# Homework Help: 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 Earthâ€™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