Gravitational Acceleration Formula

The acceleration of gravity is denoted by ‘g’. Anything that falls on the earth has some acceleration associated with it. However, if a body falls freely under the gravitational influence, we refer to this acceleration as the acceleration due to gravity, which means that the value of ‘a’ becomes equal to ‘g.’

So, the value of acceleration due to gravity, i.e., g = 9.8 ms\[^{-2}\]. However, for easy experimental calculations, we use the value of g as 10 ms\[^{-2}\].

The formula of acceleration due to gravity can be expressed in various system units. Like MKS system, CGS. It has a certain formula derived from Newton’s universal law of gravitation. Also, the dimensions of this physical quantity are expressed as the dimensional formula of acceleration due to gravity.

Here, we will understand the acceleration due to gravity formula. Also, we will derive acceleration due to gravity dimensional formula. 

Besides this, we will derive an expression for acceleration due to gravity as well.

Formula of Gravitational Acceleration

Now, let us derive the gravitational acceleration formula.

From the Universal law of gravitation, when two bodies of mass ‘‘m’ and ‘M’ are separated by the square of distance ‘r,’ the force of attraction between these two masses is expressed as:

F = G\[\frac{mM}{r^{2}}\]

Here,

• m is the mass of a smaller object

• M is the mass of a bigger object, (let’s say the earth) 

• G is the Universal Gravitational Constant, whose value is 6.673 x 10\[^{-11}\] Nm\[^{2}\]kg\[^{-2}\]

Also, we notice that ‘F’ is inversely proportional to the square of the distance between two bodies, i.e.,

F α \[\frac{1}{r^{2}}\]

The smaller object gets pulled towards the center of the bigger object.

Also, we know that the F = mg, now, substituting the value of F as:

mg = G\[\frac{mM}{r^{2}}\]

We get our required gravitational acceleration formula, which is:

Acceleration of Gravity Formula

The acceleration of gravity varies with the following:

• Variation with density

• Variation with height

Variation of Acceleration Due to Gravity with Depth

We know, that the acceleration due to gravity formula on the earth’s surface is G\[\frac{M}{Re^{2}}\]

Also, the density, ρ = \[\frac{Mass}{Volume}\] = \[\frac{M}{V}\]

Since the earth is spherical, so its volume is \[\frac{4}{3}\]πRe\[^{3}\]

Now, replacing the value of M with ρ . \[\frac{4}{3}\]πRe\[^{3}\] in the above formula of acceleration due to gravity:

                                  = G \[\frac{\rho . \frac{4}{3} \pi Re^{3}}{Re^{2}}\]         

So, the formula of gravitational acceleration varying with density is:

Now, let us find the acceleration due to gravity formula for ‘g’ varying with height:

Formula of Acceleration Due to Gravity g at Height h

Acceleration due to gravity ‘g’ varies with the height away from the surface of the earth. Its formula is as follows:

g\[_{h}\] = g(1 + \[\frac{h}{R}\])\[^{-2}\]

Here,

g = Acceleration due to gravity at the surface of the earth (in ms\[^{-2}\] or cms\[^{-2}\])

g\[_{h}\] = Acceleration due to gravity at ‘h’

R = The radius of the earth, i.e., R\[_{e}\]

h = Height from the surface of the earth (in metres)

Now, for finding the dimensional formula of gravitational acceleration, let’s take a look at the method to derive the dimensional formula for acceleration due to gravity.

Acceleration Due to Gravity Dimensional Formula

We know that the unit of G in MKS is Nm\[^{2}\]kg\[^{-2}\], which is equal to kg\[^{-1}\]m\[^{3}\]s\[^{-2}\].

Besides this, the units of M and r are kg and m, respectively.

Since we have r\[^{2}\], so its unit becomes m\[^{2}\].

Now, we know that mass equals to [M]

Distance ‘r’ equals to [L], and

Time ‘t’ refers to [T] 

Now, replacing the MKS system units with the dimensional formula:

G = [M\[^{-1}\]L\[^{3}\]T\[^{-2}\]]

r\[^{2}\] = [L\[^{2}\]]

M = [M]  (the dimensional formula of M is ‘M’)

Now, putting these values in the above formula of acceleration due to gravity.

g = G\[\frac{M}{r^{2}}\]

= G\[\frac{M}{r^{2}}\]

= [M-1L3T-2] \[\frac{[M]}{[L^{2}]}\]

Solving the Equation and Cancelling the Common Terms:

=   [M\[^{0}\]L\[^{-1}\]T\[^{-2}\]]

Thus, we get the dimensional formula for acceleration due to gravity as [M\[^{0}\]L\[^{-1}\]T\[^{-2}\]], from which the unit of acceleration due to gravity formula, i.e., ms\[^{-2}\] can be derived. 

Before we calculate acceleration due to gravity, let us look at some acceleration due to gravity examples:

Calculate Acceleration Due to Gravity

Let’s solve a question related to the Acceleration of Gravity Formula.

We know that the moon’s radius is 1.74 x 10\[^{6}\] m. Let’s suppose that the mass of the moon is 7.35 x 10\[^{22}\] kg. Now, let us calculate the acceleration due to gravity on the surface of the moon (g\[_{m}\]).

We know that on the surface of the moon, the distance to the center of mass will be equal to its radius.

Therefore,  r = 1.74 x 10\[^{6}\] m

Mass of moon, i.e., m = 7.35 x 10\[^{22}\] kg

Since we know that the value of the universal gravitational constant G = 6.673 x 10\[^{-11}\] Nm\[^{2}\]kg\[^{-2}\]

On the surface of the moon, the acceleration due to gravity can be stated as:

g\[_{m}\] = G\[\frac{M}{r^{2}}\]

Substituting the values mentioned in this equation:

g\[_{m}\] =  (6.673 x 10\[^{-11}\]) . \[\frac{7.35 \times 10^{22}}{(1.74 \times 10^{6})^{2}}\]

On solving the above equation, we get the value of ‘g\[_{m}\]’ as;

g\[_{m}\] = 1.620 ms\[^{-2}\]

Hence, the value of the acceleration due to gravity on the surface of the moon (g\[_{m}\]) is 1.620 ms\[^{-2}\].

Conclusion

In the nutshell, the gravity of Earth is denoted by g, which is the net acceleration that is imparted to bodies due to the combined influence of gravitation and centrifugal force. In SI units, the gravitational acceleration is measured in metres per second squared or in newtons per kilogram.