What is the relationship between gravitation, centripetal and centrifugal force on the Earth?

Question

I'm trying to analyze a situation wherein a ship is moving across the surface of the earth. I am trying to analyze this situation in a reference frame that is rotating with the earth (NED frame).

I am confused about the relationship between gravitational, centripetal, and centrifugal forces acting on the boat. Gravitation is simple enough and would exist even if the earth was not spinning.

My question is, the earth is spinning, and we learned in physics class that an object undergoing circular motion experiences a center seeking (centripetal) acceleration. Wouldn't this therefore increase the force experienced by the boat beyond just the mass attraction between the boat and planet (gravitation)?

When I search online though, I really only find mention of the centrifugal force reducing the effect of the gravitational force, most notably on the equator where the distance from the rotational axis is maximized. Why isn't centripetal force included in the calculations?

Answer 1

When confused, ditch centrifugal.

Let us go back to the basics. The original Newton's 2nd Law (N2L) states that $$ \begin{align} \tag1\sum\vec F&=\frac{\mathrm d\vec p}{\mathrm dt}\\ \tag2\sum\vec F&=m\vec a \end {align} $$ The latter version, more common, but only holding if the mass is held constant, and the motion is non-relativistic.

However, teachers tend to forget to emphasise that there is an ideological split involved in here. On the left, are meant to be bona-fide, tangible, forces, causes, whereas on the right are effects, observations.

In particular, the Earth pulls on the boat with the gravitational force, abbreviated as the weight of the boat, and that is a real force. For now, you should ignore that Einstein showed later that this is fictitious.

There is another force, the bouyant force, that props the ship up from sinking into the ocean. It looks a bit like the normal reaction force if the ship were on land.

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Now, we know that the Earth is spinning. That means, for a boat that is simply sitting still in the sea, it is actually orbiting the Earth's rotational axis. This is an observed effect. To do this, the resultant acceleration must be the centripetal acceleration. This appears on the right hand side of the N2L.

There is no single force responsible for the centripetal acceleration.

Instead, the real forces must conspire to produce this observed centripetal acceleration, by cancelling insufficiently in just the right amounts to make this work.

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Consider a bob going around a horizontal circle when pulled by a string. The only horizontal force acting on the bob is the tension in the string. There is no "tension in the string + centripetal acceleration". There is only "the tension in the string supplied the required centripetal acceleration."

Answer 2

The expression 'centrifugal force' can in most circumstances be understood as: 'a centripetal force is required'.

In the case of the spinning Earth:
At the equator the measured gravitational acceleration is a slightly lower value than the measured gravitational acceleration at the poles.

At the Equator: 9.7805 $m/s^2$
At the poles: 9.8322 $m/s^2$

About one third of the difference is due to the fact that an object located on the Equator is about 21 kilometers further away from the Earth's geometric center than an object at (either one of) the poles. (Due to its rotation the Earth has an equatorial bulge.)

But the main part of the difference is as follows: to be co-rotating with the Earth at the Equator requires a centripetal acceleration. Providing that centripetal acceleration goes at the expense of the gravitational acceleration.

The acceleration that is required to circumnavigate the Earth's axis along the equator, at one revolution per sidereal day, is 0.0339 $m/s^2$.

From that we can infer that the newtonian gravitational acceleration at the Equator must be 9.8144 $m/s^2$

We can only infer that value, due to the equivalence of inertial and gravitational mass. Because of that equivalence some things are not accessible to measurement. A single measurement cannot inform you whether some of the gravitational acceleration is expended in providing required centripetal acceleration.

Throughout the history of mechanics the equivalence of inertial mass and gravitaional mass has been among the most important factors.

Historically:
Newton had formed the hypothesis of an inverse square law of gravity. In terms of Newton's law of Universal Gravity: the gravitational acceleration experienced by a planet is proportional to the gravitational mass of that planet. So: does that mean that Newton had to know the mass of each planet? Newton had the following insight: if inertial mass and gravitational mass are equivalent then in order to verify the concept of a universal inverse square law of gravity it is not necessary to know the mass of each planet.

The amount of required centripetal force for a planet to remain in orbit is proportional to the inertial mass. If gravitational mass and inertial mass are equivalent then the mass drops out of the calculation.

Returning to measurement:

A very interesting physical effect that arises from the Earth's rotation is called the Eötvös effect, after the Hungarian scientist Loránd Eötvös.

Eötvös developed instruments for gravimetry, his instrument were very sensitive.

Quoting from the wikipedia article:

In the early 1900s, a German team from the Geodetic Institute of Potsdam carried out gravity measurements on moving ships in the Atlantic, Indian, and Pacific oceans. While studying their results, the Hungarian nobleman and physicist Baron Roland von Eötvös (Loránd Eötvös) noticed that the readings were lower when the boat moved eastwards, higher when it moved westward.

When you have a velocity with respect to the Earth (such as the velocity of a ship), in eastward direction, you are circumnavigating the Earth at an angular velocity slightly larger than the angular velocity of the Earth itself.

That slightly larger angular velocity has a correspondingly larger requirement for centripetal acceleration. That is what Eötvös recognized in the measurement data.

The required centripetal acceleration goes at the expense of gravitational effect.

I noticed that in another answer the theory of General Relativity was mentioned in a byline.

Obviously General Relativity is outside the scope of this question, but let me make a few remarks.

