Hi Everyone,
In a uniform circular motion, the acceleration and net force is towards the center, but why does the object dont go to the center if it is accelerating in that direction?
thanks,
Ron

 

The force is towards the center because the object moving tends to escape from its circular orbit due to its inertia. This force towards the center is the force which keep the object rotating.

 

The Zen answer is -- it is all about the balance of forces in the universe.

 

This subject was discussed at great length in this thread

http://forum.allaboutcircuits.com/showthread.php?t=23728&page=6

 

Hi Everyone,
In a uniform circular motion, the acceleration and net force is towards the center, but why does the object dont go to the center if it is accelerating in that direction?
thanks,
Ron

This is precisely analogous to energy storage in an CL circuit. The acceleration force (EMF) is always at right angles to the current flow (ELI the ICE man) so no energy is dissipated. It's all imaginary!

(Actually, I think imaginary is an unfortunate term, but it's ingrained in our math)

Eric

 

This is precisely analogous to energy storage in an CL circuit. The acceleration force (EMF) is always at right angles to the current flow (ELI the ICE man) so no energy is dissipated. It's all imaginary!

(Actually, I think imaginary is an unfortunate term, but it's ingrained in our math)

Eric

It came from pondering the solution to the algebraic equation

x^2 + 1 = 0

which has no solution in the domain of real numbers. What would you call it instead, assuming you knew nothing about electrical engineering?

 

It came from pondering the solution to the algebraic equation

x^2 + 1 = 0

which has no solution in the domain of real numbers. What would you call it instead, assuming you knew nothing about electrical engineering?

Probably nothing, since I didn't understand imaginary numbers at all until I started playing around with R.F.!

 

It came from pondering the solution to the algebraic equation

x^2 + 1 = 0

which has no solution in the domain of real numbers. What would you call it instead, assuming you knew nothing about electrical engineering?

However, I have given the matter some thought over the years. The problem I have with "imaginary" is the subtle implication that such numbers have less of a physical implcation than "real" numbers, which we know is far from the truth. The first time I got knocked on my keister playing with an ignition coil, the "reality" of this "imaginary" reactance was driven home quite soundly!


Maybe our collective brainpower on AAC might come up with a better term than Imaginary! We could all become famous!

Eric

 

However, I have given the matter some thought over the years. The problem I have with "imaginary" is the subtle implication that such numbers have less of a physical implcation than "real" numbers, which we know is far from the truth. The first time I got knocked on my keister playing with an ignition coil, the "reality" of this "imaginary" reactance was driven home quite soundly!


Maybe our collective brainpower on AAC might come up with a better term than Imaginary! We could all become famous!

Eric

Again with the "legends in our own minds" theme. Ahh...cha...cha...cha
Another thread successfully jacked!

 

Hi Everyone,
In a uniform circular motion, the acceleration and net force is towards the center, but why does the object dont go to the center if it is accelerating in that direction?
thanks,
Ron

An interesting application of this phenomenon is its ability to permit the calculation of the approximate altitude above the earth that a satellite is likely to be in order to remain in geo-stationary orbit.

Two equations are involved.

The first equation is the attraction force between two bodies seperated by a distance r.

\(F = \frac{Gm_1m_2}{r^2}\)

The second equation is the angular momentum of the satellite.

\(F=m_2r\omega^2\)

Where:

m1 = mass of the earth
m2 = mass of the satellite

Setting these two equations equal to each other and solving for r, we get the expression:

\(m_2r\omega^2 = \frac{Gm_1m_2}{r^2}\)

\(r^3 = \frac{Gm_1m_2}{m_2\omega^2}\)

The mass of the satellite cancels out to form the expression:

\(r^3 = \frac{Gm_1}{\omega^2}\)

\(r= \sqrt[3]{\frac{Gm_1}{\omega^2}}\)

The value of G (the gravitational constant) is:

\(G=6.67E10^{-11}\)

The mass of the earth is:

\(m_1=5.98E10^{24}\)

The value of ω-squared:

\(\omega^2=\left(\frac{2\pi}{24*60*60}\right)^2\)

Plugging all the values into the equation we get:

\(r=\sqrt[3]{754E10^22}\)

\(r = 42250 km\)

Subtracting the radius of the earth (6371 Km) yields the approximate altitude:

altitude = 35879 Km

 

An interesting application of this phenomenon is its ability to permit the calculation of the approximate altitude above the earth that a satellite is likely to be in order to remain in geo-stationary orbit.

Interestingly, this assumes the Earth to be a point source of gravity....a pinhead with the mass of the Earth. Actually LOCAL perterbations of gravity can have significant effects, creating "bumps" in the normal elliptical orbital path. When I worked for the phone company, we had to do a small amount of satellite tracking....I'm glad someone ELSE had to figure out the bumpy data for us!

Eric

 

Interestingly, this assumes the Earth to be a point source of gravity....a pinhead with the mass of the Earth. Actually LOCAL perterbations of gravity can have significant effects, creating "bumps" in the normal elliptical orbital path. When I worked for the phone company, we had to do a small amount of satellite tracking....I'm glad someone ELSE had to figure out the bumpy data for us!

Eric

You're right about the variation in gravity. I was mainly interested in showing that the altitude could be estimated using the principle of circular motion.

hgmjr

 

Here is a simple treatise on the OP's original question from a simple place:

http://webpages.ursinus.edu/lriley/courses/p111/labs/node5.html