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Questions on a sphere with a hole cut out of it look hard, because the shape has no simple symmetry. They are not hard once you know two shell results and one trick: fill the hole in and then subtract it. This post explains the shell theorem, the field inside a solid sphere, and the superposition method, with two worked examples and the mistakes examiners look for.
What does the shell theorem say?
NCERT's Gravitation chapter states two results for a hollow spherical shell of uniform density:
For a point mass outside, the force is as if the entire mass of the shell were concentrated at its centre.
For a point mass inside, the force is zero.
NCERT adds a caution: unlike a metal shell in electrostatics, a mass shell does not shield bodies outside it from pulling on a particle inside. Only the shell's own pull cancels.
The JEE (Advanced) 2026 brochure lists "gravitational potential and field", and the JEE (Main) 2026 syllabus lists the variation of g with depth, which rests on these results. Check your exam year's syllabus.
What is the field inside a solid sphere?
OpenStax University Physics states the consequence: inside a sphere, only the mass within radius r contributes, and it acts as if at the centre. For uniform density ρ this gives
g = (4/3)πGρr.
So g grows in a straight line from zero at the centre to its surface value at r = R, then falls as 1/r² outside. A handy form: g(r) = g_s × r/R inside, where g_s = GM/R² is the surface value. Our post on the variation of g with height and depth uses the same result for a mine.
How do you handle a cavity?
Use superposition. A sphere with a spherical hole equals a full sphere plus a smaller sphere of negative mass that fills the hole. Find each field as a vector and add them. Do not subtract the masses and then use one GM/r² formula from the big centre, because the hole is not centred there.
Worked example 1: a point outside
A uniform sphere of radius R had mass M0 before a spherical cavity of radius R/2 was cut from it. The cavity's centre is R/2 from the sphere's centre O, so it touches the surface. Find the field at a point P on the line through both centres, 2R from O on the cavity's side.
Full sphere: GM0/(2R)² = GM0/(4R²), towards O.
Filling sphere: its mass is M0/8, because volume goes as radius cubed. Its centre is 2R − R/2 = 3R/2 from P. Field: G(M0/8)/(9R²/4) = GM0/(18R²), towards the cavity.
Net: both pulls point the same way along the line, so subtract: GM0/R² × (1/4 − 1/18) = 7GM0/(36R²).
In terms of the remaining mass M = 7M0/8, this is 2GM/(9R²). If the full sphere had a surface field of 9.8 m/s², the field at P would be 9.8 × 7/36 ≈ 1.91 m/s².
Worked example 2: inside the cavity
Inside a uniform sphere the field is a vector pointing to the centre: g = −(4/3)πGρ r, where r is the position from O. Inside the cavity, add the full sphere's field and the negative sphere's field, which is +(4/3)πGρ(r − a), where a is the position of the cavity's centre:
g = −(4/3)πGρ r + (4/3)πGρ(r − a) = −(4/3)πGρ a.
The position r has cancelled. The field is the same everywhere in the cavity, pointing parallel to the line from the cavity's centre to O. With a = R/2 and (4/3)πGρ = GM0/R³, its size is GM0/(2R²): half the full sphere's surface value, or 4.9 m/s² in the example above.
Why is superposition hard for students?
The same reasoning appears in electrostatics. Singh (2006) studied Gauss's law with 541 introductory students and found that using it well implicitly requires understanding superposition and symmetry, and that undergraduates had many common difficulties with these ideas. Practising the cavity problem in gravity builds the same skill for charged spheres with holes.
Which mistakes cost marks?
Subtracting the cavity's mass and then using G(M0 − M_cavity)/r² from the big centre.
Assuming the field inside a cavity is zero. That holds only for a concentric hole.
Adding magnitudes when the two fields point in different directions.
Applying the shell theorem to non-spherical or non-uniform shells.
A practice routine
Derive g = (4/3)πGρr inside a uniform sphere.
Sketch g against r from 0 to 3R for a solid sphere and for a shell.
Redo worked example 1 for a point on the side away from the cavity. (Answer: GM0/R² × (1/4 − 1/50) = 23GM0/(100R²).)
Prove the cavity field is uniform for a cavity of any size and position inside the sphere.
If the shell results feel like rules rather than reasons, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Next time you see a sphere with a hole, write "full minus filling" as your first line.
