Isaac Newton’s Tower Drop Puzzle: The Counter-Intuitive Solution
- When an object falls from a high tower, it lands slightly to the east of the vertical drop rather than directly underneath, according to a classical physics puzzle...
- The tower drop problem was central to the birth of classical physics.
- To understand where the object lands, observers must consider the Earth's rotation from west to east.
When an object falls from a high tower, it lands slightly to the east of the vertical drop rather than directly underneath, according to a classical physics puzzle originally posed by Isaac Newton in a letter to Robert Hooke in 1679. The counter-intuitive phenomenon occurs because the falling object maintains its initial angular momentum, spinning faster about Earth’s axis as its distance to the center decreases.
Historical Origins and the Foundation of Classical Physics
The tower drop problem was central to the birth of classical physics. Isaac Newton posed the problem in a 1679 letter to Robert Hooke, noting that the solution was “quite contrary to ye opinion of yevulgar.” The correspondence between Hooke and Newton rekindled Newton’s interest in gravity and motion, ultimately leading to his formulation of the mathematical foundations of classical physics in the Principia. The problem is featured in the book Wonders in Motion, The Story of Dynamics, written by David Acheson, an emeritus professor of maths at Oxford, on sale at the Guardian Bookshop.
Mechanics of the Tower Drop Deflection
To understand where the object lands, observers must consider the Earth’s rotation from west to east. While an initial thought might suggest the object falls slightly to the west as the Earth rotates underneath it, that argument ignores the fact that the object starts out rotating with the Earth alongside the tower itself. Another common assumption is that the object lands exactly at the bottom of the tower, which is also incorrect. The object actually lands slightly to the east because it spins about its axis slightly faster as it falls, operating by essentially the same mechanism that makes a spinning ice-skater spin faster when pulling in their arms. The physical effect is small, but not absurdly so. If the Eiffel tower were on the equator, the deflection to the east would be about 11cm.
Conservation of Angular Momentum in Falling Objects
The mathematical result derives directly from the conservation of angular momentum, which is a measure of the momentum of an object rotating about an axis. While gravity pushes a falling object toward Earth’s center of mass, the object simultaneously spins in a circle around that rotational center. As the object falls, its angular momentum stays the same, meaning the expression wr2—where w is the object’s angular velocity and r is its distance to the rotational axis (in this case, the centre of the Earth)—stays the same. Because the value of r decreases during the fall, the value of w must increase, causing the object to spin faster than the Earth is spinning and land slightly to the east of the position directly underneath the tower.