I had a phone-call from a friend the other morning, who was driving her kids to school, aged 8 and 11. They wanted to know, "if you could weigh the Earth, how heavy would it be?" I recalled the mass to be about 6 x 10^21 tonnes, which when I checked is about right, and so I said, "it's 6 thousand, million, million, million tonnes... No, million, million, million; not million million," and then I said, "that's the trouble, once numbers get bigger than a few thousand, we can't imagine what they mean!" Then the 11 year old asked, "how do you know how heavy it is?" and I said, "you can measure it from its gravity," which seemed to suffice for that moment. It's a good question, though, and the answer provides a sense of perspective regarding the planet.

In the case of the Earth, we can estimate its mass because we know the acceleration due to gravity at some point near the Earth's surface, g = 9.8 m s^-2. This may be equated with the gravitational constant, G = 6.67 x 10^-11 m^3 kg^-1 s^-2 and the (mean) radius of the Earth, r = 6.37 x 10^6 m. Thus:

GmM/r^2 = mg,

where M = Earth's mass and m = some smaller mass close to the Earth's surface. By cancelling the terms, m, and solving for M, we get:

M = gr^2/G

= 9.8 m s^-2 x (6.37 x 10^6)^2 m^2/6.67 x 10-^-11 m^3 kg^-1 s^-2 = 5.96 x 10^24 kg (i.e. about 6 x 10^21 tonnes).

Another approach to the problem is to use the "satellite method", which in the present case refers to the Earth-Moon system, but is used by astronomers to determine the masses of the other planets, the Sun, distant stars in binary systems, the Milky Way galaxy and even entire clusters of galaxies. We can express (according Newton's Law):

F(gravity) = GMm/r^2,

where G is the gravitational constant, M is the Earth's mass and m is the mass of the satellite (Moon), with r being the distance between the centres of the two bodies. We can further express for a simple circular orbit, the centrifugal force (which acts in opposition to the gravitational force):

F(centrifugal) = mv^2/r,

where v is the angular velocity of the satellite. For a stable stationary orbit to exist, the two forces must be equal and opposite, and so we can write that F(gravity = F(centrifugal), and hence:

GMm/r^2 = mv^2/r. By, once more, cancelling the terms, m, and rearranging, we get:

M = v^2 r/G.

Assuming a circular orbit, the mean angular velocity, v is the circumference of the orbit divided by the time (t) taken for the satellite to make that orbit, i.e. v = 2 pi r/t, and so if we substitute for v, we find:

M = 4 pi^2 r^3/G t^2.

Since the mean distance between the Earth-Moon centres is 384,000 km and the orbital period is 27.32 days ( = 2.36 x 10^6 seconds),

M = 4 pi^2 (3.84 x 10^8 m)^3/6.67 x 10^-11 m^3 kg^-1 s^-2 (2.36 x 10^6 s)^2 = 6.02 x 10^24 kg.

Thus the methods agree pretty well. In a posting "Carbon in the Sky" (January 6th 2007), I worked out that the mass of the Earth's atmosphere is about 5.3 x 10^18 kg, and so we can deduce that the relative mass of the atmosphere to the total mass of our blue planet Earth is 1/1,136,000 (i.e. less than one millionth of it), a value that might easily be thought insignificant...

but not from our point of view!

Related Reading.

(1) http://www.astronomycafe.net/qadir/q1223.html

(2) Nelkon and Parker, Advanced Level Physics, 4th Edition, Heinmann Educational Books, London, 1978.

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