The acceleration in earth's gravity is $9.807 m/s^2$ at my location. But is this value uniform in the entire earth? For example in the inner core or outer core. What is the acceleration in the center of the earth?


3 Answers 3


he acceleration in earth's gravity is $9.807m/s^2$. But is this value uniform in the entire earth?

It's not even uniform over the surface of the Earth. The value of 9.80665 m/s2 is (roughly) an average value at the surface of the Earth. Gravitation at the surface varies slightly from that average value, with the highest acceleration at the surface being at the North Pole and the lowest atop a mountain near the equator. Inside the Earth, the gravitational acceleration increases slightly with increasing depth1 until one reaches the core-mantle boundary. At that point, it begins dropping with increasing depth, eventually reaching zero at the center of the core.

1 That's a bit of an oversimplification. Nonetheless, gravitational acceleration inside the Earth remains above the Earth surface value all the way down to the core-mantle boundary. It peaks sharply at the core-mantle boundary, above 10 m/s2, but also peaks slightly near the Moho discontinuity. Gravitation in the mantle initially decreases very, very slightly with increasing depth, but then increases toward the global maximum at the core-mantle boundary. There is no place in the mantle where gravitational acceleration is less than that at the surface.


From the surface and outwards, gravity decreases quadratically according to Newton's law of gravitation:

$$g=G\frac m {d^2}$$

From the surface and inwards, gravity decreases linearly (assuming constant density), according to:

$$g=\frac34 \pi G \rho d$$

$G$ is the gravitational constant, $g$ gravitational acceleration, $\rho$ Earth's density and $d$ distance from the centre.

This corresponds to the green linear curve on this picture (from Wikipedia ):

enter image description here

Linearity is only the case for constant density, though, which isn't entirely the case throughout our planet Earth. The real case is shown as the more bumpy curve.

And why is that?

The explanation is that as you descend into the depth of the Earth, you suddenly have some soil above your head. Naturally, the gravitational force from this amount of mass pulls upwards. Only the mass below you still pulls downwards.

Assuming constant density (and perfect sphere-shape) it turns out due to symmetry that the ring (or shell) of mass above your head cancels itself out when considered all around the planet. The only non-balanced force left is that from the sphere below you - as if you were standing on the surface of a smaller planet.

You can calculate how this mass changes with depth, by considering the volume formula of a sphere. Inserting that into Newton's law of gravitation gives us the linear version of the formula shown above.

When moving farther away from the surface of Earth, the amount of mass below us remains the same. Therefor all parameters in the law remains constant, except for distance, and the perfect quadratic decrease is seen in real life as well.

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    $\begingroup$ Maximum gravity is not found at the surface. From the graph you yourself posted, it reaches a maximum at the core-mantle boundary, which is over halfway down toward the center of the Earth. $\endgroup$ Nov 8, 2018 at 13:06
  • $\begingroup$ Still not good: Naturally, the gravitational force from this amount of mass pulls upwards. No, it doesn't. The gravitational acceleration due to stuff at $r>R$ is nothing per Newton's shell theorem. The only contributor to gravitational acceleration at some point inside the Earth is all the stuff at $r<R$. $\endgroup$ Nov 8, 2018 at 13:27
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    $\begingroup$ @DavidHammen That is not correct. Stuff above you definitely pulls upwards in you - an apple on a tall tree exerts a tiny gravitational force upwards in you. The Moon in the sky pulls upwards in you. I think you misunderstand the sentence. Each particle in any soil above you pulls upwards. Particles next to you pull sideways. Particles below you pull downwards. Etc. The shell theorem then says that when we consider them all in the entire shell all around the planet, these forces cancel out. As explained in the following sentence. $\endgroup$
    – Steeven
    Nov 8, 2018 at 13:51
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    $\begingroup$ Apparently nobody reminded that $g$ on Earth usually also includes the effect of centrifugal force. In other words, since Earth is rotating, it isn't an inertial reference frame. This changes the apparent value of $g$ decreasing it with decreasing latitude, from poles to equator. Centrifugal effect is added to variation due to non-spherical form of Earth - it amounts to $0.034 \mathrm{m/s^2}$, twice the effect of Earth's flattening. $\endgroup$
    – Elio Fabri
    Nov 9, 2018 at 11:00

The number $9.807 m/s^2$ is the acceleration on the surface of the earth. If you get closer to the middle of the earth, let's say at a distance $r$ from the center, this value decreases to $\frac{r}{R_E} \times 9.807 m/s^2$, where $R_E$ is the radious of the earth. The point is that the part of the earth oustide of you does not contribute to the acceleration. Therefore, right in the center of the earth, you effectively feel no gravitation. (Remark: Do not go there to check it out, you will not feel too well in the inner core, it's hot.)

A further remark: Look at this picture from Wikipedia. You see that the earth has the shape of a potato and not a perfect sphere. Therefore, even on the surface of the earth, there are some variations to the acceleration, although they are very slight.

  • $\begingroup$ Oh sorry for my mistake, I corrected it. $\endgroup$
    – Luke
    Nov 8, 2018 at 11:22
  • $\begingroup$ Well the Earth itself does not have the shape of a potato, nor a pear for that matter. The Earth really is an oblate spheroid to a good approximation. I think the potatoes and pears show up after you subtract one or more of the lowest order terms in spherical harmonics, and it depends if you are plotting a gravitational iso-potential, or the shape residual itself. $\endgroup$
    – uhoh
    Nov 8, 2018 at 11:46
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    $\begingroup$ Uh, no. Gravitational acceleration inside the Earth does not decrease linearly inside the Earth. It instead increases slightly all the way down to the core-mantle boundary, a bit over half way to the center of the Earth. $\endgroup$ Nov 8, 2018 at 12:26
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    $\begingroup$ Yeah, of course I made a very strong simplification I should have named: That the earth is homogeneous inside. I considered this exact enough to answer the original question. $\endgroup$
    – Luke
    Nov 8, 2018 at 13:15
  • $\begingroup$ further fun with potatoes: How did early estimates of a “potato radius” set 1 eV ~ GMμ/R and get 200 to 300 km? $\endgroup$
    – uhoh
    Nov 9, 2018 at 1:56

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