I have read that gravitational field intensity and acceleration due to gravity are two different physical quantities that have the same direction, magnitude, and units. So, if all the units, magnitudes, and directions are the same for the two quantities (and also, physically, both essentially mean the acceleration produced in a point mass at a point) then what is the difference between them?

Edit (Comment)

As a couple of answers point out, physically, the two quantities are different as the former (the field intensity) is the quantity that describes the physical entity that the gravitational field is at a certain point whereas the latter (the acceleration due to gravity) describes the acceleration (a characteristic of the motion) of the particle put in the field. The answer by @jaydesai10 makes this difference rather transparent via pointing out that the field intensity will always be present at a point (as long as the source of the gravitational field is around) but the acceleration due to gravity will be relevant only when an actual particle is put in the field to experience the field.

I would like to point out that the curious fact that these two (meant to be different) quantities generically turn out to be the same in their values is thus a non-trivial fact--in other words, a law of nature. The popular name for this law is, of course, the (weak) equivalence principle. The fact that this is, in fact, non-trivial can be seen via comparing the situation to the situation in electrostatics where the electric field intensity at a point and the acceleration produced (due to the electric field) in a charged particle put at that point are clearly two different quantities. The relation between these two quantities depends on the ratio of the electric charge and the inertial mass of the particle in question. And this ratio, as we know, turns out to be different for different particles--unlike the gravitational scenario where the ratio of the gravitational mass and the inertial mass is independent of the particle in question (which is essentially the statement of the weak equivalence principle).

  • 2
    $\begingroup$ Could you mention/quote which book says that? $\endgroup$
    – jinawee
    Jan 30, 2014 at 22:35

6 Answers 6


The two quantities are on opposite sides of Newton's second law equation $\vec F=m\,\vec a$

The force on a mass $m$ in a gravitation field $\vec g (= g \,\hat d)$ is $\vec F = m \,\vec g = m\, g\,\hat d$ where $g$ is the magnitude of the gravitational field strength and $\hat d$ is the unit vector in the down direction.

Assuming no air resistance then using this force and Newton's second law you can find the acceleration of the mass in free fall.

$\vec F =m\, \vec a \Rightarrow m\, g\,\hat d = m\,\vec a = m\,a\, \hat d \Rightarrow \vec a = a \,\hat d = g \,\hat d$ where $a$ is the magnitude of the acceleration.

So the acceleration of free fall $\vec a$ has the same magnitude as the gravitational field strength $g$ and is in the same direction $\hat d$.

To differentiate between the two quantities you can use $\rm N\, kg^{-1}$ as the unit of gravitational field strength and $\rm m\, s^{-2}$ as the unit of acceleration although dimensionally they are the same.


Gravitational Intensity and Gravitation acceleration , even though have same dimensions are different physical quantities.

Gravitational Intensity of a mass body A at a given point is defined as the force on a unit mass body. It is just a physical quantity that is defined to help us find out the force exerted by the mass body A on any given mass in its field. So, if at any particular instant the gravitational intensity at a given is E , it does not imply that the gravitational acceleration of any mass at that point is equal to E, it's gravitational acceleration corresponds to the resultant force on it due to other bodies . For eg. Consider Moon, to find out the force exerted by the sun on it, we find out the gravitational intensity of the Sun at that point , but that does not mean that it is the gravitational acceleration.

Even for an isolated system of two bodies, gravitational intensity at a point is the property of the gravitational field associated with the mass A in consideration above, while the gravitational acceleration is the property of the other mass which is present in mass A 's gravitational field, that is even if the other body isn't present, gravitational intensity will be defined whereas gravitational acceleration will not be.

  • $\begingroup$ And what does the electric field have to do with gravity? $\endgroup$ Feb 3, 2014 at 4:21
  • $\begingroup$ -1. Please clarify your definitions of "gravitational intensity" and "gravitational acceleration". With the definitions I'd go with (g-field as the sum of all fields produced by all point masses) the statement "it does not imply that the gravitational acceleration of any [point] mass at that point is equal to E" is false. (the field produced by the point mass itself is ignored) $\endgroup$
    – user12029
    Feb 3, 2014 at 6:23
  • $\begingroup$ This answer is wrong . Indeed , the gravitational intensity and gravitational acceleration on Moon are same due to sun . Why are you including earth in one case and not in the other case ? $\endgroup$
    – Abbas
    Sep 13, 2022 at 6:12

1) Gravitation, due to mass alone, is acceleration, m/sec^2.

2) Gravity includes Earth's spin and explains how plumb lines are still normal to the (smooth) surface. Spin also enters gravitomagnetic effects like frame dragging.

3) Gravitational potential explains tides and changes the observed rate of clocks versus their altitude, J/kg. The gravitational field, acceleration of a test mass near a massive object, is the negative gradient of the gravitational potential. In GR, it is replaced by the metric tensor.

4) Gravitational field is phenomenological, N/kg.


I think that the difference in the gravitational field and the gravitational acceleration comes directly from their definition. Quoting from HC Verma’s book

It is assumed that first a body creates a gravitational field around itself and then this field exerts a force on another body.

The author clearly says that the gravitational field gives rises to the gravitational acceleration or in other words, without gravitational field, gravitational acceleration cannot exist but if there is no gravitational acceleration of a body that doesn’t mean that there is no gravitational field at that point in space.


Gravitational field vector is indeed same as gravitational acceleration vector . However , a physical quantity is not only characterised by it's vector properties. Their magnitude , direction and units are same .

However , their concepts are different . The gravitational acceleration is physically $dv/dt$ whereas gravitational field is a field . It can contain and transport energy and momentum . Fields are not just a mathematically defined object , but a entity having much more physical significance . They have their own laws of dynamics and behaviour .

Overall , gravitational field intensity and acceleration vectors are same , but they are separate physical quantities .


Dimension of Acceleration due to Gravity: $LT^{−2}$

Dimension of Gravitational Field Intensity: $MLT^{-2}M^{-1} = LT^{-2}$

As dimensions of both are same, both are same physical quantities. All those physics books saying them different are wrong.

  • $\begingroup$ Despite its popularity in India, HC Verma is one of worst books on the Earth just to confuse your concepts. Avoid it at any cost. $\endgroup$ Jan 31, 2014 at 8:23
  • 1
    $\begingroup$ Then according to you all the quantities which have no dimensions must also be equal? Is $\pi = e$? Moreover many quantities have the same dimensions but are entirely different in nature $\endgroup$ Jan 20, 2017 at 1:34
  • $\begingroup$ and user931 HC’s book is one of the best book in the world. If you cannot understand that book it’s your mistake. Buy it at any cost $\endgroup$ Jan 20, 2017 at 1:37
  • 1
    $\begingroup$ @AaryanDewan That's not what user931 is saying. To take your example, he is only saying that $\pi$ and $e$ are of the same type, namely dimensionless numbers. $\endgroup$ Jan 20, 2017 at 1:37
  • 1
    $\begingroup$ Okay @user1583209 . Take a look at this example. Here group pair has the same dimensions but are entirely different. (Work, Energy and enthalpy) (Trigonometric ratios and relative density ) (Entropy and Boltzmann constant) (Stress and coefficient of elasticity) $\endgroup$ Jan 20, 2017 at 1:43