How far is to the center of the earth?

Question

Its a scince quetion

Answer 1

The radius of Earth (or any other planet) is the distance from its center to a point on its surface at mean sea level. Like most planets, Earth is not a perfect sphere, but instead is somewhat flattened at the North and South Poles, and bulges at the equator, which means that its radius and corresponding radius of curvature differs depending on where you measure it (and, in the case of curvature, even which direction is being faced!). This elliptic shape is known as an oblate spheroid/ellipsoid.

Given all of the advancements in measuring technology (including satellites) and tailoring to regional topography, many different reference ellipsoid models have made their way into general usage over the years, providing slightly different values.

However, local variations in terrain negate any chance of pronouncing an absolutely "precise" radius/radius of curvature—one can only find a mathematically precise value based on a given model (with the plethora of—some seemingly outdated—models accommodating regional terrain and accumulated data found from them).

Equatorial radius: a

The Earth's equatorial radius, or semi-major axis, is the distance from its centre to the equator and equals 6,378.135 km (≈3,963.189 mi; ≈3,443.917 nmi).

Polar radius: b

The Earth's polar radius, or semi-minor axis, is the distance from its center to the North and South Poles, and equals 6,356.750 km (≈3,949.901 mi; ≈3,432.370 nmi).

Radius at a given geodetic latitude

The Earth's radius at geodetic latitude, \phi\,\!, is:

R(\phi)=\sqrt{\frac{(a^2\cos(\phi))^2+(b^2\sin(\phi))^2}{(a\cos(\phi))^2+(b\sin(\phi))^2}};\,\!

Radius of curvature

The Earth's equatorial radius of curvature in the meridian is:

\frac{b^2}{a}\,\!= 6335.437 km

The Earth's polar radius of curvature is:

\frac{a^2}{b}\,\!= 6399.592 km

The Earth's radius of curvature in the (north-south) meridian at \phi\,\! is:

M=M(\phi)=\frac{(ab)^2}{((a\cos(\phi))^2+(b\sin(\phi))^2)^{3/2}};\,\!

The Earth's radius of curvature in the prime vertical, which is perpendicular, or normal, to M at geodetic latitude \phi\,\! is:

N=N(\phi)=\frac{a^2}{\sqrt{(a\cos(\phi))^2+(b\sin(\phi))^2}};\,\!
N is defined as the radius of curvature in the plane which is normal to both the surface of the ellipsoid at, and the meridian passing through, the specific point of interest.

The Earth's mean radius of curvature (averaging over all directions) at \phi\,\! is:

R_a=\sqrt{MN}=\frac{a^2b}{(a\cos(\phi))^2+(b\sin(\phi))^2};\,\!

The Earth's radius of curvature along a course at geodetic bearing, \alpha\,\!, at \phi\,\! is:

R_c=\frac{{}_{1}}{\frac{\cos(\alpha)^2}{M}+\frac{\sin(\alpha)^2}{N}};\,\!

Quadratic mean radius: Qr

The ellipsoidal quadratic mean radius provides the best approximation of Earth's average transverse meridional radius and radius of curvature:

Q_r=\sqrt{\frac{3a^2+b^2}{4}};\,\!

It is this radius that would be used to approximate the ellipsoid's average great ellipse (i.e., this is the equivalent spherical "great-circle" radius of the ellipsoid).
For Earth, Qr equals 6,372.795477598 km (≈3,959.871 mi; ≈3,441.034 nmi).

Authalic mean radius: Ar

Earth's authalic ("equal area") mean radius is 6,371.005076123 km (≈3,958.759 mi; ≈3,440.067 nmi). This number is derived by square rooting the average (latitudinally cosine corrected) geometric mean of the meridional and transverse equatorial, or "normal" (i.e., perpendicular), arcradii of all surface points on the spheroid, which can be reduced to a closed-form solution:

A_r=\sqrt{\frac{a^2+\frac{ab^2}{\sqrt{a^2-b^2}}\ln{\left(\frac{a+\sqrt{a^2-b^2}}b\right)}}{2}}=\sqrt{\frac{A}{4\pi}};\,\!
(where A is the authalic surface area of Earth. This
would be the radius of a hypothetical perfect sphere
which has the same, geometric mean oriented surface
area as the spheroid.)

[edit] Volumetric radius: Vr

Another, less utilized, sphericalization is that of the volumetric radius, which is the radius of a sphere of equal volume:

V_r=\sqrt[3]{a^2b};\,\!

For Earth, the volumetric radius equals 6,370.998685023 km (≈3,958.755 mi; ≈3,440.064 nmi).

(Note: Earth radius is sometimes used as a unit of distance, especially in astronomy and geology. It is usually denoted by RE.)

[edit] Meridional Earth radius

Another radius mean is the meridional mean, which equals the radius used in finding the perimeter of an ellipse. It can also be found by just finding the average value of M:

M_r=\frac{2}{\pi}\int_{0}^{90^\circ}\!M(\phi)\,d\phi\;\approx\left[\frac{a^{1.5}+b^{1.5}}{2}\right]^{1/1.5};\,\!

For Earth, this works out to 6367.446988834 km (≈3,956.548 mi; ≈3,438.146 nmi).

Therefore, the values defined below are based on a "general purpose" model, refined as globally precise as possible, to 5 metres.

Answer 2

About 4000 miles or 6400 kilometers.

Answer 3

It is approximately 4000 miles.

Answer 4

6470 kms

Answer 5

i think 6440 kms