Account for energy stored in a gravitational field and in a stretched or compressed spring

Learning goal: Explain gravitational potential energy as energy stored in the field of a system of attracting masses, calculate it near Earth's surface and for widely separated bodies, read the spring constant and the stored energy from a force vs. distance graph, and calculate the elastic potential energy of a spring stretched or compressed a given distance.

Energy belongs to the system, not to one object. In earlier courses you calculated gravitational potential energy with mgh and spoke of the ball "having" potential energy. In this course the picture is refined. When a ball and Earth pull on each other, both change speed (Earth by an amount far too small to notice), yet neither one is handing its own energy to the other. The energy is exchanged with the gravitational field that the pair shares, and the amount held there is what we call gravitational potential energy. A single mass alone in space still has its own gravitational field, but there is no second object in that field to interact with, so there is no gravitational interaction energy to speak of. Gravitational potential energy belongs to the system of two or more attracting masses. Pulling the masses apart takes work against their attraction, and that work is stored in the field; letting them come together draws the stored energy back out, usually as kinetic energy. A ball thrown upward is slowing because its kinetic energy is flowing into the ball-Earth field; as it falls, the field returns that energy as kinetic energy. Saying "the ball has 29 J of potential energy" is a convenient shorthand, but the honest statement is "the ball-Earth system has 29 J more energy stored in its field than when the ball was on the floor".

Near Earth's surface: ΔU = mgΔh. When an object moves through a height change Δh that is tiny compared with Earth's radius (6.37 × 10⁶ m), the gravitational field strength g is very nearly constant at 9.8 N/kg, and the change in the system's gravitational potential energy is ΔU = mgΔh. Two conditions matter. First, the formula gives a change; you are free to call any level "zero" and only differences have physical meaning. Second, it is an approximation that fails when Δh is large enough for g to change noticeably. Worked model: a 2.0 kg book is lifted 1.5 m from the floor to a shelf. ΔU = (2.0 kg)(9.8 N/kg)(1.5 m) = 29.4 J. The person's muscles transferred 29.4 J into the book-Earth field. When the book later falls, the field transfers 29.4 J back out, and (ignoring air resistance) the book reaches the floor with 29.4 J of kinetic energy.

Large separations: U = −GMm/r. For a satellite hundreds of kilometers up, or the Moon, g is not constant and mgΔh overestimates the energy. The general expression for the gravitational potential energy of two point-like masses M and m whose centers are a distance r apart is U = −GMm/r, with G = 6.67 × 10⁻¹¹ N·m²/kg². Here the zero of energy is chosen at infinite separation, so every bound system has negative U; less negative means more energy stored in the field. The change when a body moves from r₁ to r₂ is ΔU = GMm(1/r₁ − 1/r₂). Worked model: how much energy must go into the field to raise an 800 kg satellite from Earth's surface (r₁ = 6.37 × 10⁶ m) to a 400 km altitude (r₂ = 6.77 × 10⁶ m)? Using M = 5.97 × 10²⁴ kg: ΔU = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(800)(1/(6.37 × 10⁶) − 1/(6.77 × 10⁶)) = 2.95 × 10⁹ J. The near-surface formula gives mgΔh = (800)(9.8)(4.0 × 10⁵) = 3.14 × 10⁹ J, about 6 % too high, because g falls off as the satellite climbs. (Putting the satellite into orbit also requires kinetic energy, which this calculation does not include.)

Elastic potential energy and the force vs. distance graph. A spring, bungee cord or rubber band stores energy when stretched or compressed from its relaxed length. For an ideal spring kept within its proportional (straight-line) range, the restoring force is proportional to the change in length: F = kΔx, where k is the spring constant in newtons per meter (N/m) and Δx is the distance stretched or compressed from the relaxed length. Plot force on the vertical axis and distance on the horizontal axis and you get a straight line through the origin. The slope of that line is the spring constant k. The area under the line, from zero out to a stretch Δx, is the work done stretching the spring, which is the energy stored in it: a triangle of base Δx and height kΔx, so ΔE_elastic = ½ k Δx². The area, not F × Δx, is the right calculation because the force is not constant: it grows from zero as the spring is pulled, and as a spring oscillates back and forth the force on the attached mass keeps changing, largest at the extremes and zero at the relaxed length. Worked model: a spring with k = 50 N/m is stretched 0.20 m. The force at that stretch is F = (50 N/m)(0.20 m) = 10 N. The stored energy is ½(50)(0.20)² = 1.0 J, which is exactly the area of the triangle under the graph from 0 to 0.20 m: ½ × 0.20 m × 10 N = 1.0 J. Stretching it to 0.40 m stores ½(50)(0.40)² = 4.0 J, four times as much for twice the stretch, because the energy depends on the square of Δx.

Misconception: doubling the stretch doubles the energy. It doubles the force but quadruples the energy, because the extra stretch happens against a larger force. The triangle on the graph shows this directly: doubling the base also doubles the height.

