# Why does gravity decrease at the equator ?

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1. Mar 30, 2015

### rohit199622

I know that radial force comes into the play but the direction of gravity (9.8) is also the same as radial acceleration so instead it should increase on equator

2. Mar 30, 2015

### SteamKing

Staff Emeritus
Should it? In which direction is the radial acceleration oriented with respect to the earth? Is this in the same direction as gravitational acceleration?

Remember, gravity accelerates objects toward the center of the earth.

3. Mar 30, 2015

### Staff: Mentor

4. Mar 30, 2015

### A.T.

5. Mar 30, 2015

### Staff: Mentor

The picture is the latter, although the accompanying text also describes the rotational effects.

6. Mar 30, 2015

### rohit199622

Okay so .. Whenever we rotate something like stone attached a thread or anything we give centripetal force so that it doesnt go away but in case of rotation due to earth there is no centripetal force being provided gravity is a different force all together this is how I understand it ..I got confused just because there was a formula thing in my text book to calculate gravity change at equator and is my explanation right ?

7. Mar 30, 2015

### Bandersnatch

Gravity is the only fundamental force at play here. It supplies the centripetal acceleration.

Take an object in a circular orbit for example. The centripetal force needed to move in circles is supplied by the force of gravity. There's nothing 'left' of the gravitational force to make the object move closer or farther than the radius of the orbit. The object is in free-fall (accelerometer would read 0).

On Earth's surface, since it rotates much slower than orbiting objects move, the force required to keep it in a circular orbit is much lower. All the excess force of gravity left over is what keeps pushing you into the ground, and is what accelerometers measure as $g$.

8. Mar 30, 2015

### Delta²

To make it crystal clear can someone give the formula of the radial acceleration? I know the formula for the gravitational acceleration is $g=G\frac{M_{earth}}{(R_{earth}+h)^2}$

9. Mar 30, 2015

### SteamKing

Staff Emeritus
In the case of a stone being swung on a thread, "we" don't give it anything.

Remember, according to Newton's laws of motion, once a body is put into motion, it tends to remain in motion, unless acted upon by an external force. Swinging the stone on a thread, the stone wants to keep traveling in a straight line tangent to the circle. This causes a tension to develop in the thread, which is why it appears that the stone wants to pull away from the center of rotation.

It's those "formula things" you should pay attention to. There is a centripetal force applied to objects on the equator, but this force is relatively small. It's the same centripetal force which develops when swinging a stone on a thread.

10. Mar 30, 2015

### SteamKing

Staff Emeritus
This formula is for calculating the value of g at altitude h above the surface of the earth.

In circular motion, the tangential velocity v = r ω, where r is the radius of the circular path of motion and ω is the angular velocity.

The centripetal acceleration α = v2 / r = r ω2

For the earth at the equator, r = 6 371 000 meters and the period of rotation = 24 hr = 86 400 sec.

Therefore, ω = 2 π radians / 86 400 sec = 7.272 × 10-5 radians / sec

The centripetal acceleration at the equator = r ω2 = 6 371 000 * [7.272 × 10-5]2 = 0.034 m/s2

Since the value of g is approximately 9.81 m/s2, one can see that the rotation of the earth introduces a small change in the value of g.

Last edited: Mar 30, 2015
11. Mar 30, 2015

### Delta²

Just to clarify it a bit more, this small change is a phenomenical change in g, its not because rotation produces some sort of anti gravitational field (just saying now), its because thiss 0.034m/s^2 goes as centripetal acceleration and what is left 9,81-0,034 is what we feel as the reaction force from the surface of the earth. is that correct?

12. Mar 30, 2015

### rohit199622

Thanks so much I get it now

13. Mar 30, 2015

### SteamKing

Staff Emeritus
It's not due to a reaction from the earth; it's due to our bodies (or whatever) wanting to continue traveling in a straight line as the earth turns beneath our feet.

It's something like being in an elevator which starts to go down; you feel momentarily lighter as the elevator car starts to accelerate downward. Once the car has stopped accelerating and has reached a constant velocity, then the only acceleration which you perceive is that due to gravity.

14. Mar 30, 2015

### A.T.

That is correct in the non-rotating frame.

In the co-rotating frame you have an inertial centrifugal force, due to rotation of the frame, not of the Earth. But I wouldn't call that an "anti gravitational field", because it acts away from the axis, not away from the center. So it is opposite to gravity only on the equator.

15. Mar 30, 2015

### Tom_K

You said the evil word!

Part of the gravitational acceleration is needed to keep everything rotating with the surface of the earth, rather than travelling in a straight line, as they would according to Newton’s first law of motion.

http://www.regentsprep.org/regents/physics/phys06/bcentrif/casette.gif

If the gravitational acceleration is 9.83 m/s^2 at a certain point on the earth’s surface, and 0.03 m/s^2 is needed to keep things moving with the rotating surface, then 9.8 m/S^2 is left to provide the acceleration factor in an object’s weight. So objects at the equator weigh a bit less, everything else being equal.

16. Mar 30, 2015

### A.T.

In the inertial frame where everything is rotating. Not in the co-rotating frame where everything is static

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