The neutral point between two masses
Between two masses there is a point where the two fields are equal in magnitude and opposite in direction, so the resultant field strength is zero. It lies closer to the smaller mass.
GM₁/x² = GM₂/(d − x)² ⇒ x/(d − x) = √(M₁/M₂)What this actually means
Between the two bodies the fields point in opposite directions, so setting their MAGNITUDES equal is enough. G cancels immediately, then take the square root of both sides to turn a quadratic into a linear equation. That square-root step is what keeps the algebra clean.
For Earth and Moon, √(M_E/M_M) = √(6.0 × 10²⁴ / 7.4 × 10²²) = 9.0, so x = 9d/10. The null sits about 3.4 × 10⁸ m from Earth, roughly 90% of the way to the Moon. It hugs the lighter body, which makes intuitive sense: you have to get very close to the weakling before it can match the giant.
Squaring both sides instead throws up a second root that lies OUTSIDE the two masses, where the fields point the same way and cannot cancel. Reject it and say why. 'Taking the physically valid root' is a real mark in some schemes.
The potential at this point is NOT zero. Field strength cancels because it is a vector; potential is a scalar and both contributions are negative, so they add to something distinctly negative. Mixing these up is a classic MCQ trap.
What IS true at the null point is that φ is a MAXIMUM (least negative) along the line joining the masses, because g = −dφ/dr and g = 0 means the gradient is zero. That fact powers the 'minimum speed to travel from the Moon to the Earth' question.
Assuming the potential is also zero at the neutral point, or keeping the unphysical root outside the two masses.
Prove it — watch it be true
- Turn on the two-mass mode so both bodies and their fields are shown.
- Slide the test mass probe along the line joining them and watch the two field arrows fight, one shrinking as the other grows.
- Snap to the computed neutral point and confirm the resultant g readout goes to zero, and that the point sits close to the smaller mass.