December 12, 2002

One Hundred Interesting Mathematical Calculations, Number 3
#### One Hundred Interesting Mathematical Calculations, Number 3: The Strength of Gravity

#### Weighing the Earth

Newton's law of gravitation says that the gravitational force F exerted on one object (call it m_{1}) by another object (call it m_{2}) is given by the equation:

F = Gm

_{1}m_{2}/(d^{2})

The force is equal to the gravitational constant G multiplied by the product of the two objects' masses and then divided by the square of the distance between the two objects.

When we measure distances in meters and masses in kilograms, then the force F turns out to be measured in units called Newtons. What is a Newton? If we have a force of one Newton, and apply it to an object that masses 1 kg for one second, then, if the object were stationary at the start of the second, at the end of the second it would have a velocity of 1 m/s--one meter per second--in the direction the force pushed it. (In more familiar units, 9.8 Newtons = 2.2 pounds: 4.55 Newtons = 1 pound; 0.224 pounds = 1 Newton.)

Through many careful and ingenious experiments, scientists have determined the value of the gravitational constant G. In this metric system of measurement, G = 6.67 x 10^{-11}.

This gravitational force can be very weak. Consider the gravitational force exerted on a 1 kg object by another 1 kg object located 1 meter away is. Using Newton's graviational equation and substituting in "1" for all the masses an distances, we get:

F = G x 1 x 1 / (1

^{2}) = G = 6.67 x 10^{-11 }m kg/(s^{2})

How strong is this force? If we apply such a force to a one-kilogram object for one second, it will give the object a speed of 6.67 x 10^{-11} m/s. Recall that there are 86,400 seconds in a day, and 365 days in a year. Multiplying, we see that a speed of 6.67 x 10^{-11} m/s is a speed of 0.002 meters/year--or two millimeters a year.

Suppose we take an 88 pound boy--a boy that masses 40 kg, and weighs 392 Newtons. By "weighs 392 Newtons," we mean that 392 Newtons is the strength of the earth's gravitational pull on the boy. How large is the earth to exert such a pull?

We know that the boy stands 6,400 km from the center of the earth, that the earth is roughly a sphere, and that when we calculate the gravitational attraction by a spherical mass we can do our calculations as if all the mass of the sphere were concentrated at its center. (This is one of the many things that Sir Isaac Newton discovered).

So we take our equation:

F = Gm

_{1}m_{2}/(d^{2})

and we substitute in the things that we know:

- G is 6.67 x 10
^{-11} - m
_{1}is the mass of the boy, is 40 kg - m
_{2}is the mass of the earth, which we don't know - F is the gravitational attraction of the earth on the boy, which is 392 Newtons
- d is the distance between the boy and the center of the earth, which is 6,400 km or 6.4 x 10
^{6}meters.

So our equation, after substitution, is:

392 = (6.67 x 10

^{-11})(40)(m_{2})/((6.4 x 10^{6})^{2})

Let's use the commutative law to move the thing we don't know--the mass of the earth, m2--our in front of the rest of the right-hand-side:

392 = (m

_{2})(6.67 x 10^{-11})(40)/((6.4 x 10^{6})^{2})

Let's multiply together the (6.67 x 10^{-11}) and the (40):

392 = (m

_{2})(2.67 x 10^{-9})/((6.4 x 10^{6})^{2})

Let's square the distance in the denominator:

392 = (m

_{2})(2.67 x 10^{-9})/(4.10 x 10^{13})

Let's do the division:

392 = (m

_{2})(6.51 x 10^{-23})

And now if we divide both sides by (6.51 x 10^{-23}), we get m_{2} by itself on the right-hand-side and we have our answer:

6.02 x 10

^{24}

6,020,000,000,000,000,000,000,000 kilograms is the mass of the earth. That's 6,020,000,000,000,000,000,000 tons. If we remember that there are six billion people alive on the earth, that is about 1,000,000,000,000,000--one quadrillion--tons of earth-stuff per person.

