Why do we weigh less at the equator?

Click For Summary
SUMMARY

The discussion centers on the phenomenon of weight variation at the equator versus the poles, specifically addressing the role of centrifugal force and non-inertial reference frames. Participants clarify that at the equator, individuals experience a fictitious centrifugal force due to the Earth's rotation, which reduces the normal force measured by scales. The equations governing these forces are established, with emphasis on the gravitational force and the fictitious force in non-inertial frames. Understanding these concepts is crucial for grasping why weight differs based on geographic location.

PREREQUISITES
  • Understanding of gravitational force and its effects
  • Basic knowledge of non-inertial reference frames
  • Familiarity with centrifugal and centripetal forces
  • Ability to interpret physics equations related to motion
NEXT STEPS
  • Study the principles of non-inertial reference frames in physics
  • Learn about the effects of centrifugal force in rotating systems
  • Explore gravitational variations across different geographic locations
  • Investigate the mathematical derivation of forces in circular motion
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of motion and forces, particularly in relation to Earth's rotation and gravitational effects.

bigsaucy
Messages
38
Reaction score
0
Hello all,

I was just bugged by the fact that you weigh more at either of the poles than you do at the equator.

I reason that because at the equator you feel the Earth spinning about it's axis, you would feel an additional centripetal force toward the centre of the Earth ontop of the force of gravity. Whereas at the poles, there is no additional centripetal force acting on you, therefore you should feel lighter, clearly this isn't the case.

Why is this so? Please Help!
 
Physics news on Phys.org
Yes you are almost right. Except when you are at the equator you will feel an additional centrifugal force away from the center of the earth, not an additional centripetal force. You may not have run into non-inertial reference frames in a physics course yet. When you are standing at a point on the equator, you are actually centripetally accelerated. You are actually in a non-inertial reference frame. The procedure for solving problems in a non-inertial reference frame is to add a fictitious force equal to your mass times your acceleration in the opposite direction as the acceleration. So at the pole you will have

F_g - F_N = 0

in the non-inertial reference frame at the equator you have

F_g - m \frac{ v^2 }{r} - F_N = 0

So the weight that a scale will measure is

F_N = F_g - m \frac{v^2}{r}

And actually it is really not necessary to use non-inertial reference frame. You can simply say that you are accelerated at the equator and you have

F_g - F_N = ma_c = m \frac{v^2}{r}

and you get the same exact equation. But yes you had the right idea. But if there was an extra force directed toward the center of the earth, you would actually weigh more.
 
I have learned about non-inertial reference frames however I don't quite understand it completely and I feel that that is what is hindering my understanding of this 'conundrum'. Could you please explain it further?

Thanks

My main concern is that how can there be a a centrifugal force without the normal force exerting it?
 
bigsaucy said:
I have learned about non-inertial reference frames however I don't quite understand it completely and I feel that that is what is hindering my understanding of this 'conundrum'. Could you please explain it further?

Thanks

My main concern is that how can there be a a centrifugal force without the normal force exerting it?

It is not a real force. It is a fictitious one. It is similar to the following. Consider yourself in an elevator. You cannot see outside, there is no windows. You feel your feet pressed against the floor. This sensation that you feel seems to be a gravitational force, like you are accustomed to on earth. However, you also think of another possibility. Perhaps the elevator is accelerating upwards at 9.8 m/s^2, and you are in deep space away from any massive body. The two situations would have exactly the same effect on you and in fact they are equivalent. So being in an accelerated reference frame causes you to feel this fictitious force. This is why we can add this fictitious force equal to -ma in a non inertial reference frame.
 
Last edited:
thank you, I'm starting to understand now, much appreciated!
I'm new to this forum so if there is something that I'm supposed to do to show my appreciation like subscribing or a thumbs up, let me know.
 
bigsaucy said:
thank you, I'm starting to understand now, much appreciated!
I'm new to this forum so if there is something that I'm supposed to do to show my appreciation like subscribing or a thumbs up, let me know.

Becoming a PF contributor is always much appreciated :-p Plus you get to have a super cool avatar and even a signature at the end of your posts!
 
AlexChandler said:
It is not a real force. It is a fictitious one. It is similar to the following. Consider yourself in an elevator. You cannot see outside, there is no windows. You feel your feet pressed against the floor. This sensation that you feel seems to be a gravitational force, like you are accustomed to on earth. However, you also think of another possibility. Perhaps the elevator is accelerating upwards at 9.8 m/s^2, and you are in deep space away from any massive body. The two situations would have exactly the same effect on you and in fact they are equivalent. So being in an accelerated reference frame causes you to feel this fictitious force. This is why we can add this fictitious force equal to -ma in a non inertial reference frame.

I wonder how you can use Einstein's elevator and still claim it's a fictitious force. I refer to centrifugal force, of course; centripetal force is definitely fictitious.
 
AC130Nav said:
I wonder how you can use Einstein's elevator and still claim it's a fictitious force. I refer to centrifugal force, of course; centripetal force is definitely fictitious.

I'm not claiming that it is fictitious. It is a fictitious force by the definition of "fictitious force". "Fictitious force" is simply what we call apparent forces that act on masses in a non-inertial reference frame.

http://en.wikipedia.org/wiki/Fictitious_force

And the centripetal force in this case is definitely not a fictitious force, it is simply the gravitational force...
 
AlexChandler said:
I'm not claiming that it is fictitious. "Fictitious force" is simply what we call apparent forces that act on masses in a non-inertial reference frame.

http://en.wikipedia.org/wiki/Fictitious_force

And the centripetal force in this case is definitely not a fictitious force, it is simply the gravitational force...

Centripetal "force" in all cases is simply a greater gravity well (which is admitedly a force) or the resistance of deformation as in the case of a centrifuge or a sci-fi space station. Whoever invented the term (I won't bother to look it up), invited misunderstandings such as this one.
 
Last edited:

Similar threads

Replies
17
Views
2K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 8 ·
Replies
8
Views
12K
Replies
6
Views
6K
Replies
2
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
3
Views
3K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K