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Gravity Around the Globe


We all know that the constant "g" refers to 9.8 m/s^2, the acceleration of objects due to the field of gravity on Earth. But what if I were to tell you that "g" may not be as constant as you thought it was?

Some important values: 

MEarth = 5.972 x 10^24 kg
REarth = 6.371 x 10^6 m
hEverest = 8,848 m
hMariana = -10,994 m

And the equation that will be used: 

g= GME / R^2

where R is the radius of the earth plus the height of each location: (RE+ h)

Now let's find the value of "g" atop the summit of Mount Everest, which is 8,848 meters above sea level.

Image result for mount everest
Fun fact (if you didn't already know it): Mount Everest is not the tallest mountain on earth! You heard me correctly, the tallest mountain is actually Mauna Kea in Hawaii. From base to summit, Mauna Kea is about 10,200 meters high, a whole 3,352 meters taller than Everest! However, since Mauna Kea is partially underwater, and its peak only rises to 4,205 meters above sea level, Mount Everest often takes credit as the highest mountain on Earth.

g= GME / (RE + hEverest)^2
g= (6.67 x 10^-11)(5.972 x 10^24) / (6.371 x 10^6 + 8,848)^2
g= 9.78645 m/s^2

So the value of "g" at the top of Mount Everest is actually about 0.014 m/s^2 smaller than the 9.8 m/s^2 that we all know and love. This makes sense though, because the peak of Everest increases the radius, so "g" becomes weaker. 

Next, let's find "g" at Challenger Deep in the Mariana Trench, the deepest known part of the Earth's crust. It's 10,994 meters below sea level. 

Image result for mariana trench challenger deep

g= GME / (RE + hMariana)^2
g= (6.67 x 10^-11)(5.972 x 10^24) / (6.371 x 10^6 - 10,994)^2
g= 9.8476 m/s^2 

At Challenger Deep, the difference in "g" is a whopping 0.04 m/s^2. Once again, this seems logical because is it closer to the earth's core, so the radius decreases, and "g" should go up. 

The force of gravity can also be affected by locations on the surface of the earth, not just the distance from the center of the earth. This is due to the centrifugal force created by the earth's rotation. The earth rotates around its axis through the North and South Poles, so the poles hardly spin while the equator spins much quicker. Since centrifugal force = mv^2 / r, if v increases (like at the equator), the force will increase, and if v decreases (like at the poles), so will the force. Centrifugal force goes outward from the earth's surface, opposing the force of gravity that draws inward to the earth's core. Therefore, the centrifugal force cancels out some of the force of gravity. At the equator, there is more centrifugal force, so there is less gravitational force. At the poles where there is less centrifugal force, there is more gravitational force. 

Another factor of gravity is the earth's shape. Believe it or not, Earth is not spherical, but flat. 

Image result for flat earth

I'm joking. Sorry to all the flat-earthers out there... and everyone reading this really. 

No, the earth is not perfectly spherical (or flat), but it is in the shape of an oblate spheroid. 


The equator actually bulges out a little due to centrifugal force. The radius of the earth at the equator is about 6,378 km, while at the poles the radius is about 6,257 km. As we saw earlier, when the radius increases, the force of gravity decreases, and vice versa. At the equator, the force of gravity lessens, while at the poles, the force of gravity increases.

Taking into account both the rotation and shape of the earth, at the equator, "g" is 9.78 m/s^2 and at the poles it's 9.83 m/s^2.  

So there you go! While a value of 9.8 m/s^2 may be approximately constant around the globe, it's not always true. 

http://curious.astro.cornell.edu/about-us/42-our-solar-system/the-earth/gravity/94-does-your-weight-change-between-the-poles-and-the-equator-intermediate'

http://wtamu.edu/~cbaird/sq/2014/01/07/do-i-weigh-less-on-the-equator-than-at-the-north-pole/


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