The maximum-potential point between two bodies
Along the line joining two bodies, φ reaches a maximum (least negative) value at the point where the resultant field is zero, because g = −dφ/dr and a zero gradient means g = 0. A projectile only needs enough energy to reach that point; beyond it the far body pulls it in.
What this actually means
This is the tidy payoff for having both the vector and the scalar picture. The neutral point is defined by the fields cancelling; the potential maximum is defined by the φ–r curve turning over. They are the SAME point, because g is minus the gradient of φ.
It is a maximum, not a minimum, because φ plunges towards minus infinity at each body and has to come back up in between. Least negative equals largest, so the top of the hump is the highest potential anywhere on the line.
The classic question: what is the minimum speed for a meteorite leaving the Moon to reach the Earth? Answer by energy conservation from the Moon's surface to the maximum-potential point P, arriving there with zero kinetic energy. ½mv² = m(φ_P − φ_Moon).
The follow-up asks why reaching the Earth needs no more speed than reaching P. Because from P onwards the resultant field points towards the Earth, so the Earth's pull does positive work and accelerates the meteorite the rest of the way. P is the top of the hill; after that it is downhill all the way.
Sketch the shape correctly: two deep wells joined by a hump that never rises above the axis. The peak sits much closer to the smaller body, and the whole curve stays negative throughout.
Calling the turning point a minimum, or claiming φ is zero there because the field is zero.
Prove it — watch it be true
- Switch the φ–r graph into the two-body view along the line joining the masses.
- Drop the live tangent on the hump and confirm the gradient is zero exactly where the g–r graph crosses zero.
- Use the work-done two-marker tool from the smaller body's surface to that peak to read the minimum energy per kilogram needed.