Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Would the effect of gravity on me change if I were to dig a very deep hole and stand in it? If so, how would it change? Am I more likely to be pulled downwards, or pulled towards the edges of the hole? If there would be no change, why not?

share|improve this question
add comment

5 Answers

up vote 22 down vote accepted

The other answers provide a first-order approximation, assuming uniform density (though Adam Zalcman's does allude to deviations from linearity). (Summary: All the mass farther away from the center cancels out, and gravity decreases linearly with depth from 1 g at the surface to zero at the center.)

But in fact, the Earth's core is substantially more dense than the outer layers (mantle and crust), and gravity actually increases a bit as you descend, reaching a maximum at the boundary between the outer core and the lower mantle. Within the core, it rapidly drops to zero as you approach the center, where the planet's entire mass is exerting a gravitational pull from all directions.

This Wikipedia article goes into the details, including this graph.

And there are other, smaller, effects as well. The Earth's rotation results in a smaller effective gravity near the equator, the equatorial bulge that results from that rotation also has a small effect, and mass concentrations have local effects.

share|improve this answer
add comment

Assuming spherically symmetric mass distribution within Earth, one can compute gravitational field inside the planet using Gauss' law for gravity. One consequence of the law is that while computing the gravitational field at a distance r < R (with R being the radius of the Earth), one can ignore all the mass outside the radius r from the center

\begin{equation} \oint_{S_r} g_r \cdot dA = -G \int_{B_r} \rho dV \end{equation}

where gr is the gravitational field at distance r from Earth's center, ρ is Earth's density, Sr is the sphere of radius r centered on Earth's center of mass and Br is the volume enclosed by Sr. Assuming that ρ only depends on the distance r from the center of the Earth, we can simplify this as follows

\begin{equation} \oint_{S_r} g_r \cdot dA = -4\pi G \int_0^r \rho(s) ds \end{equation}

\begin{equation} g_r = -\frac{G}{r^2} \int_0^r \rho(s) ds \end{equation}

Setting Mr to denote the portion of Earth's mass enclosed within Sr, we can rewrite the last formula as

\begin{equation} g_r = -\frac{GM_r}{r^2} \end{equation}

Now, letting ρr denote the average density of the portion of the Earth enclosed within Sr, we have

\begin{equation} g_r = -\frac{4 \pi G \rho_r r}{3} \end{equation}

The conclusion is that the gravity inside Earth depends roughly linearly on the distance from the center of the planet and density variations account for the deviations from linearity.

An interesting way to visualize this is to think of an over 12,700 kms long elevator from Hamilton, New Zealand to Cordoba, Spain. During the travel (which at average speed of 200km/h would take almost three days) passengers would feel gradual and roughly linear decrease in weight and in the middle of the journey would experience weightlessness followed by gradual increase in weight as they near the surface on the other side. Also, around the midpoint of the journey the floor and the ceiling would swap.

share|improve this answer
add comment

Acceleration due to gravity at depth d below the earth's surface is given by:

$g(d) = G M_e \dfrac{R_e - d}{R_e^3}$

Where,

G = Universal gravitational constant
Me = Mass of the earth
Re = Radius of the earth
d = depth below the earth's surface

enter image description here

share|improve this answer
add comment

This is a really interesting problem which have had a first solution from ancient greeks (I heard Carlo Rovelli to recount this). Indeed, when you dig a hole in the Earth, gravity starts to become proportional to distance and, if you will dig in the way to open up a hole to the other side of the Earth, a mass will oscillate in the well you created.

share|improve this answer
add comment

Actually it is not true about a mass oscillating if dropped down a hole dug right through the Earth, because the Earth is rotating. It would only work in one case where the hole were placed directly through the axis of rotation. Then there is the issue of increasing atmospheric pressure with depth, which would increase drag and lower the object's terminal velocity to quite un-spectacular levels, and rapidly damp out the oscillating motion. Rather than the science-fiction 'pop out the other side' result, the object would slow to a near stop on its first passage to the centre, overshoot by only a few hundred metres and reverse direction for a few increasingly feeble movements until viscous forces dominate and it floats around aimlessly in convection.

share|improve this answer
add comment

protected by Qmechanic Feb 22 '13 at 13:39

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.