General Relativity frames the equivalence of inertial mass and gravitational mass at a whole different level.

Still: some things from the newtonian theory carry over to GR.
There is the notion of a gravitationally bound system. There is a hierarchy of nested levels.
The Earth-Moon system is a gravitationally bound system
Next level up: the Solar system
Next level up: our Galaxy; the stars of our Galaxy are all orbiting the center of mass of the Galaxy.

For the planets of the solar system: Retrograde motion is apparent motion.

Incidentally: when the circumference of the Earth is known the angular velocity of the Earth can be inferred by capitalizing on the earlier mentioned Eötvös effect.

On a non-rotating planet there is no Eötvös effect: two ships, moving in opposite direction (with equal velocity wrt the surface), will measure the same gravitational acceleration.

On a rotating planet: the difference in measured value between moving in eastward direction and moving in westward direction is proportional to the angular velocity of the planet. Obtain a measurement of that difference, combining with known circumference allows you to infer the rotation rate of the planet.

Answer 3

Rotating reference frames can be tricky. Some of the trickery is that we typically only use them when it simplifies the math. This can make it tricky to separate the effects of the frame from the nature of the problems that lead us to prefer that frame in the first place.

The most useful thing I've found in these situations is to recognize that we can always reframe a problem into any arbitrary rotating frame we like. We can figure out the physics of your ship in a frame centered on the earth and rotating at 100 revolutions per minute. The results may involve some pretty strange numbers, but they will be right. We could also compute things in a frame centered on the ship and spinning sideways at 20 rpm. Now these aren't really useful rotating frames. They don't really offer any insight. The frame you are thinking of, rotating fixed with the earth, is much more useful. But its good to remember that at the highest level, you any rotating frame is valid.

This is useful because of the centrifugal acceleration term that appears in a rotating frame. We know that objects move in a straight line (absent of forces) in an inertial frame. In a rotating frame, they don't move in a "straight line." They move along a curve because the rotating frame is moving underneath the object. In an inertial frame, we say $F=ma$, but in a rotating frame, we have to alter the equation to account for this rotation of the frame. We end up with $F=m(a_r-\omega r^2)$, where $r$ is the distance to the axis of rotation, and I subscripted $a_r$ just to make clear its a different value than the $a$ in the previous equation. That's just a mathematical derivation from the definition of how our rotating coordinates move with respect to the inertial ones. The $\omega r^2$ term is a centrifugal acceleration - center fleeing. If an object follows a straight line in an inertial frame, it is effectively fleeing the axis of rotation.

Often we rephrase this acceleration in a force-like way. If I rearrange the equation to get $F+m\omega r^2=ma_r$ we get something that looks like a force. This is a "centrifugal pseudoforce." I'm pedantic about calling it a "pseudoforce" not a "force" because it isn't a force. There's no equal and opposite reaction for the centrifugal pseudoforce. But it looks like a force and, in many cases, we an intuit about it as if it were a force. And we do. But always remember that it's a pseudoforce (or an acceleration).

So now let's bring in the particular rotating frame you are interested in. You are looking at ECEF (Earth Centered, Earth Fixed) or NED (North East Down), both of which are rotating frames. You correctly intuited that this is a useful frame. Your ship's velocity is a sane vector. The velocity of the oceans is 0 in these frames. This makes a lot of things simple, which is good.

So we've covered centrifugal acceleration (or pseudoforce), and you've indicated that gravity is straight forward. The last one is centripetal force. This one's tricky because "centripetal force" isn't so much a force as it is a class of forces which have the property of pointing towards the center of rotation (centripetal is center seeking). But now that we understand that the rotating frames are distinct from the motion of the objects, we can make sense of it. There is a centrifugal acceleration in this rotating frame. Thus, if there is no force acting on the ship, it would "flee" the center of rotation. But we know that isn't true. We know its velocity in the Down direction is 0. So we know there must be some force acting on the ship to keep its downward velocity at 0.

In your case, in the Down direction, you have two forces. Gravity is pulling down. Buoyancy is pushing up. The sum total of gravity plus buoyancy must be non-zero in order for the ship to keep its downward velocity of 0. In particular, the sum must exactly equal $m\omega r^2$ in the center-seeking (centripetal) direction. If it summed to anything else, then the object would have a non-zero downward velocity and would be moving up or down.

Answer 4

In this case, the centripetal force is the gravitational force. "Centripetal" and "gravitational" refer to two different aspects of the force:

• Gravitational refers to the source of the force, i.e. the force comes from the gravitational interaction between two bodies.
• Centripetal refers to the direction of the force; a centripetal force is any force directed towards the axis that a circularly moving object revolves about.

In general, every force has both a direction and a source. Depending on the situation, any given force could be either gravitational, centripetal, both, or neither. For a ball on a string swinging around in a circle, the source is the tension in the string and the direction is centripetal. For an object sitting motionless on a sloped surface, there are three forces: their sources are gravitational, normal, and frictional; and their respective directions are downward, perpendicular to the surface, and parallel to the surface. (Somewhat confusingly, we can use "normal" to refer to both the source and direction of a force!)

But the key thing to remember is that "centripetal" never refers to the source of the force itself. For an object accelerated linearly downward by gravity, you wouldn't consider both the gravitational force and the downward force as two different forces! In the same way, an object in circular orbit doesn't have separate gravitational and centripetal forces - they are one and the same.