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"A tunnel is dug through the Earth and a ball is dropped in. Show that it moves in SHM and find its period." It is a classic because it ties gravitation to oscillations. This post goes past the usual answer: a tunnel that does not pass through the centre, the speed at the middle by two methods, and how real Earth differs from the textbook one.
Where is this in the syllabus?
The tunnel itself is not in the NCERT text, but both pieces are. The NCERT Class 11 chapter Gravitation (Reprint 2026-27) uses the shell argument to show that at depth d, g(d) = g(1 − d/R). The Oscillations chapter defines SHM by a = −ω²x. The NTA Syllabus for JEE (Main) 2026 lists the variation of g "with altitude and depth" and SHM, as does the NEET (UG) 2026 syllabus. Our post on showing that a motion is SHM covers the diameter case briefly; here we go further.
Why is the motion simple harmonic?
At distance r from the centre of a uniform Earth, only the mass inside radius r pulls, so the field is g(r) = g r/R, pointing to the centre. The force on a ball of mass m is F = −(mg/R) r.
Now take a straight tunnel that misses the centre: a chord. Let x be the ball's distance from the chord's midpoint. The component of the force along the tunnel is (mg r/R) × (x/r) = (mg/R)x, pointing back to the midpoint. The walls take the rest. So along any straight tunnel, a = −(g/R)x, which is SHM with ω = √(g/R).
Worked example 1: the numbers
Take R = 6371 km, the mean radius in NASA's Earth fact sheet, and g = 9.8 m/s².
ω = √(9.8 / 6.371 × 10⁶) = 1.24 × 10⁻³ rad/s
T = 2π/ω = 5066 s ≈ 84.4 min
One-way trip (end to end is half a cycle): T/2 ≈ 42.2 min
Because ω does not depend on the tunnel, every straight tunnel takes the same 42 minutes, whether it runs 100 km or right through the centre.
How fast is the ball at the centre?
Method 1, SHM: for a diameter tunnel the amplitude is R, so v_max = ωR = √(gR) = √(9.8 × 6.371 × 10⁶) ≈ 7.9 km/s.
Method 2, energy: inside a uniform sphere the potential at the centre is −3GM/2R and at the surface −GM/R, the same maths as in our post on the potential inside a charged solid sphere. The drop is GM/2R = gR/2 per kilogram, so ½v² = gR/2 and v = √(gR) again.
Worked example 2: a chord tunnel
A straight tunnel joins two points 2000 km apart, measured along the chord (sample numbers). The amplitude is half the chord, a = 1.0 × 10⁶ m.
Trip time: still 42.2 min
Top speed, at the midpoint: ωa = 1.24 × 10⁻³ × 1.0 × 10⁶ ≈ 1240 m/s
Speed 500 km from the midpoint: ω√(a² − x²) = 1.24 × 10⁻³ × √(10¹² − 2.5 × 10¹¹) ≈ 1070 m/s
Depth of the midpoint below the surface: R − √(R² − a²) ≈ 79 km
What does the real Earth change?
Earth is denser towards its core, so g does not fall linearly with depth. Using density data from seismology, Klotz (2015), American Journal of Physics found the fall along a diameter takes 38 minutes, against 42 for uniform density, and that the trip time for other straight tunnels lies between 42 minutes for short ones and 38 for long ones. Exam questions assume uniform density, no friction and no rotation; say so in your answer.
What mistakes do students make?
Calling the one-way trip T. End to end is half a period.
Using g as constant. A constant g gives no restoring force proportional to x, so no SHM.
Thinking a short tunnel is faster. The period is the same; a shorter tunnel just means a smaller amplitude and lower top speed.
Using R as the amplitude for a chord. The amplitude is half the chord length.
A practice routine
Derive a = −(g/R)x for a chord tunnel, with a diagram.
Find the top speed for a 500 km chord. (Check: about 310 m/s.)
Get v at the centre by the energy method without notes.
Read our post on gravity inside a sphere with a cavity and predict the motion in a tunnel through a cavity.
When the link between gravitation and SHM is the part you cannot see, the Fermi Concept Library covers every concept in physics, so you can revise g at depth and SHM side by side.
Try practice step 2 now; it takes two lines once the chord derivation is clear.
