Why the Moon's Own Gravity Does Zero Work On It

Vectors can be added and subtracted, but multiplying two of them together needs a genuinely new rule, and there are two distinct ways to do it -- one, the scalar product, produces an ordinary number from two vectors, and it is this one needed to properly define work. The scalar (or dot) product of vectors A and B, written A.B, is defined as A.B = AB cos θ, where A and B are the vectors' magnitudes and θ is the angle between them; since A, B, and cos θ are all plain numbers, A.B is a scalar, carrying no direction of its own even though it is built from two directional quantities. Geometrically, B cos θ is exactly the projection of B onto A, so A.B is the magnitude of A multiplied by the component of B lying along A -- equivalently, the magnitude of B multiplied by the component of A along B. Written out in components, using the unit vectors î, ĵ, k̂ (each satisfying î.î = ĵ.ĵ = k̂.k̂ = 1 and î.ĵ = ĵ.k̂ = k̂.î = 0, since perpendicular unit vectors have cos 90° = 0), the scalar product becomes simply A.B = AxBx + AyBy + AzBz, obeying both the commutative law (A.B = B.A) and the distributive law (A.(B+C) = A.B + A.C) exactly like ordinary multiplication. A single, crucial consequence follows directly from the cosine in its definition: A.B is exactly zero whenever A and B are perpendicular (cos 90° = 0), regardless of how large either vector actually is -- the property that turns out to govern exactly when a force does zero work on a moving object.

A familiar kinematic relation from straight-line motion, v² - u² = 2as, hides the work-energy theorem inside it: multiplying both sides by m/2 gives (1/2)mv² - (1/2)mu² = mas, and since Newton's second law makes ma equal to the force F, the right side becomes simply Fs. Generalising to full vectors (using the scalar product to handle the displacement's and force's directions properly), this becomes (1/2)mv² - (1/2)mu² = F.d -- the left side is the change in the quantity 'half the mass times speed squared,' called kinetic energy (K), and the right side, force dotted with displacement, is called work (W). This gives Kf - Ki = W, the work-energy theorem: the change in an object's kinetic energy equals the work done on it by the net force. Work itself, for a constant force F causing a displacement d, is precisely W = F.d = (F cos θ)d, where θ is the angle between the force and the displacement. This definition makes three distinct situations give exactly zero work: zero displacement (a weightlifter holding a mass steady does no work on it, however tiring); zero force (an object sliding on a frictionless table needs no horizontal force to keep moving); and, most subtly, a force perpendicular to the displacement, since cos 90° = 0 -- a block sliding on a smooth table is never affected by gravity's own work, because gravity acts straight down while the motion is horizontal, and, remarkably, the very same reasoning shows that the Earth's gravitational pull on the Moon does no work on it at all, if its orbit is treated as a perfect circle, since the gravitational force always points radially inward while the Moon's velocity is always tangential, exactly perpendicular to it at every instant.

Kinetic energy, K = (1/2)mv² = (1/2)m(v.v), is a scalar quantity, always positive (or zero), unlike work, which can genuinely be negative. Physically, it measures the work an object is capable of doing simply because it is moving -- a notion humans have used intuitively for a very long time, harnessing a fast-flowing stream's kinetic energy to grind corn, or a sailing ship's kinetic energy gained from the wind. This measure of 'capacity to do work' turns out to be exactly consistent with the work-energy theorem in reverse: firing a bullet into a block of wood, the bullet's kinetic energy does work against the wood's resistance as it burrows through, and the emergent kinetic energy (after the wood has done negative work on the bullet, slowing it) can be calculated directly, without ever needing to know the complicated, varying resistive force inside the wood at all -- exactly the kind of situation where tracking energy directly is far more powerful than trying to solve the detailed dynamics step by step.

