How Do They Calculate the Mass of the Earth?

image001In science, there could be a wealth of observations, experiments and knowledge behind the simplest fact. For instance, the mass of the earth is 5.972 × 1024 kg. But where does that number come from? That simple question opens a can of worms for those with a phobia of math or physics. But for the rest of us, it reveals a pleasant sequence of questions, each opening up a new topic, like a set of Russian dolls.

At first the problem seems deceivingly simple. If you know the circumference of the Earth, you know its radius from which you can get calculate its volume. Then with an estimate of average density, you can obtain its mass. Begin with an educated guess:  an average of the densities of the common rock granite and iron gives a value (5250 kg/m3, almost 95% accurate, by fluke). The circumference of 40 000 km attributed to another clever but lucky measurement by Eratosthenes, when converted to a volume and then multiplied by the density yields an Earth-mass of 5.67 X 1024 kg, off by 12%.
But what are the more accurate methods of measuring the mass of the earth (me) and its radius (re)? Its mass can be obtained from a formula based on two experimentally determined constants, g = the gravitational acceleration on Earth and G = gravitational constant. image015

Of course my questions begs other questions: from where does the expression originate?How do you get a better value for the radius and how do you get G and g?
Calculating the Radius of the Earth

image005Jean-Félix Picard, a contemporary of Sir Isaac Newton, first measured the size of the Earth to a reasonable degree of accuracy in a survey conducted in 1669–70. He measured one degree of latitude along the Paris Meridian using triangulation along thirteen triangles that stretched from Paris to a clock tower near Amiens. From his measurements, one degree of latitude was equivalent to 110.46 km. A full circle or 360 degrees produced a circumference of 39765.60 km, corresponding to a terrestrial radius of 6328.9 km. That was not a fixed value because as they realized in the 17th century, our planet is not a perfect sphere.

The value of G

Newton realized that any pair of objects, whether planets or metal spheres,  attract one another with a force proportional to the product of their masses. Before we get to what inspired the relationship, the more general formula that led to  image015

is the Law of Universal Gravitation:image025

( a nice mnemonic for it is , G-memory!) where F is the force of attraction between the mass a planet, in this case Earth, and that of an object, mo.  Since the weight (mog) of that object is the same force we can equate the two expressions, cancel  mand solve for me.



from Wikipedia article on  Cavendish’s experiment

Rather than focusing on planets whose masses were unknown, Cavendish used two large balls of lead and their attraction to small balls connected in a torsion balance. The force between the large balls (M) in the diagram caused a small angle Θ  to be formed between their connecting rod’s position and the original location when the large and small balls were separated by a distance r . He also measured the period of oscillation, T. image020

The small balls’ moment of inertia, I,  is mL2/2, where m = mass of the small ball and L is the length of the torsion balance beam.  He also used two equivalent expressions ( image023) for the turning force(torque) of the torsion wire where k = the wire’s torsion constant. When these expressions are substituted into image026 we obtain




which gave  Cavendish 6.75 x 10-11 N m2/kg2, a value only off by 1.2 % from the currently accepted 6.67 x 10-11 N m2/kg2. The latter was also obtained from a torsion balance using gold, platinum, and glass small spheres at the U. S. National Bureau of Standards from 1925 to 1928.

But what about g, the value for gravitational acceleration?

This involves a much simpler experiment and mathematics. When an object is dropped from rest, since its initial velocity is zero, its final velocity will be the product of g and time, t.

v = gt

Its potential energy, mgh, with respect to the surface will be all converted to kinetic energy, 0.5mv2.

So mgh = 0.5mv2.

g = v2 /(2h), but if we substitute gt for v from the previous expression:

g = 2h/ t2  

All we need is a free fall apparatus which measures the time taken for a steel ball to fall from a height, h. Using several data points, height is plotted against the square of time

( h = (g/2) t2  ) and the slope of the line is g/2.  Gravitational acceleration varies on different locations on earth, but the average value is 9.81 m/s

The Source of…image026
From the measurements of other astronomers, Kepler realized that the relative ratio  of any two planets’ periods, t1 and t2 , was related to the ratio of their orbital radii, r1 and r2 , by what is dubbed Kepler’s 3rd law:image030
image028For example Earth and Venus’ orbital periods are 224.7 and 365.2 days long. We are 1 astronomical unit from the sun while Venus is 0.723 times as far, a fact that can be obtained trigonometrically from observing the planet at its position of greatest elongation, in other words when it’s at the vertex of a 90o angle between us and the sun. The sine of the observed angle, 46.3o , is the ratio of the opposite side (Venus-sun relative distance) to the hypotenuse(sun-earth distance). If we use a hypotenuse of 1, we obtain a value of 0.723.
(224.7/365.25)2  = 0.378 = (0.723)3 = 0.378.
Another way of expressing Kepler’s 3rd Law is that the square of any planetary period, T, is some constant, C, multiplied by cube of its orbital distance, R:   T2 = CR3.
When an object like a planet moves at constant speed, its direction constantly changes. So its velocity, a vector quantity, is changing or accelerating.  The centripetal force towards the sun is given by F = mv2 /R. We’ll derive that shortly.  In one period, the planet will cover a circumference of 2πR in time T, so its velocity, v = distance/time = 2πR/T. Substituting that into the centripetal force expression, we get

F =  image033.

Since T2 = CR3 ,   F =  image034 , and this made Newton cognizant of the inverse square relationship between force and orbital distance. He also realized that 4π2 /C represented the product of a universal constant, G, and a second mass attracted to the first. Hence, image026
Why is Centripetal Force Equal to mv2 /R?

drawing by the author

In the smallest of time intervals, the velocity of an object moving in a circle changes because its direction changes. But its speed is constant. So if in time Δt, an object goes from v1 to vf, their absolute values are the same. If you draw the velocity vectors as tangents to the circle, each one makes a 90 degree angle with the radius. Extending the vectors (see dotted lines) we get a quadrilateral ABCD, where

180 – Φ +90 + 90 + Θ = 360. Thus Φ = Θ!
In a small time interval, Δt, the arc DB will be a straight segment forming a triangle. The distance covered in time Δt, will equal v Δt = DB.  Since the triangle is isosceles, just like DEF and since both triangles’ equal sides form the same angle, they are of the same shape and the sides are proportional. image038Since Δv/Δt = a = acceleration, substituting and cross- multiplying we get a = v2/r. Multiplying both sides by the mass, m,
F = ma = mv2/r.





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