If you could travel to the center of the Earth (or any planet), would you be weightless there?

  • $\begingroup$ Assuming we're ignoring gravity due to other bodies in the planetary system...? $\endgroup$ Jan 4, 2011 at 8:39
  • $\begingroup$ @codeMonk it would only matter if you consider the 2nd order tidal kind of effects from those bodies. $\endgroup$ Aug 1, 2011 at 17:19
  • $\begingroup$ This question and my answer are closely related, though I don't think this question is an exact duplicate. $\endgroup$ Sep 12, 2012 at 0:57

5 Answers 5


Correct. If you split the earth up into spherical shells, then the gravity from the shells "above" you cancels out, and you only feel the shells "below" you. When you are in the middle there is nothing "below" you.

Refrence from Wikipedia Gauss & Shell Theorem.

{I am using some simplistic terms, but I don't want to break out surface integrals and radial flux equations}

Edit: Although the inside of the shell will have zero gravity classically, it will also have non zero gravity relativistically. At the perfect center the forces may balance out, yielding an unstable solution, meaning that a small perturbation in position will result in forces that exaggerate this perturbation.

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    $\begingroup$ @Noldorin: actually you feel the gravity of all of the Earth's mass at a radius less than your own (otherwise you'd start floating every time you step into a basement). If I remember correctly the gravitational acceleration inside a solid sphere of uniform density is proportional to $r$. $\endgroup$
    – David Z
    Jan 3, 2011 at 19:17
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    $\begingroup$ @mbq - the equilibrium will be stable not unstable. The forces drop to zero linearly from the surface to the center of mass of the Earth. There won't be a strong restoring force for small displacements but there will be a force that increases linearly towards 1g at the surface. What about this makes you say that it is "very unstable?" To me that implies that any small deviation will result in a force away from the center which is the opposite of what would happen. $\endgroup$
    – inflector
    Jan 3, 2011 at 21:50
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    $\begingroup$ @mbq: it's not an unstable equilibrium at all. Quite on the contrary it serves as almost perfect harmonic oscillator. This can be best seen by digging a tunnel through the Earth and putting a ball into it. $\endgroup$
    – Marek
    Jan 3, 2011 at 23:12
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    $\begingroup$ @JustJeff: sure it seems like that would be the case, but if you go through and do the calculation you find that it's actually not. Assuming the shell is perfectly spherical, you won't feel any gravitational force inside it even if you are close to the inner surface. Basically the reason is that the force from the small bit of mass directly above you is balanced out by the force from the large remainder of the mass below you. For details, look at the Wikipedia articles linked in jalexiou's answer. (The one on the shell theorem is probably a little easier to follow) $\endgroup$
    – David Z
    Jan 4, 2011 at 0:27
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    $\begingroup$ Everybody quit deleting comments! It makes the thread very hard to follow. $\endgroup$
    – Colin K
    Feb 28, 2011 at 14:39

The simplest way to think about it is that there is mass all around you in the center of the Earth so you get an equal gravitational "pull" from all directions. The pulls cancel out so you get no acceleration.

If one assumes constant density for the Earth (which isn't strictly speaking true but it is close enough for this illustration) the gravitational acceleration drops linearly from 1g at the surface to 0 at the center of the Earth. So you'd get a zero if you stepped on a scale at the center of the Earth.

The more complicated explanation is that acceleration due to gravity is the derivative of the gravitational potential. This potential is a minimum at the center of the Earth and grows quadratically up to the surface. It then continues to increase at a lower rate. Since at the exact center is flat (like the bottom of a valley), the derivative which is a measure of the rate of change is zero, and there is no acceleration.

Interestingly, even though you would be weightless there, the effects of gravity are highest at the center of the Earth. You get more gravitational time dilation, for example, than you do at the surface.

