Module 1: Mathematical Foundations · Practice · assumes Foundation 10
Four worked examples on the approximation JEE uses most — including the one where Newton’s mechanics falls out of Einstein’s.
What this set is for
Foundation 10 called (1 + x)n ≈ 1 + nx the most used approximation in JEE physics and then demonstrated it once. These four use it properly — on gravity at altitude, on a subtraction that would otherwise lose its answer to rounding, and on relativity, where it recovers the whole of classical mechanics from a formula that looks nothing like it.
The last example runs the other way, keeping a term the approximation usually discards, to see what was being thrown away. Same three-part structure throughout.
On this page
1. Gravity a little way up
Problem. The gravitational acceleration at height h above the Earth’s surface is g(h) = g0(1 + h/R)−2, where R = 6400 km. Derive a first-order formula valid for h « R, and find the fractional decrease in g at an altitude of 6.4 km — roughly the cruising height of an airliner.
The physics
The exact law is an inverse square, which is awkward: it means the weight of an aircraft depends on its altitude in a way that no simple formula captures. But 6.4 km against an Earth radius of 6400 km is a thousandth part, and over such a small fraction of the distance the inverse square cannot curve much. Expect the answer to be a small fraction of a percent, and expect the exact curve to be indistinguishable from a straight line over this range.
Left: an approximation that degrades gently. Right: a term that was set to zero, restored.
The mathematics
The expression is already in the standard form, with x = h/R and n = −2:
(1 + x)n ≈ 1 + nx, so (1 + h/R)−2 ≈ 1 − 2h/R
g(h) ≈ g0(1 − 2h/R)
The fractional change follows at once:
Δg/g0 ≈ −2h/R = −2(6.4/6400) = −0.002
a decrease of 0.2%
How good is it? The exact value is (1.001)−2 = 0.998003, against the approximate 0.998000 — agreeing to five decimal places. The discarded terms are of order (h/R)² = 10−6, which is a million times smaller than the quantity being computed and far below the precision of any measurement of g.
Note the factor of 2. A student who reasons that gravity falls off with distance and so drops by about 0.1% is out by a factor of two, because the inverse square means each part per thousand of extra distance costs two parts per thousand of field. That factor is the entire content of the approximation.
The contemplation
Foundation 10 worked this at Everest’s summit and found 0.28%; here it is 0.2% at an aircraft’s altitude. The repetition is worth noticing rather than skipping, because it is the same three lines both times and it will be the same three lines every time this shape appears. That is what makes an approximation worth learning as a technique rather than as a result: the physical situations differ, the algebra does not. And notice what has been bought — an inverse square, which is genuinely non-linear, has become locally a straight line. Every “uniform gravitational field” you will ever assume rests on exactly this step.
2. A difference that a calculator loses
Problem. A satellite orbits at radius r = 1.0000 × 107 m. A second satellite orbits 20 m higher. Find the difference in the gravitational field strength between the two orbits, given GM = 3.986 × 1014 m³/s².
The physics
Two field strengths, each around 4 m/s², differing by a tiny amount because 20 m is nothing against ten thousand kilometres. The obvious approach is to compute both and subtract — and that approach fails, not because the physics is wrong but because two nearly equal numbers subtracted leave almost nothing but the rounding errors of the two. This is a situation where the approximation is not a shortcut; it is the only route to a trustworthy answer.
The mathematics
The direct route, and why it disappoints. The two fields are GM/r² and GM/(r + 20)²:
g1 = 3.986×1014 / 1014 = 3.98600000 m/s²
g2 = 3.986×1014 / (1.000002×107)² = 3.98598406 m/s²
A calculator carrying eight significant figures gives a difference of about 1.6 × 10−5. One carrying six figures gives zero. The answer lives entirely in digits most devices have already discarded.
The approximation. Factor out and expand instead:
g2 = GM/(r + Δr)² = (GM/r²)(1 + Δr/r)−2 ≈ g1(1 − 2Δr/r)
so Δg = g2 − g1 ≈ −2g1 Δr/r
= −2(3.986)(20 / 107) = −1.594 × 10−5 m/s²
Four significant figures, from arithmetic you could do on paper. The subtraction was performed algebraically, before any numbers were substituted, so the two large quantities cancelled symbolically and never had to cancel numerically.
The contemplation
Here is a case where the approximate method is more accurate than the exact one, which sounds impossible and is not. The exact formula is perfectly correct; what fails is evaluating it numerically, because subtracting two numbers that agree to seven digits leaves a result determined by the digits that were rounded away. The approximation avoids the subtraction entirely by doing it in symbols first. This is worth remembering beyond exams: whenever a physical question asks for a small difference between large quantities, the instruction is to expand rather than to compute and subtract.
