Why the test mass must be small
The test mass must be small because it produces its own gravitational field, which would distort the field being measured. A small mass keeps that disturbance negligible.
What this actually means
This is a stock two-mark explain question and the answer is two clauses: the test mass has its own field, and that field would disturb the source and change the very thing you are measuring.
Physically, a heavy test mass pulls the source body towards it. The source shifts, the separation changes, and the g you measure is no longer the g that existed before you arrived. Measurement should not vandalise the thing measured.
Say 'negligible', not 'zero'. The disturbance never truly disappears; it just becomes small enough to ignore. Strictly the definition uses a limiting process, g = lim(m→0) F/m, though H2 does not require you to write that.
The identical argument reappears in Electric Fields for the small positive test CHARGE, so learning it once buys you the mark twice. Swap mass for charge and field for electric field and the sentence transfers unchanged.
Do not confuse this with the reason m cancels in the derivation. m cancels because the definition divides by it. The test mass must be SMALL for a completely separate reason, namely non-disturbance.
Saying only 'so it does not affect the field' without stating that the test mass has a field of its own.
Prove it — watch it be true
- Switch to two-mass mode and set the second mass to something tiny, then look at the field vectors near the planet.
- Raise the second mass towards planetary scale and watch the field pattern warp and a neutral point appear.
- Take the second mass back down and see the pattern relax to the undisturbed radial field.