How much would gravity be 10km below surface, compared to what is at sea level?


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    $\begingroup$ Well, that depends on the size and density of the object in question. I assume you're talking about Earth? $\endgroup$ – F1Krazy Jun 5 '17 at 15:06
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    $\begingroup$ This is answered straightforwardly in the en.wikipedia.org/wiki/Gravity_of_Earth --> Depth section. $\endgroup$ – adonies Jun 5 '17 at 15:53
  • $\begingroup$ If I could just read it! Wikipedia is blocked in where I live. $\endgroup$ – UKS Jun 5 '17 at 15:54
  • $\begingroup$ Maybe you could check this: study.com/academy/lesson/… and use @Ross Millikan's suggestion in answer to compute the new value. $\endgroup$ – adonies Jun 5 '17 at 16:27

In a uniform sphere, the gravitational acceleration increases linearly from the center to the surface, so for earth it would be 636/637 of sea level. The fact that the earth is more dense near the center makes the change even less, so for most purposes you can ignore it. The next level of approximation would be to compute the mass inside of your radius (so delete the top 10 km all around the earth) and compute the gravity from that mass and the radius of 6360 km.


Assuming you are talking about the Earth, a 10km change in altitude is negligible. This graph shows the calculated free fall acceleration at different distances from the center of the earth using 3 different density models. Note that 10km is well below the level of precision of the graph.


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