3. Newton inside Einstein
Problem. In special relativity the kinetic energy of a particle is K = (γ − 1)mc², where γ = (1 − v²/c²)−1/2. Show that for v « c this reduces to K ≈ ½mv². Then estimate the error in the classical formula for a satellite moving at 8 km/s.
The physics
Two formulas for the same quantity, written in symbols that share almost nothing. One involves the speed of light and a square root; the other is the expression from school. They must agree in the regime where both are believed to apply, because objects moving slowly have a kinetic energy and it does not depend on which theory you consult. Showing that they do is the point — and the mechanism is the binomial approximation, applied to a fractional power.
The mathematics
Put x = v²/c², which is small when v « c. Then γ = (1 − x)−1/2, and applying the approximation with n = −½ to the quantity (−x):
γ ≈ 1 + (−½)(−x) = 1 + ½x = 1 + v²/(2c²)
Substituting into the kinetic energy, the two ones cancel:
K = (γ − 1)mc² ≈ [v²/(2c²)] mc²
= ½mv²
The c² cancels completely, which is why the classical formula contains no trace of the speed of light despite being an approximation to a formula built around it.
The size of the error. At 8 km/s, v/c = 8000 / (3×108) = 2.67×10−5, so x = v²/c² ≈ 7.1×10−10. The first discarded term is of order x², relative to the retained term of order x, so the fractional error is roughly x itself — about one part in a billion. For a 1000 kg satellite carrying 3.2×1010 J of kinetic energy, that is an error of about 20 J: less than the energy of a dropped textbook.
The contemplation
Classical mechanics is not a rival to relativity that happened to lose. It is relativity’s first-order approximation, and this calculation is the proof — three lines, and ½mv² emerges from a formula that mentions the speed of light. Foundation 10 claimed that every physical theory is a model with a domain, and that Newton’s mechanics fails near light speed. Here is the same statement from inside: the failure is not a breakdown but a departure, growing as v²/c² grows, and utterly negligible for a satellite. What you were taught at school was never wrong. It was bounded, and nobody mentioned the boundary.
4. What the small-angle approximation threw away
Problem. The period of a pendulum released from amplitude θ0 is more accurately T = T0[1 + ¼sin²(θ0/2) + …], where T0 = 2π√(L/g). Find the fractional increase in period at amplitudes of 0.2 rad and 60°, and comment on when T0 may be used.
The physics
Foundation 10 derived T0 = 2π√(L/g) using sin θ ≈ θ, and that formula contains no amplitude at all — it says a pendulum swinging through 5° and one swinging through 50° take exactly the same time. That is a strong claim, and it is false; it is what the approximation asserts, not what pendulums do. The formula above restores the leading correction, and the question is how large it is.
The mathematics
At 0.2 rad (about 11.5°), the angle is small enough to apply sin x ≈ x inside the correction itself:
sin²(θ0/2) ≈ (θ0/2)² = θ0²/4
ΔT/T0 ≈ θ0²/16 = 0.04/16
= 0.0025, or 0.25%
At 60° = π/3 rad, the angle is no longer small, so the correction must be evaluated directly rather than approximated further:
sin(30°) = 0.5, so sin²(θ0/2) = 0.25
ΔT/T0 ≈ ¼(0.25) = 0.0625, or 6.25%
Note that using θ0²/16 at 60° would have given 6.85% — the small-angle step inside the correction has itself started to fail, which is a reminder that an approximation applied inside an approximation needs its own justification.
When may T0 be used? A pendulum clock losing 0.25% gains about four minutes a day, which is intolerable for a clock and irrelevant for a laboratory measurement of g to two figures. The answer depends entirely on the tolerance of the task, which is the honest position rather than an evasion.
The contemplation
Every other example in this set discarded a term. This one picked one back up, and out of it came a physical effect the approximation had denied entirely — that a pendulum’s period depends on how far it swings. The dependence was never absent from the pendulum; it was absent from the model, because sin θ ≈ θ had removed the only place amplitude could have entered. That is what an approximation does, and it is worth stating plainly: it does not make a quantity small, it makes it zero, and the effect vanishes from your description whether or not it has vanished from the world.
One approximation, four jobs
The same three-line step turned an inverse square into a straight line, rescued a subtraction that arithmetic could not perform, produced the whole of classical kinetic energy from a relativistic formula, and — run backwards — restored an effect it had previously erased.
Foundation 10 argued that the binomial expansion, the small-angle results and the tangent line are one idea in three costumes: close enough to a point, every smooth curve is straight. Nothing in these four examples required more than that.
Assumes: Foundation 10 · Type: Worked examples · Reading: 20 min · Difficulty: Intermediate