# How can I measure the mass of the Earth at home?

1. How can I measure the mass of the Earth at home?

2. How was the mass of the Earth first measured?

• Archimedes could have done it with a lever: "give me a place to stand on, and I will move the earth" (not a place at home however). – Gerard Nov 8 '10 at 10:22
• Archimedes could have done it with a ladder, and some fishing twine (if he could get hold of some that was thin and strong enough. See fourmilab.ch/gravitation/foobar – sigoldberg1 Nov 9 '10 at 22:53
• I added two new tags of "experimental-physics" and "education" to your question. I hope you don't mind. For me, a complete success with this question is for a couple of kids to get together and actually do the experiment, maybe asking questions along the way. Then I would add the tag "experiment" to their question, to indicate that it is dealing with a real, live, experiment. – sigoldberg1 Nov 10 '10 at 21:25

Yes we/you can.

I recall seeing a famous video of a homemade version of the Cavendish torsion balance experiment from the early 1960's, made I think for the PSSC high school course. Basically, the physicist hung a torsion balance from a high ceiling by a long (>10 m?) piece of computer data tape (chosen because it would not stretch). He carefully minimized air currents. The torsion masses were two .5 kg bottles of water on a wooden bar (no magnetic interference). Mass, in the form of boxes of sand, say 20kg was piled around on the floor as static mass and then reversed in position with respect to the suspended masses. There was a clear plastic box around the balance (with a hole in its top for the suspending tape to pass through) also to minimize the effect of air currents, since the lateral force on each bottle is about G*m1*m2/r^2 = (6.7e-11)*0.5kg*20kg/(0.1m)^2 N ~ 6.7e-8 N, i.e. a lateral force on each bottle equivalent to that generated by a weight of about 7 micrograms, about that of a 1 mm^3 grain of sand. This is visible to us because the long arms of the torsion balance convert this small force into a torque on the suspending filament, and the restoring torque is itself very small.

I found an Italian dubbed version of the video on Youtube. See http://www.youtube.com/watch?v=uUGpF3h3RaM&feature=related and a slightly longer version at http://www.youtube.com/watch?v=V4hWMLjfe_M&feature=related. I believe the demonstrator was Prof. Jerrold Zacharias from MIT and the PSSC staff. If anyone can point me to the original undubbed black and white film loop, I'd appreciate it.

It looked really crude but qualitatively it worked. The mirror moved upon reversal of the mass positions. Yeah, experimental physics!! Calculate it out. Use your laser pointer. Glue mirrors. Calibrate. Give it as an experiment in class. Make a (music?) video. Put it on Youtube and embed it here. Social physics.

I also found some other do it your self experimenters with crude equipment, experimental tips (try fishing line) and different masses. See http://funcall.blogspot.com/2009/04/lets-do-twist.html http://www.hep.fsu.edu/~wahl/phy3802/expinfo/cavendish/cenco_grav.pdf and http://www.fourmilab.ch/gravitation/foobar/, which uses a ladder, some cobblestones, monofilament fishing line and has videos. For the experiment in this last reference, you don't need mirrors, since you can see the balance masses move directly because their excursion is so large. See also http://www.youtube.com/watch?v=euvWU-4_B5Y

For all these experiments there is no calibration of the restoring force of the twisted filament (which Cavendish did from the free torsion period of the balance), the balance beam of one appears to be styrofoam, (so I would worry about subtle charge effects), and the beam hits the support of the fixed masses so that it bounces and we do not see the harmonic angular acceleration we might expect. This last problem is apparently well known to amateur experimenters in this field. Another exposition and video is at http://www.juliantrubin.com/bigten/cavendishg.html

The best summary and historical exposition I found is at https://en.wikipedia.org/wiki/Torsion_bar_experiment . I did not realize that the experiment was originally designed by John Michell, a contemporary, whose designs and apparatus passed to Cavendish upon his death. See https://en.wikipedia.org/wiki/John_Michell. Newton had considered the deviation from vertical that a stationary pendulum would have near a terrestrial mountain in the Principia (1686). Although he considered the deviation too small to measure, it was measured 52 years later at Chimborazo, Ecuador in 1738, which was the first experiment showing that the Earth was not hollow, apparently a live hypothesis at the time. The same experiment was repeated in Scotland in 1774. See https://en.wikipedia.org/wiki/Schiehallion_experiment . Mitchell devised the torsion balance experiment in 1783, and started construction of a torsion balance. Cavendish did his experiment in 1797-1798. To me this is all quite inspiring.

Editorial (I'll move this positive rant to meta soon) - given the obviously widely varied audience on this site, I would very much like to see more questions like this one relating to amateur or home experiments. The analysis of the data and possible sources of errors in these experiments is often subtle, and is very instructive. To have real physicists and other clever students publicly criticize some aspect of an experiment provides something that many students may never get otherwise. The social network framework will help many newcomers from different countries learn what real science is in a way that yet another dose of imperfectly understood theory never will. And it's fun too.