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    $\begingroup$ I think this answer adds significantly to jalexiou's because it points out that a simple symmetry argument leads to weightless at the center of Earth. Therefore, the specifics of the law of gravity (its inverse-square-ness) are not really important to this question. $\endgroup$ Jan 4, 2011 at 1:02
  • $\begingroup$ Also since the Earth-Moon revolves around a point 4000km from the centre of the earth you would be weightless in a donut around the centre $\endgroup$ Sep 13, 2012 at 14:51
  • $\begingroup$ I'd like you to elaborate on your claim of increased time dilation at the center. $\endgroup$
    – Fernando
    Oct 26, 2012 at 12:56
  • $\begingroup$ @Fernando The rate of your proper time relative to a distant observer is about $1 + V/c^2$ where $V$ is the gravitational potential (which is negative). Compared to someone far from the Earth (but at the same distance from the Sun), someone at the surface of the Earth would lose about two seconds per century, and someone at the centre, about three. $\endgroup$ Mar 14, 2013 at 20:19

I like answers that appeal to symmetry, so I answer this one with a question: If you were at the center, which way would you fall? That tells us you could stay floating there.

  • $\begingroup$ -1: this argument forgets the fact that earth is spinning and rotating around the sun. $\endgroup$
    – Shing
    Oct 20, 2018 at 18:07

In the following the term "charge" refers either to mass or to electric charge and the term "Inverse Square Law" refers either to Newton's Gravitational Law or to Coulomb's Law, respectively.


A. The Inverse Square Law for Spheres with uniform surface charge density

Proposition A: Let a sphere of radius $\:\rm{R}\:$ with uniform surface charge density $\:\rho_{s}\:$ and empty interior. Then:

(a1) the force exerted upon a point charge $\:\xi\:$ in the interior or on the surface of the sphere, as in Fig. 01, is zero (cancels out). In terms of potentials, the whole sphere (surface + interior) is an equi-potential region.

(a2) the force exerted upon a point charge $\:\xi\:$ in the exterior of the sphere, as in Fig. 03, is equal to the force exerted by a point particle at the center of the sphere with charge equal to its total surface charge $\:\Xi_{s}=\rho_{s}\cdot4\pi{\rm{R}}^{2}\:$. In terms of potentials, the potential outside the sphere equals that created by its total surface charge $\:\Xi_{s}\:$ concentrated on its center.

Figure01 Figure03 Figure03

An intermediate conclusion in the proof of this Proposition is that the magnitude of the force exerted by the "cup" AKBMA of the sphere on the point charge $\:\xi\:$ in Fig. 02 is proportional to $\:\sin^{2}\omega\:$, where $\:\omega\:$ is the angle by which any point of the cyclic edge AMBA of the cup observes the line segment $\:b\:$ (that between the charge $\:\xi\:$ and the center of the sphere). More exactly this force is by magnitude:

\begin{equation} \vert \mathbf{f}_{AKBMA}\vert=k \cdot \dfrac{\Xi_{s}\cdot \xi}{{\rm{b}}^{2}}\sin^{2}\left(\dfrac{\omega}{2}\right)=\left(k \cdot \dfrac{4\pi\rho_{s}\xi{\rm{R}}^{2}}{{\rm{b}}^{2}}\right)\sin^{2}\left(\dfrac{\omega}{2}\right)=constant\cdot \sin^{2}\left(\dfrac{\omega}{2}\right) \tag{A-01} \end{equation}

But this force is cancelled by the force exerted by the "cup" CLDNC of the sphere which is equal in magnitude, but opposite direction:

\begin{equation} \mathbf{f}_{CLDNC}=\;-\;\mathbf{f}_{AKBMA} \tag{A-02} \end{equation}

So, if we remove these two "cups" the force doesn't change. But if we enlarge "cup" AKBMA by moving its cyclic edge AMBA to the left, this last will coincide with the cyclic edge CNDC of the left cup CLDNC. Then removing the two cups is as if we remove the whole sphere leaving the net force unchanged, that is, zero.

