Acceleration Due to Gravity

Acceleration due to gravity on the surface of the earth

For a symmetrical object, the gravitational force acts towards the centre of the object. The value of g is almost constant near the surface of the earth, but as we move far from the surface of the earth, the strength of gravity decreases as the height increases.

Figure 2. An object with mass m under the influence of a bigger body, such as a planet with mass M.

Based on experimental data, it has been observed that the acceleration due to gravity is inversely proportional to the square of the distance of the object from the centre of mass of the larger object.

\[g \propto \frac{1}{r^2}\]

Here, r is the distance of the object from the centre of the earth. The acceleration due to gravity is not only proportional inversely to r^2 but also directly proportional to the mass of the body attracted to, in this case, the earth.

For instance, the acceleration due to gravity on the earth is different from the acceleration due to gravity on the moon. Thus, we have another proportionality, as follows:

\[g \propto M\]

We assume that the mass of the object is significantly less with respect to the mass of the planet or body to which it is attracted. Algebraically, this is written as:

\[m << M\]

Here, m = mass of the object and M = mass of the bigger object or planet.

Combining both these proportionalities, we get:

\[g \propto \frac{M}{r^2}\]

\[g = \frac{GM}{r^2}\]

Based on experimental data, the value of G for the earth has been found to be G = 6.674⋅10^-11 Nm² kg^-2.

Suppose the object is not on the surface of the earth but at a height h from the surface. In that case, its distance from the centre of mass of the earth will now be:

\[r = R + h\]

Here, R is the radius of the earth. Substituting for r in the earlier equation, we now get:

\[g = \frac{MG}{(R + h)^2}\]

(&)

Hence, we can see that as h increases, the strength of gravity decreases.

Acceleration due to gravity below the surface of the earth

The acceleration due to gravity does not follow the quadratic relationship when the object is below the surface of the earth. In fact, acceleration and distance are linearly dependent on each other for r < R (below the surface of the earth).

If an object is at r distance from the centre of the earth, the mass of the earth responsible for the acceleration due to gravity at that point will be:

\[m = \frac{Mr^3}{R^3}\]

This can be easily deduced using the formula for the volume of a sphere.

We have assumed the Earth to be a sphere, but in reality, the radius of the earth is at its minimum at the poles and at its maximum at the equator. The difference is quite small, and so we assume the earth to be a sphere for simplified calculations. The acceleration due to gravity follows the proportionality explained earlier:

\[g \propto \frac{m}{r^2}\]

Substituting for m, we get:

\[g = \frac{GMr}{R^3} g \propto r\]

We can now see that as G, M, and R are constants for a given object or planet, the acceleration linearly depends on r. Hence, we see that as r approaches R, the acceleration due to gravity increases according to the above linear relation, after which it decreases according to &, which we derived earlier. In practice, most of the real-world problems include the object being outside the surface of the earth.

Geometric interpretation of acceleration due to gravity

The acceleration due to gravity has a linear relation with r until the surface of the earth, after which it is described by the quadratic relation we defined earlier.

This can be geometrically seen with the help of the graph above. As r increases, g reaches its maximum value when r=R=radius of the earth, and as we move away from the surface of the earth, the strength of g decreases according to the relation:

\[g \propto \frac{1}{r^2}\]

The equation describes a parabola, which is quite intuitive, given the definition we saw earlier.

We also note that the value of acceleration due to gravity is 0 at the centre of the earth and almost 0 when far away from the surface of the earth. To demonstrate the application of this concept, consider the following example.