Most real forces are not constant -- they vary as an object moves, and calculating work in that case needs a genuinely new technique. Splitting a journey into many tiny steps of displacement Δx, small enough that the force F(x) is nearly constant over each one, each step contributes an approximate work ΔW = F(x)Δx, exactly the area of one thin rectangle under a force-versus-displacement graph; adding up all these thin rectangles across the whole journey approximates the total work done. Letting these steps shrink toward zero width turns this sum into something exact: the total area under the F(x) curve, computed by the mathematical technique of integration, W = the integral of F(x) dx from the initial to the final position. This same idea rebuilds the work-energy theorem for a variable force too: the rate of change of kinetic energy, dK/dt, works out to m(dv/dt).v = F.v (again using Newton's second law), which equals F(dx/dt); rearranging gives dK = F dx, and integrating both sides from the initial to the final position proves Kf - Ki = the integral of F dx = W, exactly the same relationship already found for a constant force, now shown to hold universally. The work-energy theorem, though extremely useful, is genuinely an INTEGRATED, scalar version of Newton's second law -- it deliberately leaves out information about time (a compression distance can be calculated without knowing how long it took) and about direction (a full vector calculation would need three separate component equations, not one scalar one), trading away that detail for a much simpler calculation whenever only the overall energy change matters.

A stretched bowstring, released, sends an arrow flying at great speed -- the energy clearly came from somewhere, and it was 'stored' in the string's own stretched configuration, ready to be released. This notion of stored, position-dependent energy is potential energy, and it can be made mathematically precise for a special class of forces: a force F(x) is called conservative if a potential energy function V(x) can be defined such that F(x) = -dV/dx (the force equals the negative slope of the potential energy). For the gravitational force near the Earth's surface, mg, this gives V(h) = mgh directly, the negative of the work done BY gravity in raising an object to height h. A genuinely important physical consequence follows: since the work done by such a force is the integral of F(x)dx = Vi - Vf, this work depends ONLY on the starting and ending positions, never on the specific path taken between them -- and, as a direct result, the work done by a conservative force over any closed path that returns to its starting point is always exactly zero, since then Vi = Vf. Not every force qualifies: friction, for instance, is explicitly non-conservative, since the work done against it over a closed path is generally NOT zero (dragging an object around a loop always loses energy to friction, regardless of returning to the start), and no potential energy function can be meaningfully assigned to it. One further subtlety is worth holding onto: the actual VALUE chosen for zero potential energy is always arbitrary, fixed only by convenience (the ground for gravity, an unstretched spring's natural length for a spring) -- but once chosen, it must be used consistently throughout any one calculation.

Combining the work-energy theorem (ΔK = F(x)Δx) with the definition of a conservative force (-ΔV = F(x)Δx) gives, by simple substitution, ΔK + ΔV = 0, or Δ(K+V) = 0 -- the sum K+V, called the total mechanical energy, never changes at all, so long as only conservative forces are doing work. Dropping a ball from height H demonstrates this cleanly: at the top, its energy is entirely potential, EH = mgH; partway down, at height h, its energy has split between potential and kinetic, Eh = mgh + (1/2)mvh²; and at the ground, its energy is entirely kinetic, E0 = (1/2)mvf² -- and since mechanical energy is conserved throughout, all three expressions are equal, immediately recovering the familiar free-fall result vf = √(2gH) without needing kinematics at all. A more elaborate application: a bob on a light string, given exactly enough horizontal speed v0 at the lowest point A to complete a full semicircular loop with the string just barely going slack (tension exactly zero) at the highest point C, can be analysed purely through conservation of mechanical energy together with Newton's second law at the critical point C (where gravity alone must supply the centripetal force, mg = mvC²/L). Taking potential energy as zero at A, the mechanical energy there is simply (1/2)mv0²; at C, it is (1/2)mvC² + 2mgL (since C sits a height 2L above A); equating these two, and separately using the tension-zero condition at C to pin down vC = √(gL), together give the initial speed needed: v0 = √(5gL) -- a genuinely striking result, reachable only by combining an energy argument with a force argument, neither one alone being sufficient.

