This is a common physics exercise:

Suppose the earth is a sphere of radius $6370$ km. If a person stood on a scale at the north pole and observed the scale reading (his weight) to be $mg$, what would the scale read if he stood on it at a point on the equator?

I decided to try to solve it myself, first with intuition, and then a diagram.

My intuition tells me that weight (as seen on a scale) is the normal force $N$. On the North Pole, $N = W$, where $W$ is the weight. However, on the Equator, there is a centrifugal force. Like on a merry-go-round, one is pull away from the Earth. That means $N$ is "lightened" as such: $N = W - C$. From there, simple calculations are made to get a numerical answer. According to other sources, this is correct.

However, I tried solving this "without ever being on a merry-go-round". I draw a diagram. Draw the earth as a circle, with a person as a point. To the center, there is $W$, because $W$ is in the direction of the mass. In addition, centripetal force ($C$) is pointed to the center of a spinning object (i.e. same way as $W$). Opposing both is $N$. So, I get $N = W + C$. This is different than the above equation.

Where do I go wrong in my second solution?

  • $\begingroup$ Comment to the post (v2): It should probably be stressed that the assumption that the shape of the spinning Earth is a perfect sphere (rather than, say, an oblate ellipsoid) is unrealistic, cf. e.g. this and this Phys.SE posts. $\endgroup$
    – Qmechanic
    Jan 26, 2017 at 17:12

3 Answers 3


The problem you are having is that you are thinking of "the centripetal force" as a force that you have to account for in drawing your diagrams. Not entirely you fault. The phrase does read that way.

Basic rule: there is no thing called "the centripetal force". Instead that phase is a label for some combination of real force to take on the role of causing the inward acceleration that must be present for curving motion to occur. What combination of forces varies from problem to problem and even from time to time in a single situation. What that combination is is something that you have to discover in each problem.

But you do have one hint. The simple kinematics of the motion tell you what the magnitude of acceleration must be.

In this case $F_c = |\text{gravity times mass}| - |\text{normal reaction force}|$. This is a specific case of the general rule that centripetal force is the sum of the inward pointing forces (or components of forces) minus the sum of the outward pointing ones.1

Secondly, you need to know that a bathroom scale measures the normal force between the person and the ground. This gives us a different understanding of "weight" than $mg$ (and it is one that works right when we say a astronaut in orbit is "weightless").

So, $$\text{weight} = \text{normal reaction force} = mg - m\frac{v^2}{r}\,,$$ is the expected reading on the scale at the equator.2

To expand a little on the "why" of this approach, I want to start by talking about equilibrium.

Once you have identified that some object is in equilibrium then you know something about the forces acting on it in aggregate: they add up to zero. That doesn't tell you about any single a priori, but if you know all the other force (or know that there is only one force) then it lets you finish.

The "centripetal force" is very much like that: because the object is in curving motion you know that it is not in equilibrium but accelerating, and furthermore you know that the component of its acceleration that points to the (instantaneous) center of curvature is $a_c v^2/r = r\omega^2$.

And that tells you that the total of all the radial forces acting on it must be $ma_c$ (and again, this is something you know about a collection of forces). But before you can use this knowledge to your advantage you must identify which forces (or components thereof) should be combined before being set to that total.

So an early step in working problems that involve "centripetal force" is asking "Which of these force point toward the center and which point toward the outside?" The former enter into the sum with a positive sign and the later with a negative sign.

1 You should note that because these are the only forces acting on the person at the equator and they do not cancel out because that person is not in equilibrium. But you knew that: they are moving in a circle, right?

2 Except that we have forgotten to account for the Earth's non-spherical mass distribution. But that is another story.

  • 1
    $\begingroup$ Of course there's such a thing called centripetal force. Perhaps you mean to imply that it's not an actual force exerted on one object by another but is instead the net force acting along a radius to keep a body in circular motion. It can be calculated and is quite useful in problem solving. I suspect you know this, suggest you re-word. $\endgroup$
    – bpedit
    Jan 26, 2017 at 2:33
  • $\begingroup$ @bpedit I've been using the framework in this answer for several semester in intro classes; it works. The point is that which actual forces contribute to the centripetal force (and with what sign even) varies from case to case. Of course there is some combination of forces that keep the thing moving in a curve, but without considerable case students can easily think that 'the centripetal force' is on the same footing as the gravitational force, the normal force and the frictional force. It's not, and we should take care to explain that. $\endgroup$ Jan 26, 2017 at 5:10
  • $\begingroup$ The OP here got into trouble for exactly that reason. He wrote down the gravitational force with the right sign. He recognized that the normal would point outward and would adapt to the other forces (well done!). Then he wrote down an additional force pointing to the center rather than recognizing that it represented the net of the previous two forces. I've seen the exact confusion repeatedly and this is my formula for heading it off at the pass. $\endgroup$ Jan 26, 2017 at 5:12
  • $\begingroup$ Agreed it can be confusing for novice students. I taught Physics for 24 years. But when you state there in no such thing and then proceed to give a formula for it, I think there's room for semantic improvement. Just sayin'. $\endgroup$
    – bpedit
    Jan 26, 2017 at 5:55

At the North Pole you are not moving in a circle, so you are not accelerating. The net force on you is zero : $W-N=0$ where $W$ is the gravitational pull on you and $N$ is the normal reaction provided by the ground. (Actually you are travelling with the Earth in its orbit around the Sun, so you are accelerating towards the Sun. But we will ignore that acceleration, which is about $0.006 m/s^2$ - relatively small.)

At the Equator you are circling around the centre of the Earth, so you are accelerating. (This is about $0.03 m/s^2$.) The net force on you is non-zero and is called the centripetal force $C$. This is not a force in itself, it is just the name given to the amount of force required to keep an object moving at a given speed $v$ in a circle of given radius $r$ where $C=mv^2/r$. So here we have $W-N=C$.


On the North Pole, $N=W$, where $W$ is the weight.

This is where you went wrong, from the start. Weight and the normal force are vectors, not scalars. For an object at rest with respect to the rotating Earth at the North Pole, $\vec N + \vec W = 0, or $\vec N = -\vec W$, from the perspective of an Earth-centered inertial frame of reference or from the perspective of an Earth-centered, Earth-fixed frame of reference.

At the equator, the Earth-centered inertial frame of reference says that $\vec N + \vec W = \vec F_{\text{net}} = - m r\Omega^2 \, \hat r$, where $\hat r$ is the unit vector pointing away from the center of the Earth at the location of the object being weighed. The Earth-centered, Earth-fixed frame of reference says that the net force is zero if one accounts for the centrifugal force: $\vec N + \vec W + \vec C = 0$, where $\vec C = m r \Omega^2 \,\hat r$.

Note that in both cases we get $\vec N + \vec W = - m r\Omega^2 \, \hat r$. The two perspectives are consistent with one another.

Another way to look at it, a force diagram at the equator:

   Inertial perspective               Rotating perspective
   ------> N                          ------> N
   <--------- W                       <--------- W
                                      -->        C
   <--        Non-zero vector sum     .          Vector sum is zero

For a body at rest with respect to the rotating Earth, the non-inertial centrifugal force and the inertial net force add to zero. Although it would be correct to say "equal but opposite", I intentionally did not say that because that phrase carries some Newton's third law baggage. The fictitious centrifugal force and the real net force are not third law pairs.


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