# Why doesn't Wolfram Alpha show low gravitational acceleration for the Hudson Bay?

I thought I would be clever and add a new answer to Hudson Bay Has Low Gravity? by using Wolfram Alpha to report gravitational acceleration for different locations, but it looks like my cleverness backfired.

From this answer I found this view, a subsection of which is cropped and shown below.

I chose three points, the lowest, purple area near the West shore of the Hudson bay, a medium green-orange in lake Michigan near Chicago, and a high area just north of Iceland.

I was surprised to see Wolfram report such a high gravitational acceleration for the Hudson Bay area at 9.85 m/s^2! I was expecting something substantially lower than 9.81 m/s^2

Question(s):

1. Why is the gravitational acceleration for the west coast of the Hudson Bay so high from Wolfram Alpha?
2. Roughly what acceleration should I be getting there?
3. Could it be related to Wolfram Alpha's use of EMG2008 12th order for a model?
4. Why would it be reporting such large deviations in altitude for locations on water?

Here is the data:

                              lat   lon    Total   vert dev  down     west    south    elev
Western Hudson Bay    'low'  60.9N 94.1W  9.85176  0.00302  9.85172  0.00967  0.02816   -22
Lake Mich. (~Chicago) 'med'  41.7N 87.3W  9.8188   0.00352  9.81874  0.01119  0.03274  +175
North of Iceland      'high' 66.6N 18.8W  9.86107  0.00259  9.86104  0.00845  0.02406  -445


Wolfram Alpha links and raw data screen captures:

• I have a feeling that this question would most likely be sent to WA, not us. – Gimelist Apr 22 at 10:30
• @Gimelist I think individual users should speak for themselves. I also have a strong hunch the answer lies within item 3; EMG2008 12th order. The question is tagged with gravity and models and so I think gravity model should be on-topic here. Let's give it a few days and see what happens? – uhoh Apr 22 at 11:12
• I agree that this is off topic, and have flagged it as such. For the future, I think it's also helpful to limit each post to a single question. – g.z. Apr 30 at 1:23
• @g.z. If it was a random site I can understand, but Wolfram Alpha is a distinctly reputable site and has probably implemented EMG2008 12th order properly, which means that the question is about gravity models of Earth which is on-topic. I've been pretty good at keeping my ~1,900 Stack Exchange questions narrowly focused. The four indexed items are so closely related that in this particular case it would be unwieldily to ask them as four separate questions, linking each to the other three. – uhoh Apr 30 at 2:29
• @uhoh: I just retracted the close vote. I've been playing with WA & obtaining g for various land based sites with Chicago's latitude (N & S): 41.7N & 47E, 125E; 41.7S & 72W, 145E, 173E. Chicago has an elev of 175 m (your WA data). In Chile, 41.7S, 72W the elev is 200 m & the values for g are very similar: 9.8188 v 9.81877. Also 41.7S, 145E, elev 144 m has a g of 9.81932. WA might be using a formula based on a whole world "line of best fit analysis" derived from EMG2008 12th order. – Fred Apr 30 at 14:39

Wolfram Alpha reports values between 9.78 and 9.88 m/s^2 depending on the location, which is odd because it should stay within the range 9.76 - 9.83 all over the world. It also seems to be using the ocean floor as a reference elevation if the point falls in water.

Using Natural Resources Canada's tool, it reports a gravity of 981933.3 mGal (9.819333 m/s^2) for Hudson Bay at 60.9N, 94.1W at sea level, which seems more likely for that location.

Keep in mind that while the area near Hudson Bay has a negative gravity anomaly, this does not imply that it should have weaker gravity than Chicago (9.8026 m/s^2), which is much closer to the equator. It means that Hudson Bay's gravity is lower than what it should be at that latitude. Gravity is even weaker near the equator (around 9.78) because as you get closer to the equator, the centrifugal force is stronger and slightly counteracts the force of gravity, also the mass of the Earth is slightly further away from you, thus has less gravitational pull on you. These 2 effects alone create gravity variations up to 0.5% (around 0.05 m/s^2)

Gravity anomalies are relative to a theoretical normal gravity based on an idealized ellipsoidal rotating Earth, and these anomalies are small compared to the variation caused by cetrifugal forces and the oblateness of the Earth. Anomalies range between -50 and +50 mGal (+-0.0005 m/s^2). Normal gravity at the Equator for sea level is about 9.78 m/s^2, and near the poles, around 9.83 m/s^2. Altitude also has a significant effect, at 5000 meters, gravity is about 0.01 m/s^2 weaker than at sea level.

• +1 This is a great answer! Thank you for looking into this and sorting out what's going on here, and for taking time to explain further about gravitational anomalies. – uhoh Jun 27 at 0:55

Short answer because there was no ice cap there.

Why does the hudson bay have lower gravity, isostatic rebound, there used to be an icecap right there, several miles of ice. That pushed the continental crust further down in to the mantle, like a boat with a heavy load. Mantle is denser than continental crust. When the icecap melted the rock started to rebound but it takes a LONG time to return to equilibrium. Mantle is not a liquid is does not rebound quickly. Basically the weight of the ice displaced a portion of the mantle and much of that mantle is still displaced. until it completely rebounds there is less mass there than in places with no such displacement. We can even measure how fast it is rebounding, several millimeters per year.

As a side note that is also why there is an ocean on top of continental crust right there, the land is still depressed below sea level. It should take the area about ~1,000,000 year to rebound but the glaciers only meted about 11,000 years ago. So in about 989,000 years there will no longer be a gravitation depression there because the rock will have returned to equilibrium.

But why does the western shore not have it, because there was no glacier there.