# Gravity and acceleration

1. Oct 30, 2004

### UrbanXrisis

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

"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:

2. Oct 30, 2004

### 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

Last edited by a moderator: Oct 30, 2004
3. Oct 30, 2004

### 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?

4. Oct 30, 2004

### vsage

Last edited by a moderator: Oct 30, 2004
5. Oct 30, 2004

### UrbanXrisis

6. Oct 30, 2004

### vsage

7. Oct 30, 2004

### UrbanXrisis

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

like this?

8. Oct 31, 2004

### vsage

9. Oct 31, 2004

### UrbanXrisis

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

like this?

10. Oct 31, 2004

### vsage

Looks good to me.

11. Oct 31, 2004

### UrbanXrisis

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

12. Oct 31, 2004

### 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.

Last edited by a moderator: Oct 31, 2004