Also in Fig. 02 we have

\begin{equation} \mathbf{f}_{AMBDNCA}=\;\mathbf{0} \tag{A-03} \end{equation}

B. The Inverse Square Law for Spheres with uniform volume charge density

Proposition B: Let a sphere of radius $\:\rm{R}\:$ with uniform volume charge density $\:\rho_{v}\:$. Then:

(b1) the force exerted upon a point charge $\:\xi\:$ in the interior of the sphere located at a radial distance $\:\rm{r}\:$ from is center is equal, according to Proposition A, to that exerted by the volume charge density of a sphere of radius $\:\rm{r}\:$, $\:\Xi_{v}\left(\rm{r}\right)=\rho_{v}\cdot \dfrac{4}{3}\pi{\rm{r}}^{3}\:$, concentrated on the center. The magnitude of this force is:

\begin{equation} \vert f_{inside} \vert =k \cdot \dfrac{\Xi_{v}\left(\rm{r}\right)\cdot \xi}{{\rm{r}}^{2}}=k \cdot \dfrac{\rho_{v}4\pi\xi{\rm{r}}^{3}}{3{\rm{r}}^{2}}=constant \cdot \rm{r}\:,\quad \rm{r}\le \rm{R} \tag{B-01} \end{equation}

(b2) the force exerted upon a point charge $\:\xi\:$ in the exterior of the sphere and at radial distance $\:\rm{r}\:$ from is center is equal, according to Proposition A, to that exerted by the volume charge density of a sphere of radius $\:\rm{R}\:$, $\:\Xi_{v}\left(\rm{R}\right)=\rho_{v}\cdot \dfrac{4}{3}\pi{\rm{R}}^{3}\:$, concentrated on the center. The magnitude of this force is:

\begin{equation} \vert f_{outside} \vert =k \cdot \dfrac{\Xi_{v}\left(\rm{R}\right)\cdot \xi}{{\rm{r}}^{2}}=k \cdot \dfrac{\rho_{v}4\pi\xi{\rm{R}}^{3}}{3{\rm{r}}^{2}}=constant \cdot \rm{r}^{-2}\:,\quad \rm{r}>\rm{R} \tag{B-02} \end{equation}


Suppose that Earth is a perfect sphere with uniform volume mass density. Then:

Proposition C:

(c1) A body located at the center of the Earth is weightless.

(c2) Imagine a tunnel of small cross section running along a whole diameter, so passing through the center of the Earth. A body placed in the tunnel at a radial distance $\:{\rm{r}}_{0}\:$ from the center will execute a simple rectilinear harmonic oscillation with center the center of the Earth, since in the case of gravity the force is always attractive to the center and according to equation (B-01) proportional in magnitude to the distance from this center of attraction.


You would not be weightless at the center of the Earth. In other words, the Earth does not follow a geodesic. Let me explain.

The Earth is not spherical, it is an oblate spheroid. The acceleration of a uniform non-spherical body in a spherical gravitational field does not follow an inverse square law. The acceleration of the center of mass does not equal the acceleration at the center of mass. An accelerometer fixed at the center of the Earth would read approx 1.75 pgal (1.75e-14 m/$\mathrm{s^2}$), not zero.

  • $\begingroup$ What does "pgal" mean (Google was unhelpful)? And if the acceleration is non-zero at the center of mass, due to the non-uniformity of the Earth's shape, am I correct in assuming there's a point where the acceleration is zero? How far from the geometric center is that point, and in what direction? $\endgroup$ Sep 12, 2012 at 0:56
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    $\begingroup$ An object released at the geocenter would accelerate in the direction opposite to the Sun.<br> $\endgroup$
    – Nick
    Sep 13, 2012 at 10:18
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    $\begingroup$ The point where the acceleration is zero is about .15 m closer to the sun. If the Sun and Earth were stationary, that point would be .22 m closer to the Sun. $\endgroup$
    – Nick
    Sep 13, 2012 at 12:06
  • $\begingroup$ @KeithThompson "gal" is acceleration in cm/s^2, so 'g' = 9.8m/s^2 = 980 gal. pgal is pico-gal. It doesn't get used much in physics but geologists use it for gravimetric measurements $\endgroup$ Sep 13, 2012 at 14:49
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    $\begingroup$ Very interesting. And kind of embarrassing that I'd have answered zero acceleration at the centre and would also always have done so: but one can instantly see that you're right! $\endgroup$ Jun 10, 2014 at 12:03

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