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Uniform Circular Motion vs Circular Orbit due to Gravity 
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#1
Aug1314, 09:32 AM

P: 10

In uniform circular motion, (eg, a mass on the end of a string moving in a horizontal circle) centripetal force is the only thing causing acceleration. we have the kinematic relationship V=RW
or velocity is proportional to radius. I.e a bigger radius means greater linear speed? For the motion of a satellite in earth orbit in uniform circular motion, we have centripetal force due to gravity alone. Equating the force of gravity to centripetal force we obtain V=square root of :(GM/R) i.e. speed is now INVERSELY proportional to orbital radius. I dont understand whats going on here. How can a centrally directed force, being the only thing causing acceleration, lead to two different relationships between V and R? 


#2
Aug1314, 09:56 AM

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There are two different things at work here. A gravitational orbit can only have one radius for a given tangential speed 
when Gm_{1}m_{2}/r^{2} = v^{2}/r. A mass on a string can have any speed and the tension will vary accordingly. For a mass on a string of length r, the centripetal force will be the same as the g force in an orbit of radius r. Nothing to worry about. You have two independent variables with string but just one with an orbit. 


#3
Aug1314, 10:28 AM

P: 10

But isnt tension constant in a horizontal circle? I dont get what the two variables are in the mass on a string case. I did the calculation you mention. Indeed, for a length of string equal to R we get the same results for velocity.
Also, for a given angular velocity (and hence period), V=RW describes both cases. But if you double R, in one case the speed increases and in the other it decreases. Which is my original problem. Put another way, if things in smaller radii orbit move faster in gravity, why do points on the earths equator move faster than those further in? I know its a pretend rigid body, but the force is gravitation in both cases, no? Sorry, Im still confused. 


#4
Aug1314, 11:30 AM

P: 683

Uniform Circular Motion vs Circular Orbit due to Gravity
On Earth however, if the gravitational force is too large for our "orbit" then we will still "orbit" because there is the normal force which balances the excess gravitational force. So we don't need to be moving at the speed that your orbitalequations would say, because we are no longer required to move at such a speed that ALL of the gravitational force causes centripetal acceleration. (Maybe I could've said it better, but I hope it makes sense?) 


#5
Aug1314, 11:43 AM

P: 10

Thanks guys, but I keep thinking I get it and then I end up getting stuck again.



#6
Aug1314, 11:45 AM

HW Helper
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W=square root of :(GM/R^3) 


#7
Aug1314, 11:59 AM

P: 683

In other words, if I increased the angular velocity, the tension would increase appropriately. However for orbits, if you increase the angular velocity, then the gravity does not increase (and so you lose your orbit) 


#8
Aug1314, 01:25 PM

Mentor
P: 11,997

Note that geostationary orbits are different from geosynch orbits. Geostationary orbits are where the satellite remains in the same spot in the sky relative to an observer on Earth. Geosynchronous orbits are orbits where the satellite's orbital period is equal to one day, regardless of where in the sky it happens to drift. You can have satellites pass over the poles of the Earth in a geosynchronous orbit. The orbital requirements for a satellite to be in a geostationary orbit are even more strict than a geosynch. There is only one plane and one radius the satellite can orbit in and still stay above the same location on the Earth. Changing the radius, speed, or inclination of the satellite would put it in a nongeostationary orbit. 


#9
Aug1414, 01:34 PM

P: 10

Yes, I do think I get it now. fundamentally, the general conclusion then is, if someone asks me what happens in uniform circular motion when the radius of the circle changes, the answer is it depends on the functional form of the force causing the motion, yes?



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