Physics · Gravitation · NEET
Yes, they are the same quantity. E = F/m = (GMm/r²)/m = GM/r², which is exactly the formula for g. So E and g have the same value and the same direction at every point. The unit N/kg is dimensionally equal to m/s². We call it 'field intensity' when we think about the field, and 'g' when we think about how fast a body accelerates. For NEET, if a question gives you g at a point, that value is also the field intensity there.
The defining formula is E = F/m, where F is the gravitational force on a test mass m. For a source point mass M at distance r, E = GM/r². The SI unit is N/kg (newton per kilogram), which is the same as m/s². Its dimensional formula is [M⁰ L¹ T⁻²], identical to acceleration.
Field intensity E = F/m is force per unit mass (a vector, unit N/kg). Gravitational potential V = W/m is potential energy per unit mass (a scalar, unit J/kg). E tells you the force direction and strength; V tells you the energy. They are linked by E = −dV/dr. Do not mix their units in NEET problems: N/kg is field, J/kg is potential.
It is a vector. Its direction is the direction of the gravitational force on the test mass, which always points toward the source mass (gravity is attractive). When many masses are present, add their field intensities as vectors. This is different from gravitational potential, which is a scalar and is added algebraically.
Both units are correct and equal. From E = F/m the natural unit is newton per kilogram (N/kg). Since 1 N = 1 kg·m/s², dividing by kg gives m/s². NEET options may use either form, so treat 50 N/kg and 50 m/s² as the same number.
A body of mass 60 g experiences a gravitational force of 3.0 N at a point. The magnitude of the gravitational field intensity at that point is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. E = F/m is independent of the test mass m, because a larger test mass feels a proportionally larger force. E depends only on the source mass M and distance r: E = GM/r².
It follows an inverse-square law: E = GM/r². If you double the distance r, E becomes one-fourth. At the Earth's surface E = g ≈ 9.8 N/kg, and it decreases as you go higher.
It is zero. At the centre, mass pulls equally in all directions, so the net force on a test mass is zero, giving E = 0. Inside a uniform Earth, E is proportional to the distance from the centre (E ∝ r).
Add them as vectors. Find the field due to each mass separately (E = GM/r² toward that mass), then take the vector sum. The point where the total field is zero is called the null point.
