concept.

Areas of Agreement / Disagreement

Participants generally express uncertainty and confusion regarding the calculations and assumptions made. There is no consensus on the correct approach to derive orbital velocity from the given equations, and multiple viewpoints on the interpretation of motion under gravity remain contested.

Contextual Notes

Limitations include varying levels of mathematical understanding among participants, which affects their ability to follow and contribute to the discussion. Some assumptions about motion and acceleration are not fully resolved, and there are unresolved mathematical steps in the derivations presented.

Who May Find This Useful

This discussion may be of interest to individuals studying physics, particularly those exploring concepts of motion, gravity, and circular dynamics, as well as those seeking to understand the mathematical relationships involved in these topics.

victorhugo
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Maintaining R as the constant hypotenuse in the triangle formed by x and y coordinates in a 'perfect' circle,
r2=x2+y2
r2=x2+y2

So knowing that in 9.8 metres above ground it will take 1 second for an object to fall, I tried to find how many metres in the X direction an object must cover in 1 second before it falls 9.8 metres by

x2=r2 - y2
& y2 = r - 9.8
r = 6 371 000
thus x = 11 174.6 m/s ??

it's incorrect because if we use the formula derived from matching force from centripetal acceleration and gravitational acceleration at height r, it does not match:
( geometric proof shows that v-u / v = Dr & D=vt
plug and simplify, v-u=v2t/r, & a = v-u/t thus a= v2/r)

Since a = g = 9.8, then V = (sqrt) ar = (sqrt) 9.8 x 6 371 000 = 7901.63... m/s
 
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victorhugo said:
Maintaining R as the constant hypotenuse in the triangle formed by x and y coordinates in a 'perfect' circle,
r2=x2+y2
r2=x2+y2

So knowing that in 9.8 metres above ground it will take 1 second for an object to fall, I tried to find how many metres in the X direction an object must cover in 1 second before it falls 9.8 metres by

x2=r2 - y2
& y2 = r - 9.8
r = 6 371 000
thus x = 11 174.6 m/s ??

it's incorrect because if we use the formula derived from matching force from centripetal acceleration and gravitational acceleration at height r, it does not match:
( geometric proof shows that v-u / v = Dr & D=vt
plug and simplify, v-u=v2t/r, & a = v-u/t thus a= v2/r)

Since a = g = 9.8, then V = (sqrt) ar = (sqrt) 9.8 x 6 371 000 = 7901.63... m/s

It is quite difficult - at least to me, to follow your notation / equations. I'd recommend using latex in order to see clearly what you are asking. Also a sketch would be helpful.
 
QuantumQuest said:
It is quite difficult - at least to me, to follow your notation / equations. I'd recommend using latex in order to see clearly what you are asking. Also a sketch would be helpful.
Sorry I'll try make it simpler:For the changing X and Y coordinates on a planet (assuming it's perfectly spherical), (R= Radius of a circle) R2=x2 + y2

(Assuming gravity = 9.8 m/s/s) If you fall in the Y position by 9.8 metres, it will take 1 second.
What I tried to do is cover enough distance in the X direction such that it never reaches the ground.
Change in Y = R - 9.8. Let's assume R is that of Earth, 6 371 000 Metres

In order for the object to still be at radius R above the centre, then it must've covered an X distance of:
X= √R2 - (R - 9.8)2 = 11 174.6 m/s

But that's not the orbital speed. The orbit speed is:
a = v2 r

R= 6 371 000
a = 9.8
v= 7 901.63... m/s
 
What made you feel that it would fall 9.8 m in one second. with a vertical acceleration of 9.8 m/s2?
 
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Chandra Prayaga said:
What made you feel that it would fall 9.8 m in one second. with a vertical acceleration of 9.8 m/s2?

Not sure, i skipped the logical thinking and went to easy algebra thinking, I thought if i plugged in,
a= d / t / t
d = at^2
= 9.8...
Now that I think about it. at t=1 it has only just reached 9.8 m/s. thus, from t=0 -> t=1, the average velocity must be 9.8/2 = 4.9, thus in 1 second it has fallen 4.9 metres
so D = V at t=0 + V at t=1 / 2...
using this average velocity over the first second, then the equation works!
thank you :)

But why is a = d / t / t, = 1/at^2 and not at^2?
 
victorhugo said:
But why is a = d / t / t, = 1/at^2 and not at^2?

Sorry but this expression equates all sorts of things and I can really make no sense of it, in order to help. If you can state clearly what you ask it would be fine.
 
Let us start with what kind of background you have (high school, undergraduate, etc), and at what level we are going to work on this problem.
In your last post: d = at2 is not correct.
 
Chandra Prayaga said:
Let us start with what kind of background you have (high school, undergraduate, etc), and at what level we are going to work on this problem.
In your last post: d = at2 is not correct.
Only High School, and starting university in 2 weeks.

In the definition of acceleration, the change in velocity per unit time, is Δ(D/T) / T
If I simplify it, I think it should be: D/T / T/1 = D/T x 1/T = D/T2
I'm guessing I'm missing something to do with it being a change in velocity [ (D/T)1 - (D/T)2 / T ] ?

Oh yea I'm only starting to learn calculus now. Since I have no math background (Although I got a perfect mark in my maths class it was all basic things like world related things like reading graphs and financial problems) I'm still wrapping my head around doing trigonometry, geometry and more advanced algebra.
 
After drawing some graphs on paper, I found i was wrong.

In Velocity / Time graph of slope = 10 (therefore acceleration = 10), the distance traveled at
t=1, is the average velocity from t=0 to t=1 ( because at t=1 it has only JUST reached a velocity of 10m/s, gradually) is 0 + 10 / 2 = 5
t=2, = 10 + 20 / 2 = 15
t=3, = 20 + 30 / 2 = 25

so at t=3s, d= 45m, for an acceleration of a=10m/s/s

Now I'm working on simplifying this consistent pattern...
 
  • #10
Looks like you are getting there. Good for you.
 

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