Designing with the energy account. A bungee drop can be designed with calculations alone. Suppose a 0.060 kg egg in a harness is dropped from a platform 3.0 m above the floor on a cord of relaxed length 1.5 m, and the goal is to stop it 0.30 m above the floor. At the lowest point the egg is momentarily at rest, so all the energy that left the egg-Earth field, mgΔh with Δh = 2.7 m, has gone into the cord: (0.060)(9.8)(2.7) = 1.59 J. The cord has stretched Δx = 2.7 − 1.5 = 1.2 m, so ½ k (1.2)² = 1.59 J gives k = 2.2 N/m. At the lowest point the elastic force is kΔx = (2.205)(1.2) = 2.6 N upward while the egg's weight is 0.59 N downward, so the net force is strongly upward and the egg accelerates back up; the cord must be attached firmly and the egg will bounce. Real cords are not ideal springs, so a test drop from a lower height comes first and the design is revised from the data.

Investigation: measure k and the stored energy. Materials: a steel spring, a second spring of different stiffness (or a length of bungee shock cord), a rubber band for comparison, a ring stand and clamp, a meter stick, a set of slotted masses up to 500 g, and safety glasses. Steps: hang the spring, record its relaxed length, then add masses one at a time and record the total stretch for each, keeping well below the point where the spring stops returning to its original length. Plot force (mg) against stretch. A result that supports the model is a straight line through the origin; its slope is k, and the area under it out to any stretch equals ½ k Δx². Repeat with the second spring or the cord, check how straight its graph is before quoting a k, and compare the k values (designing, building and testing a bungee drop from a measured k is the performance task in the lesson on elastic forces and Hooke's law). Last, load and then unload the rubber band, recording the stretch at each step both ways. What the model cannot show: a rubber band's graph curves, so it has no single k and ½kΔx² does not apply to it, and its unloading curve lies below its loading curve. The area under the loading curve is the work put in, the area under the unloading curve is the work recovered, and the area between them is the energy that warms the band during each cycle. Never release a loaded spring so that it snaps toward a face; unload masses by hand.

Worked example 1 — an energy account for lifting and lowering. A 4.0 kg toolbox is lifted 0.80 m onto a bench and later lowered to the floor. Lifting: ΔU = (4.0)(9.8)(0.80) = 31.4 J flows from the person into the toolbox-Earth field. Lowering slowly: the field gives up 31.4 J, and because the box does not speed up, that energy is transferred into the person's arms (and ultimately to thermal energy), not into kinetic energy. The field is the store; neither the box nor Earth alone is the source or the receiver.

Worked example 2 — a satellite, two ways. For the 800 kg satellite raised 400 km, the field formula gives ΔU = GMm(1/r₁ − 1/r₂) = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(800)(1.570 × 10⁻⁷ − 1.477 × 10⁻⁷ m⁻¹) = 2.95 × 10⁹ J, while mgΔh gives 3.14 × 10⁹ J. The near-surface formula overestimates because it assumes g stays at 9.8 N/kg all the way up, whereas at 400 km the field strength is already smaller. The sign is positive in both cases: energy went into the field because the masses were separated.

Worked example 3 — a compressed spring. A spring with k = 300 N/m is compressed 0.15 m in a toy launcher. Force at full compression: F = (300)(0.15) = 45 N. Energy stored: ½(300)(0.15)² = 3.375 J ≈ 3.4 J. If the graph of force vs. compression were drawn, it would be a straight line from (0, 0) to (0.15 m, 45 N), and the triangle under it has area ½(0.15)(45) = 3.375 J, the same answer.

Worked example 4 — reading a graph. A force vs. stretch graph for an unknown spring passes through (0.10 m, 8.0 N) and (0.30 m, 24 N). Slope = (24 − 8.0)/(0.30 − 0.10) = 80 N/m, so k = 80 N/m. Energy stored at 0.30 m: ½(80)(0.30)² = 3.6 J. Energy stored between 0.10 m and 0.30 m: 3.6 − ½(80)(0.10)² = 3.6 − 0.4 = 3.2 J, which is the area of the trapezoid under the line between those two stretches.

Representation task: graph and annotate. Draw a force vs. stretch graph for a spring with k = 80 N/m from 0 to 0.25 m, with labeled axes in newtons and meters. Shade the area under the line up to 0.25 m, calculate it (½ × 0.25 m × 20 N = 2.5 J) and check that ½ k Δx² gives the same value. Add a second line for a stiffer spring (k = 160 N/m) and write two sentences comparing the slopes and the areas at the same stretch. Then draw a separate energy bar chart for a ball thrown upward at three moments (release, halfway up, top), showing kinetic energy and field energy as bars whose total stays constant.

Independent performance: measure a spring constant and predict a stretch. Carry out the hanging-mass investigation with a steel spring whose force vs. stretch graph you have checked is a straight line, and write a report. Success criteria: a data table with at least five mass-stretch pairs in SI units; a force vs. stretch graph with a best-fit line and its slope stated as k in N/m; the stored energy at the largest stretch calculated both from the area and from ½ k Δx², with the two values agreeing; a prediction of the stretch for a mass you did not test, checked by measurement, with the percent difference; one paragraph explaining why gravitational potential energy is described as belonging to the mass-Earth system and not to the mass alone; and a statement of one thing the straight-line model cannot show about the real spring, or about the rubber band if you tested one.