• I thought of this recently as a "seeding" question, but I wasn't really expecting to see it. – Mark C Nov 9 '10 at 18:09
• I would love to see this video too. – j.c. Nov 9 '10 at 21:38
• I found a Spanish dubbed version, see above. – sigoldberg1 Nov 11 '10 at 12:38
• Given that the policy that many edits makes CW has been dropped I have removed the wiki status from this post. – dmckee Nov 13 '14 at 0:23
• The OP asks about how to measure the mass of Earth, but you describe how to measure the gravitational constant. This will most likely will be needed, for example when also measuring the radius of Earth and its surface gravity or the distance of the Moon and its orbital period, so this isn't yet completely answering the question. – fibonatic Nov 14 '14 at 10:24

You can make estimates of the Earth's mass $$M_\mathrm E$$ by estimating its average density $$\rho$$ and using the formula $$M_\mathrm E = \rho \cdot V$$, where of course $$V= \frac{4}{3} \pi R^3$$, so you have to know the Earth's radius $$R$$. This method is rather a gamble because you don't know the Earth's average density (suppose the Earth would be hollow?). Newton made such an estimate himself.

Newton's law of gravitation $$F_\text{grav} = G \cdot \frac{M_\mathrm E \cdot m}{R^2} = m \cdot g$$ describes the attraction between a mass $$m$$ on the Earth's surface and the Earth. While $$m$$, $$g$$, so $$F_\text{grav}$$ and $$R$$ can be determined, there are two unknowns: $$M_\mathrm E$$ and $$G$$.

So one can only determine the mass of the Earth together with this constant $$G$$: the gravitational constant.

You cannot measure the mass of the Earth at home unless you have access to a Cavendish torsion balance to measure the gravitational constant. The Cavendish experiment is described in this Wikipedia article. Approx. 100 years after Newton the first experiment was done to measure $$G$$.

The value of $$G \cdot M_\mathrm E$$ can be measured very accurately ($$3.9860042\times10^{14}\ \mathrm{m^3/s^2}$$).
The value of $$G$$ can be measured less accurately ($$6.674\times10^{-11}\ \mathrm{m^3/(kg\ s^2}$$) so that's the bottleneck for the determination of the Earth's mass.

The Earth's mass is not constant; it is estimated that every day 40 tons of dust from meteorites hit it. This is totally negligible for a mass of $$5.9736\times10^{24}\ \mathrm{kg}$$.

You can't measure the mass of Earth directly, as others have stated. You can calculate it knowing:

• The value of $$g$$, the gravitational acceleration (approximately $$9.8\ \mathrm{m/s}$$)
• The value of $$R_\mathrm e$$, the radius of the Earth (approximately $$6378.1\ \mathrm{km}$$)
• The value of $$G$$, the gravitational constant (approximately $$6.67\times10^{-11}\ \mathrm{N\ m^2/kg^2}$$)

and solving the following equation:

$$mg = \frac{GM_\mathrm em}{R_\mathrm e^2}$$ or $$M_\mathrm e = \frac{g R_\mathrm e^2}{G}$$

Now:

• to measure $$g$$ you can use a pendulum – this can be done at home.
• to measure $$R_\mathrm e$$ the simplest experiment is Eratosthenes' experiment – this cannot be done at home
• to measure $$G$$ you need to use a Cavendish balance – which cannot be done at home because it's a notoriously difficult experiment (the constant is really small, requires custom apparatus, a very long time, etc.).
• Eratosthenes's experiment can be done, however, by two kids at two homes via Skype. In 300BC it involved two big technologies of his day, i.e. the calendar and the mail. His experiment apparently started when he, as librarian at Alexandria got a letter from someone saying that on a certain day, he could see the sun reflected from the water at the bottom of a deep vertical well in his home town at noon. Eratosthenes did the experiment and the math. Today we could use Skype and kids in different places, vertical plumb lines, protractors (Greek?!), and do the math. Don't give up so easily. – sigoldberg1 Jan 27 '11 at 4:16
• @sig stress, but you stil need a copule of things which disqualify the experiment from the "home" category: a large vertical structure of known height, and a friend living at a significantly different latitude. – Sklivvz Jan 27 '11 at 7:27

As Gerard stated you can't directly measure the mass of the Earth.

You can measure the acceleration of gravity at the Earth's surface using a simple pendulum. You need to know two things: The length of the pendulum, which can be accurately and easily measured with a ruler. The period of the oscillation, which can be measured by taking a stop-watch and timing a number of small swings.

Then use the formula:

$g = \frac{4 \pi ^2 L}{T^2}$

Where L is the length and T is the period.

This is then related to the mass by:

$M = \frac{g R^2}{G}$

Where R is the radius of the earth and G is Newtons gravitational constant.

You still need to know these other quantities, as Gerard said. I don't know how you would find R or G at home.

Edit: Formatted the Formulae using Latex.

• just start using latex and it will parse automatically – Mark Eichenlaub Nov 9 '10 at 11:18
• Thank you, originally I tried it, but I didn't realise you had to wait and click outside the edit box so that the parser has time to change it in the preview. – Dom Nov 10 '10 at 15:25