Earth’s Size

I What are the early ideas on the Earth’s shape?

I What are the qualitative observations against them?

I Understand how Eratosthenes measured the Earth’s size (Fig. 2.1 of Lowry 2007).

I His estimate of the Earth’s circumference is 46,250 km. 15 % greater than the modern value, 40,030 km. Earth’s Size cont’d

I A French astronomer, Picard measured the length of one degree of meridian arc in 1671 using triangulation. He got 6372 km as the radius of the Earth: cf. modern value, 6371 km Earth’s Shape

I Jean Richer’s finding (1672) that his accurate pendulum clock, calibrated in Paris, was “losing 2m 30s” (i.e., indicated an increased period of the pendulum!) on an equatorial island of Cayenne.

I What does it imply about the shape of the Earth?

q l I Hint: T = 2π g

I What was Newton’s idea on the shape of the Earth? Oblate or prolate?

I How was the length of a degree of meridian measured? Earth’s Gravity

I Newton’s law of universal gravitation:

m M F = −G ˆr r 2

I Gravitational acceleration M a = −G ˆr G r 2

I Gravitational potential

M U = −G G r

I How to measure aG?

I How to measure G?

I see https://www.aps.org/publications/ apsnews/201605/big-g.cfm Earth’s Gravity cont’d

I What is the mass and the mean density of the Earth?

I Kepler’s thirs law: 4π2 GM = a3, (1) T 2 where G is the gravitational constant, M is the mass of a planet, T is the period of the orbital motion and a is the semi-major axis of the orbit.

24 3 I M= 5.974×10 kg, mean density = 5,515 kg/m . Earth’s Figure and Gravity1

I The figure of the Earth: The shape of an equipotential surface of gravity that is approximated as an ellipsoid that fits the mean sea level best.

I called the International Reference Ellipsoid (defined in Geodetic Reference System 1980 or GRS80).

I Equatorial radius (a) = 6378.136 km,

I polar radius (c) = 6356.751 km,

2 1/3 I Radius of the equivalent sphere (R) = (a c) = 6371.000 km.

1Sec. 2.4 of (Lowry, 2007) Earth’s Figure and Gravity (cont’d) Earth’s Figure and Gravity (cont’d)

I The polar flattening, f = (a − c)/a −3 I fref of the reference ellipsoid = 1/298.252 = 3.35287 × 10

I fhyd of the rotating fluid in hydrostatic equilibrium = 1/299.5.

I What does “fref > fhyd ” imply?

I Dynamical ellipticity, H = (C − A)/C, where C is the moment of inertia of the Earth about the rotation axis and A about any equatorial axis. −3 I H=3.27379 × 10 = 1/305.456 Earth’s Figure and Gravity (cont’d)

I For later convenience, let’s also define m, the ratio of the equatorial centrifugal acceleration to the equatorial gravity:

ω2a m = ge

I MacCullagh’s formula: Describes the most important two contributions to the gravitational potential of any body:

E (A + B + C − 3 I) U = −G − G + ... G r 2 r 3 Earth’s Figure and Gravity (cont’d) Earth’s Figure and Gravity (cont’d)

I For the sphroidal (i.e., oblate ellipsoidal) Earth, A = B and also I = A sin2 θ + C cos2 θ.

E (C − A) (3 cos2 θ − 1) U = −G + G G r r 3 2

I The 2nd term can be expressed in terms of the Legendre polynomial:

E (C − A) U = −G + G P (cos θ) G r r 3 2

I By rearranging terms, we get " # E C − A R 2 U = −G 1 − P (cos θ) G r ER2 r 2 Earth’s Figure and Gravity (cont’d)

I Represented by the Legendre polynomials (cf. spherical harmonics2), the potential field for a spheroid can be expressed as

" n=∞ n # E X R  U = −G 1 − J P (cos θ) G r r n n n=2

I Comparing this equation with the MacCullagh’s formula term by term, we find

C − A J = 2 ER2

I Higher order terms (e.g., n=3) represent deviations from the ellipsoid, which correspond to a pear-shaped Earth.

2http://scipython.com/book/chapter-8-scipy/examples/ visualizing-the-spherical-harmonics/ Earth’s Figure and Gravity (cont’d) Earth’s Figure and Gravity (cont’d)

I Geopotential: The potential of gravity on a rotating spheroid (Ug) is the sum of the gravitational potential and centrifugal potential,

1 U = U − ω2r 2 sin2 θ g G 2

I If the surface of the spheroid is an equipotential surface, we find the following from the definitions of f and m:

1 J = (2f − m) 2 3 Earth’s Figure and Gravity (cont’d)

2 I Equating this with J2 = (C − A)/(ER ), we get C − A C 1 = (2f − m) C ER2 3

2 I C ∼ 0.33ER for the Earth while a hollow spherical shell has 0.66 ER2 and a solid sphere 0.4ER2. What does this mean?