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

• victorhugo
In summary: D/T / T/1 is the change in velocity per unit time, and it is always positive. D/T x 1/T is the change in velocity divided by the time it takes to reach that point, and it is always positive. D/T2 is the change in velocity multiplied by the time it takes to reach that point, and it is always positive.
victorhugo
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

Universeer
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?

victorhugo
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...

Looks like you are getting there. Good for you.

## What is orbital velocity?

Orbital velocity is the velocity at which an object must travel in order to maintain a stable orbit around another object, such as a planet or a star.

## How is orbital velocity related to the circle equation y^2+x^2=r^2?

The circle equation y^2+x^2=r^2 is used to calculate the distance of an object from the center of its orbit. This distance, along with the mass of the object it is orbiting and the gravitational constant, can be used to calculate the orbital velocity using the formula v = √(GM/r), where G is the gravitational constant and M is the mass of the object being orbited.

## Can orbital velocity be calculated using the circle equation if the mass of the object being orbited is unknown?

No, in order to calculate orbital velocity using the circle equation, the mass of the object being orbited must be known. If the mass is unknown, other methods such as measuring the period of the orbit or observing the object's trajectory can be used to calculate orbital velocity.

## Is orbital velocity constant throughout an orbit?

No, orbital velocity is not constant throughout an orbit. As the object moves closer to the object it is orbiting, it will accelerate and its orbital velocity will increase. As it moves further away, it will decelerate and its orbital velocity will decrease.

## How is orbital velocity affected by the mass of the object being orbited?

Orbital velocity is directly proportional to the mass of the object being orbited. This means that as the mass of the object being orbited increases, the orbital velocity also increases.

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