Understanding the Relationship between Gravity and Radial Acceleration

[UrbanXrisis]

Is the radial acceleartion of Earth at the equator the same thing as g (9.8)?

I read that
"Radial acceleration results from the action of the force generated by the string that pulls the ball toward the center of the circle. In the case of a satellite in orbit, the force causing the radial acceleration is Earth’s gravity pulling the satellite toward the center of the planet."

Does that mean gravity is the radial acceleartion?

Also, would someone please take a look at:
https://www.physicsforums.com/showthread.php?t=50614

 

[vsage]

The semantics of the question confuse me a little but I'm leaning toward no. You have to consider that the acceleration you feel at the equator is the sum of two forces: centripetal (or lack thereof in this case) and that due to gravity. These two forces are opposed so the "centrifugal" force takes a little away from the force of gravity

 

[UrbanXrisis]

Well, I still need to find the radial acceleration of the Earth at the equator

here's what I have so far:

the equation to find radial acceleartion is a=v^2/r

v=the velocity the Earth is traveling at, which I looked up was 1041mi/hr at the equator. This then converts to 17.35 mi/s and then 28 km/s.

Then for r, I looked up that the radius of the Earth is 6378km. Then...
a= [28km/s)^2]/6378km
a=0.1229 km/s^2
a=122.9 m/s^2


did I do this correct?

 

[vsage]

You may want to doublecheck your unit conversions. (1mph is about 0.447 m/s according to google)

 

[UrbanXrisis]

hmmm..I got the same answers

 

[vsage]

http://www.google.com/search?hl=en&lr=&q=1041+mph+in+m/s

 

[UrbanXrisis]

a=v^2/r
a= [465m/s)^2]/6378m
a= 34 m/s^2

like this?

 

[vsage]

Recheck your conversion of the radius of the Earth into meters.

 

[UrbanXrisis]

a=v^2/r
a= [465m/s)^2]/6378000m
a= .034 m/s^2

like this?

 

[vsage]

Looks good to me.

 

[UrbanXrisis]

So if radial acceleration is greater than the acceleration of gravity, we would fly off the earth?

 

[vsage]

That sounds true to me but I'm not sure if I would have immediately thought like that. I know someone will probably correct my good (but wrong) intentions but the radial acceleration is just the sum of the radial forces. An object requires a 0.034m/s^2 radial acceleration to be stationary on the Earth according to your calculations above, and since g is 9.8m/s^2, only about 9.766m/s^2 is your radial acceleration experienced at the equator because the ground is "giving way" at 0.34 m/s^2, if you will. I hope I explained that right.