How much does an infinite stack of pennies weigh?

Have you ever wondered how much an infinite stack of pennies might weigh? Today you can find out. First, we get a stack of pennies 1 foot high and find it weighs 22 oz = 1.375 lb. That's all we need (along with a little physics and a little math). Let the radius of the Earth be R and the mass of the Earth be M.

Physics part: An object of mass m at a distance D from the center of the Earth weighs GMm/(D^2), where G is the gravitational constant and "D^2" means "D squared". A couple of things to notice in this formula are: the weight of something on the surface of the Earth is GMm/R^2 and that as the object of mass m moves further from the center of the Earth, that is farther from the Earth's surface, its weight decreases by the square of the distance. Since our 1 ft stack of pennies weighs 1.375 lb, its mass is given by m = R^2 (1.375) / GM where we have just plugged the weight into the formula above. That is, GMm/(R^2)=1.375, and solved for m.

Calculus idea: If we had an infinite stack of pennies we could consider the stack as a uniform rod and we could take a tiny piece of it, say dx high, at the distance x from the surface of the earth. Now the infinite stack is made up of such pieces where x ranges from zero to infinity. The total weight of the infinite stack is just the weight of all these little pieces. That is, the total weight W is the limiting value of the sum of the weights of all these dx high pieces as the value of dx gets smaller and smaller (goes to zero).

Calculation: the weight of the dx high piece at distance x from the Earth's surface depends on the mass of the little piece. Since mass is an inherent property of matter it remains constant no matter where the object is in the universe. In the physics part above we saw the mass of the 1 ft-stack, so we know the mass, dm,  of a dx ft-stack will be dx/1 = dx  of the original mass. That is dm= R^2(1.375)/(GM) *dx.
Then, the weight of this little piece at distance x from the surface of the Earth is
dw=GM(dm)/(R+x)^2 = R^2(1.375)/(R+x)^2 * dx.

Calculus part: Adding up all these little weights as x goes from zero to infinity means taking the integral of this expression. We can bring the constant R^2(1.375) outside the integral and simply compute the definite integral of 1/(R+x)^2 from zero to k and then let k go to infinity. From this we get
W= R^2(1.375) [ 1/R - 1/(R+k) ] which goes to R(1.375) as k goes to infinity.

Thus the weight of an infinite stack of pennies is W= (4000 x 5280/2000)(1.375) = 14520 tons!!

Special note: when k=R, that is, when the stack is as high as one Earth radius, R, W= 1/2(R*1.375).
This means half the weight of the infinite stack is found in the first 4000 miles of pennies!!

And now you know!!