Posted by DeLong at December 12, 2002 11:09 AM | TrackbackEmail this entry

Regarding the 1 kg mass exerting a force on the other 1 kg mass: a careless reader (whom we shouldn't really be concerned about) might think your 2 mm/year results even when the force is applied the whole year, which of course, is not what you meant or said. If the first 1 kg mass were continually moved to be 1 meter ahead of the second mass, then after a year the second mass would have been pulled a distance 3,320 m (that's an American comma, not European, i.e. 3.3 km). At the end of this year, the second mass would have accelerated up to the genuinely snail-paced 2 mm/s.

Constant acceleration is amazingly potent. It can turn an extraordinarily small acceleration into an appreciable speed. It can turn appreciable accelerations into relativistic speeds. Our intuition for this is impaired by air resistence and the terminal velocity it imposes.

"6,020,000,000,000,000,000,000,000 kilograms is the mass of the earth. That's 6,020,000,000,000,000,000,000,000 tons."

Isn't it rather

6,020,000,000,000,000,000,000 metric tons

and so about 6,622,000,000,000,000,000,000 short tons? (That is, you neglected to delete the last three zeros, if I am reading correctly.)

"If we remember that there are six billion people alive on the earth, that is about 1,000,000,000,000,000,000--one quintillion--tons of earth-stuff per person."

6,020,000,000,000,000,000,000,000/6,000,000,000

is 1,000,000,000,000,000, not 1,000,000,000,000,000,000, so using your (incorrect, I believe) number of tons, you get a thousand trillion (quatrillion?) not a million trillion, and using my (correct, I believe) number of tons, you get 1,000,000,000,000, or 1 trillion tons per person.

Of course, the really fascinating thing about gravity is just how weak it is, at least relative to the other fundamental forces. Two electrons placed one meter apart would accelerate away from each other at 250-odd m/s^2. Put another way, the electrosttic force between two protons in a nucleus is something like 35 orders of magnitude larger than the gravitational force between them (picking an example from a handy textbook...). The weakness of gravity is one of the great mysteries of physics.

But quantum gravity is probably a bit above the level you're aiming for with these...

Ah. *Sigh*. You're right...

Posted by: Brad DeLong on December 12, 2002 04:29 PMYou don't need to go to quantum gravity to address that, really. (But if you want to...) Anyways, Frank Wilczek's explanation involving only QCD is at

The tough question is why is the Higgs mass so small, but that's a different story that would take a bit longer to get into.

Posted by: Aaron on December 12, 2002 06:07 PMHmmm. One trillion (or one quintillion) tons of "earth stuff" per person. Thinking about this puts an interesting perspective on those left-wing critics that say we shouldn't be counting the raw earth materials we dig up towards total GDP because "we are actually depleting the earth's capital stock" when we do this.

Posted by: Curt Wilson on December 12, 2002 06:09 PMOops. Posted as if on Usenet. That should be

this.

Wilczek throws in a few technical terms, but the section on this stuff is straightforward, I think.

Posted by: Aaron on December 12, 2002 06:11 PM*Hmmm. One trillion (or one quintillion) tons of "earth stuff" per person. Thinking about this puts an interesting perspective on those left-wing critics that say we shouldn't be counting the raw earth materials we dig up towards total GDP because "we are actually depleting the earth's capital stock" when we do this.*

Um you do realize that most of this Earth stuff is not in the crust? Good luck to you if you are thinking of mining the Inner Core.

For interesting calculation number 4, you might want to show your kids how to do a net present value calculation.

The basics of utilitarianism became clear when I first learned algebra (and hence functions), but for the longest time I was mystified how to compare benefits across time. It was obvious to me that you couldn't do planning or strategy without it, but from age 10 to age 19 I had no idea how to actually compare benefits across time.

The discovery of interest rates and comparing net present values was a BIG revelation to me. It might be bone-deep natural to an economist, but there's a real feeling-of-power moment when you run into it for the first time.

Posted by: Neel Krishnaswami on December 13, 2002 09:16 AMPost a comment