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← all cardsTopic 7 · Gravitational Field · Newton's law & field strength

The four quantities: two vectors, two scalars

F = GMm/r² (vector, N) and g = GM/r² (vector, N kg⁻¹) fall off as 1/r² and are always attractive. U = −GMm/r (scalar, J) and φ = −GM/r (scalar, J kg⁻¹) fall off as 1/r and are always negative. Linked by g = −dφ/dr and F = −dU/dr.

per unit mass: g = F/m, φ = U/m · gradient link: g = −dφ/dr, F = −dU/dr

What this actually means

Build the 2 × 2 table once and the whole topic collapses into it. Columns: vector and scalar. Rows: between two masses (F, U) and at a point per unit mass (g, φ). Every gravitation formula lives in one of the four boxes.

The addition rule follows from the column. Field strength and force are vectors, so for two bodies you resolve by direction and they can cancel to zero. Potential and potential energy are scalars, so you add them algebraically and they can never cancel, since both are negative.

The r-dependence follows from the row: the force-type quantities go as 1/r², the energy-type quantities as 1/r. That single fact answers most of the graph-shape MCQs without any calculation.

Signs: F and g have no minus sign when you are quoting magnitudes, because attraction is handled by the stated direction. U and φ carry a compulsory minus sign, because their zero is at infinity. Never let a stray minus wander from one column to the other.

The bottom row of the table is the bridge between the columns: g = −dφ/dr and F = −dU/dr. Force-type equals minus the gradient of energy-type. That is the relationship graph questions are built on.

This exact structure returns in Electric Fields with charge in place of mass, so a table you build now is a table you reuse in a whole other topic.

The trap

Adding field strengths as if they were scalars, or trying to resolve potentials into components.

Prove it — watch it be true

  1. Display the φ–r and g–r graphs side by side and compare how fast each falls towards zero.
  2. Confirm φ is negative everywhere while the g magnitude is always positive and drops off faster.
  3. Drop the live tangent onto the φ–r curve and read its gradient against the g value at the same r.
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