Gravity at Earth's poles vs equator

[michelcolman]

I just saw a StarTalk video where Neil deGrasse Tyson says that gravity at the poles is lower than at the equator because, even though you are closer to the center, the mass around the equator outside the sphere below you somehow does not count (skip to 4:21 in the video). He is obviously wrong there, no doubt confused with the fact that when you're inside a uniform sphere of matter, the forces from the outer shell offset each other so they can be disregarded. This is not the case at the poles because there's no mass above you to offset the mass around the equator.

However, I then started to wonder about a different effect: even though you are closer to the center of the earth when you're at the poles, much of the mass is pulling with a wider angle so the vertical components are less. At the equator, the mass is further away but at a smaller angle. So what does this work out to? Could it be that the total gravity is higher at the equator and this perfectly offsets the centrifugal (pseudo)force? It might make sense because water tries to take a shape that minimizes potential energy...

(I'm just talking about a theoretical homogenous earth, I know there are local differences due to fluctuations in density)

 

[Baluncore]

However, I then started to wonder about a different effect: even though you are closer to the center of the earth when you're at the poles, much of the mass is pulling with a wider angle so the vertical components are less.

The spherical shell of mass above, and further from the centre cancels.
As you move towards the centre of the Earth, the inverse square law increases gravity, while the volume of the internal sphere falls in proportion to the cube of the radius. The cube beats the square, with gravity falling steadily to zero at the centre.

Go to a pole. Place your gravimeter on the ground and take a reading. Dig a pit next to the gravimeter, throwing the spoil into a pile on the other side of the gravimeter. Now read the gravimeter again and it will read less. The pit is not pulling it down, while the spoil heap is pulling it up.

Now place the gravimeter in the pit. The nearby pit-deep shell above, is pulling it up. The opposite pole is not pulling it down. That is not compensated for by being very slightly closer to the centre of the Earth.

 

[michelcolman]

I don't think you understood my question. I know that, when you're inside a homogenous spherical mass, the outer shell exerts no net gravity on you. I did the full calculation during my engineering studies, and it all cancels out so only the sphere from the center up to your position has a net influence.

However, when you are at the pole, there is no mass above you and therefore Neil was wrong when he asserted that only the sphere up to your distance counted for the amount of gravity you feel. The bulge at the equator does exert a net force because there's no mass above you to compensate for it. I have no questions about that part, I'm very certain Neil was wrong there.

What I was wondering about, is the effect of the angles of the vectors involved on an oblate earth. At the pole you are closer to the center, but the mass near the equator is at a greater angle with your vertical direction and therefore has a smaller vertical component. At the equator, you are further from the center but all the gravitational vectors are closer to your vertical direction (vertical as experienced by you standing on the equator), which increases their total component.

So I am asking, if you take all of this into account, which place feels most gravity, with or without centrifugal (pseudo)force.

1. Gravity is higher at the poles

2. Gravity is the same at the poles as on the equator, but centrifugal (pseudo)force makes you feel lighter at the equator

3. Gravity is less at the poles, but this is perfectly compensated for by centrifugal (pseudo)force

4. Something in between or outside 1-3.

I would kind of expect 3, since fluids tend to minimize potential energy. But I don't know if anyone has worked it out exactly, sources on the internet seem to contradict each other.

 

[PeroK]

The empirical answer is that it does not vary much across the surface of the Earth. Note that the gravitational field strength is the gradient of the potential. So, even if the Earth's surface is an equipotential, the acceleration may vary.

https://en.m.wikipedia.org/wiki/Gravity_of_Earth

 

[michelcolman]

The empirical answer is that it does not vary much across the surface of the Earth. Note that the gravitational field strength is the gradient of the potential. So, even if the Earth's surface is an equipotential, the acceleration may vary.

https://en.m.wikipedia.org/wiki/Gravity_of_Earth

From that wikipedia article I gather that gravity is stronger at the poles. So both Neil's explanation and the final result are wrong.

 

[Dullard]

1.

The 'oblateness' slightly reduces how much higher.

 

[michelcolman]

1.

The 'oblateness' slightly reduces how much higher.

So it is higher at the poles, but not quite as much higher as you would expect from simply plugging the smaller radius into Newton's formula of gravity. Got it, thanks!

 

[Janus]

The spherical shell of mass above, and further from the centre cancels.
As you move towards the centre of the Earth, the inverse square law increases gravity, while the volume of the internal sphere falls in proportion to the cube of the radius. The cube beats the square, with gravity falling steadily to zero at the centre.

Go to a pole. Place your gravimeter on the ground and take a reading. Dig a pit next to the gravimeter, throwing the spoil into a pile on the other side of the gravimeter. Now read the gravimeter again and it will read less. The pit is not pulling it down, while the spoil heap is pulling it up.

Now place the gravimeter in the pit. The nearby pit-deep shell above, is pulling it up. The opposite pole is not pulling it down. That is not compensated for by being very slightly closer to the centre of the Earth.

The steady decrease in g only works for a uniformly dense object, since this is not the case for the Earth, which which is denser at the core than the crust, what you actually get is an increase in g for a bit before it starts to decrease as shown here.

 

[Janus]

There is something else he said in this video that, while technically true, is misleading. He stated that a day on Venus is longer than its year. While the period of it sidereal rotation is longer than it's orbital period period, so its sidereal day is longer than it year, this rotation is retrograde, which in turn, results in the solar day, or the time takes the Sun to go from meridian to meridian, being 116.75( Earth)days, or a bit over 1/2 of its year. Since it is the day/night cycle that we associate as being a "day". This is likely what people would think this is what he meant by "day".

 

[michelcolman]

Wow, for such a highly regarded astrophysicist he sure makes a lot of mistakes. This is not the first time I caught him saying falsehoods on his podcasts:

- In another episode he said that the speed of earth is exactly right for it to stay in its orbit and that, if the mass of the sun were to be suddenly doubled somehow, the earth would spiral into the sun. False of course, the earth would simply take a new elliptical orbit with its apohelion near its position at that moment and its perihelion on the other side closer to the sun. Astrophysics 101, I would think...

- And another time, he explained how airplane wings work using a combination of two of the three "incorrect theories of flight" described on the nasa grc website ("equal transit time" and "air molecules hitting the bottom of the wing", both of which are false although admittedly the former one is incredibly popular so I can't really blame him too much for that).

Anyway, this has been interesting, thanks for the explanations!
Michel

 

[Sagittarius A-Star]

GR time dilation compensates - due to Earth flattening (due to rotation of the earth) - SR time dilation of a clock at the equator at sea level compared to a clock at a pole (with respect to the ECI frame) very accurately. So proper time on geoid is very accurately the same at the poles and at the equator. The difference is of the order $O (1/c^4)$.

Source (in German, see chapter 14.3 - I used Firefox):
https://books.google.de/books?id=SEckBAAAQBAJ&lpg=PA168&ots=EcmAK7b1UI&dq=einstein zeitdilatation nordpol äquator&hl=de&pg=PA171#v=onepage&q=einstein zeitdilatation nordpol äquator&f=false

 

[michelcolman]

Wow, I didn't expect that.

 

[A.T.]

Wow, I didn't expect that.

Gravitational time dilation depends on the difference in potential. In its rotating rest frame the surface tends towards an equipotential surface, with respect to the effective potential (graviational + centrifugal).