A spring's force follows Hooke's law, Fs = -kx, within the spring's elastic limit, where x is the displacement from its natural, unstretched length and k is the spring constant (large k means a stiff spring, small k a soft one) -- the negative sign showing the spring always pushes or pulls back TOWARD its natural length, opposing whatever displacement was applied. Since this force genuinely varies with x, finding the work done in stretching it from 0 to some extension xm needs the same integration already used for variable forces: Ws = the integral of (-kx)dx from 0 to xm = -kxm²/2, matching exactly the area of the triangular region under the straight-line Fs-versus-x graph. This spring force is conservative -- its work depends only on the start and end displacements, and is exactly zero when the block returns to where it started -- so a potential energy function exists: choosing V=0 at the natural length, V(x) = (1/2)kx², and differentiating confirms -dV/dx = -kx, matching the spring force exactly. Pulled out to xm and released, the block's total mechanical energy stays fixed at (1/2)kxm² throughout its subsequent motion, continuously trading between kinetic and potential energy as it passes back and forth through the equilibrium point -- speed and kinetic energy are greatest exactly at x=0 (where all the energy is kinetic) and zero at the extremes x=±xm (where all the energy is potential), the two parabolic energy curves mirroring each other perfectly at every point. Crash-testing a car against a mounted spring puts this to real, practical use: at maximum compression, the car's entire kinetic energy has converted into the spring's potential energy, letting the maximum compression be calculated directly -- and if friction also acts during the compression, a non-conservative force, the conservation of mechanical energy no longer applies on its own; instead, the work-energy theorem must be used directly, with the total mechanical energy change now equal to the work done specifically by the non-conservative force.

Power measures not just how much work gets done, but how fast -- the average power over a time t is simply Pav = W/t, and the instantaneous power, the limit as that time interval shrinks to zero, is P = dW/dt. Since the work done by a force F over a tiny displacement dr is dW = F.dr, dividing through by dt gives a genuinely compact vector formula for instantaneous power: P = F.(dr/dt) = F.v, the force dotted with the object's instantaneous velocity. Power is itself a scalar, measured in watts (W), where 1 watt equals 1 joule per second; the older unit horsepower (hp) equals 746 W, still used to describe vehicle engines; and a family's electricity bill, measured in kilowatt-hours (kWh), is actually a unit of ENERGY, not power, since 1 kWh is the energy delivered by 1 kilowatt of power sustained for 1 hour (1 kWh = 3.6x10⁶ J). A practical use of P=F.v: an elevator motor lifting a load at constant speed must supply power exactly equal to the total opposing force (gravity plus friction) dotted with the elevator's own velocity, letting a motor's required power rating be calculated directly from the load, the friction, and the desired speed alone.

Whenever two objects collide, the mutual forces they exert on each other, however complicated and however they vary during the brief collision time Δt, are always equal and opposite by Newton's third law, F12 = -F21; since each object's change in momentum is exactly that mutual force multiplied by Δt, this directly forces Δp1 + Δp2 = 0 -- momentum is ALWAYS conserved in a collision, regardless of what happens to kinetic energy. Kinetic energy, by contrast, is only sometimes conserved: a collision in which it is fully conserved is called elastic, one in which the two bodies stick together and move as one afterward is completely inelastic, and the far more common intermediate case, where some kinetic energy is lost (usually to heat or sound) but the bodies still separate afterward, is simply inelastic. For a head-on, one-dimensional elastic collision between a moving mass m1 and a stationary mass m2, momentum and kinetic energy conservation together solve completely for both final velocities: v1f = [(m1-m2)/(m1+m2)]v1i and v2f = [2m1/(m1+m2)]v1i. Two special cases stand out sharply: if the masses are exactly equal, the incoming object stops dead and the target moves off with the full original speed; and if one mass vastly outweighs the other (as when a fast neutron strikes a much heavier nucleus), the heavy mass barely moves at all while the light one simply bounces back at nearly its original speed -- exactly why neutrons are slowed most efficiently in a nuclear reactor by colliding with LIGHT nuclei (like deuterium or carbon) rather than heavy ones, since a light nucleus can actually absorb a large fraction of the neutron's kinetic energy in a single collision. In two dimensions, momentum conservation alone supplies two component equations (for a fixed collision plane), but four unknowns (two final speeds, two final angles) remain -- one more piece of information, typically one angle measured experimentally, is always needed to pin the collision down completely; remarkably, when two equal masses collide elastically with one initially at rest, simple algebra shows the two must fly apart at EXACTLY a right angle to each other, a striking, testable prediction familiar to anyone who has broken a rack in a game of billiards.