Derivation of orbital velocity formula

[Amr Elsayed]

Is there some mathematical derivation for calculating the orbital velocity based on altitude and acceleration without using calculus? I thought of equations of motion, but I always get problems.
Is there a way to derive it using laws of gravitational potential and kinetic energy ?

Are the pieces of information here in the video correct ? I guess it has mistaken with the position of the satellite if not affected with gravity.

 

[Dr. Courtney]

If you assume the orbit is a circle, you can do it without Calculus.

Set the force you need (for centripetal acceleration) equal to the force you have (Newton's law of gravitation).

The orbital radius is the Earth's radius plus the altitude.

Simplify.

 

[Amr Elsayed]

et the force you need (for centripetal acceleration) equal to the force you have (Newton's law of gravitation).

But I need to use : F centripetal = M*V^2 / r .. Which needs calculus to be derived. I want to understand it without calculus if possible

 

[DocZaius]

But I need to use : F centripetal = M*V^2 / r .. Which needs calculus to be derived.

That isn't true. There are a few ways to show that ##a_{centripetal}=\frac{V^2}{R}## without calculus.

 

[A.T.]

But I need to use : F centripetal = M*V^2 / r .. Which needs calculus to be derived. I want to understand it without calculus if possible

https://en.wikipedia.org/wiki/Centripetal_force#Derivation_using_vectors

 

[Amr Elsayed]

That isn't true. There are a few ways to show that acentripetal=V2Ra_{centripetal}=\frac{V^2}{R} without calculus

Can you please show me any of them using basic algebra or geometry ?

https://en.wikipedia.org/wiki/Centripetal_force#Derivation_using_vectors

I can't find derivations here except calculus derivation :/

 

[dean barry]

negligible mass satellite assumed
for stable orbit: gravitational acceleration = centripetal acceleration
so:
( G * M ) / r^2 = v^2 / r

 

[RaulTheUCSCSlug]

You actually just need very little calculus, but you can instead think of it as the limit approaching to zero rather than the derivative and that should help you.

 

[nasu]

Actually even the notion of instantaneous speed is based on calculus. So if you use instantaneous speed in the formula for acceleration you "use calculus".
Defining it as a limit rather than a derivative is still calculus. The notion of limit is one of the basic notions in calculus. And the derivative itself is such a limit.

So technically, you cannot do kinematics at all without calculus. You may have the illusion that you do without. :)
After all, Newton had to invent it before he can study motion.

 

[dean barry]

A handy bit of info:
In a stable orbit two body situation, each body has equal momentum.
Have a look at this two body data sheet also.

 

[ChrisVer]

Can you please show me any of them using basic algebra or geometry ?

Even if you want to use geometry, you will have to use it in a way that allows for calculus (i.e. allow for derivatives) ...otherwise words like "Acceleration" or "velocity" won't make any sense...
As for basic algebra, I don't know what that would mean... like addition, multiplication etc?