g = −dφ/dr : field is minus the potential gradient
Gravitational field strength is minus the potential gradient: g = −dφ/dr. Equivalently F = −dU/dr. The field points in the direction of decreasing potential, and the magnitude of g equals the gradient of the φ–r graph.
g = −dφ/dr · F = −dU/dr · |g| = |gradient of φ–r graph|What this actually means
The minus sign is not decoration. It says the field points DOWNHILL in potential, towards more negative φ, which is towards the mass. Write g = +dφ/dr and you have pointed gravity outwards.
Check the signs with numbers rather than trusting memory. φ = −GM/r, so dφ/dr = +GM/r². Then g = −dφ/dr = −GM/r², where the minus sign simply says the vector points inwards, towards decreasing r. Magnitude GM/r², as expected.
In graph questions this is a gradient-reading skill, not an algebra skill. Given a φ–r curve, the magnitude of g at any point is the steepness of the tangent there. Steep curve, strong field. Flat curve, weak field.
The flat-tangent case is worth its own note. Where the φ–r curve reaches a maximum, between two bodies, the gradient is zero, so g = 0. That is precisely the neutral point, arrived at from the energy side instead of the force side.
The reverse operation shows up too: the area under a g–r graph between two radii gives the change in potential, since Δφ = −∫g dr. H2 rarely asks you to integrate, but recognising area-means-energy stops you panicking at a g–r graph.
Dropping the minus sign, which reverses the field direction, or reading potential where the gradient is wanted.
Prove it — watch it be true
- Show the φ–r and g–r graphs together with the live tangent tool on the φ curve.
- Slide the tangent along and compare its gradient with the g value read from the g–r graph at the same r.
- Move in close where the φ curve is steep and confirm both the gradient and g shoot up together.