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Q & A: How gravitational force varies at different locations on Earth

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Most recent answer: 11/21/2016
Q:
I'm sure all countries do not have the same amount of gravitational force present because of it's relative position to the core or equator, so for my research I would like to know a few questions? What is the specific acceleration of gravity force acting upon the United States? And, what is the specific acceleration of gravity force acting upon Argentina? How is the specific acceleration of these gravitational forces calculated relative to the effect it has on the people who reside there?
- Lemuel W. (age 18)
mississippi, united states
A:

The variation in apparent gravitational acceleration (g) at different locations on Earth is caused by two things (as you implied). First, the Earth is not a perfect sphere—it's slightly flattened at the poles and bulges out near the equator, so points near the equator are farther from the center of mass. The distance between the centers of mass of two objects affects the gravitational force between them, so the force of gravity on an object is smaller at the equator compared to the poles. This effect alone causes the gravitational acceleration to be about 0.18% less at the equator than at the poles.

Second, the rotation of the Earth causes an apparent centrifugal force which points away from the axis of rotation, and this force can reduce the apparent gravitational force (although it doesn't actually affect the attraction between two masses). The centrifugal force points directly opposite the gravitational force at the equator, and is zero at the poles. Together, the centrifugal effect and the center of mass distance reduce g by about 0.53% at the equator compared to the poles.

You can use the following equation to calculate at a certain latitude, accounting for both of these effects:

g=g_{45}-{\tfrac {1}{2}}(g_{\mathrm {poles} }-g_{\mathrm {equator} })\cos \left(2\,lat\,{\frac {\pi }{180}}\right)

  • g_{\mathrm {poles} } = 9.832 m/s2
  • g_{{45}} = 9.806 m/ s2
  • g_{\mathrm {equator} } = 9.780 m/s2
  • lat = latitude, between −90 and 90 degrees

You can use this to find the apparent value of g at a location in the United States or in Argentina. (To get the gravitational force, also called weight, multiply g by the mass of the object you're interested in. Be sure to use consistent units.)

That equation assumes you're at sea level, but if you want to account for the effect of altitude when you go up in a plane you can use this additional equation:

g_{h}=g_{0}\left({\frac  {r_{{\mathrm  {e}}}}{r_{{\mathrm  {e}}}+h}}\right)^{2}

  • re is the Earth's mean radius (6,371.0088 km)
  • g0 is the standard gravitational acceleration (9.80665 m/s2)

The effect of changes in altitude due to actual elevation of the land is more complicated, because in addition to raising you farther from the center of the Earth the land also provides an additional source of gravity. Whether the local g goes up or down with surface altitude depends on how dense the Earth's crust is in that area.

After you calculate the difference in local gravitational acceleration between the United States and Argentina, let us know if you think of any ways it might affect the people living there.

Rebecca H.


(published on 11/21/2016)

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