Physics · Gravitation · NEET
No. This is the most common mistake. Field is a vector, so two opposite pulls can cancel and give zero field. Potential is a scalar and is always negative for gravity, so two negative potentials ADD up. At the null point V = −G·M1/x − G·M2/(d−x), which is a large negative number, never zero. Zero field and zero potential are two different points.
Put the field magnitudes equal and drop the vector signs, because at the null point the two fields point in opposite directions. G·M1/x² = G·M2/(d−x)². Cancel G and take the square root: √M1/x = √M2/(d−x). Rearrange to x = d / (1 + √(M2/M1)), measured from mass M1. Always check your answer sits between the two masses.
Closer to the SMALLER mass. A field falls off as 1/r². To match the strong field of the big mass, you must move far from it and sit near the small mass so its weak field grows enough to cancel. If M2 = 9·M1, the point is at x = R/4 from the small mass and 3R/4 from the big mass.
For two masses that attract, the null point lies only BETWEEN them on the joining line. Outside the pair, both fields point the same way (toward the masses) and can never cancel. So look only in the gap between M1 and M2.
The net gravitational force there is zero, so a body placed exactly at rest stays at rest for an instant. But it is an unstable point: a tiny push toward either mass makes that mass pull harder, and the body falls into it. So the null point is a balance point, not a safe resting place.
Two bodies of mass m and 9m are placed a distance R apart. The gravitational potential at the point on the line joining them where the gravitational field is zero is (G = gravitational constant):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the point on the line joining two masses where their gravitational fields are equal and opposite, so the net gravitational field (and net force on a test mass) is zero.
Measured from mass M1, the null point is at x = d / (1 + √(M2/M1)), where d is the separation. It comes from setting G·M1/x² = G·M2/(d−x)².
Potential is a scalar and is always negative for gravity, so the two contributions add instead of cancelling. Only the vector field cancels, giving zero field but a large negative potential.
At R/4 from the small mass m and 3R/4 from the large mass 9m, because the point sits closer to the smaller mass.
It checks whether you know the difference between a vector (field) and a scalar (potential) in one problem. Mixing them up is a classic error that NTA uses to separate careful students.
