Variation in \(g\) Due to the Shape of the Earth

The earth is not a perfect sphere but is bulged at the equator and slightly flattened at the poles. So, the value of \(g\) is somewhat more at the poles compared to the equator.

Variation in \(g\) Due to Altitude (height)

Consider a point mass \(m\) at a height \(h\) above the surface of the earth as shown in Figure below. The radius of the earth is denoted by \(r\). Since this point is outside the earth, its distance from the centre of the earth is \(\left(r+\right.\) \(h\) ). 

Let \(g_h\) be the acceleration due to gravity at a height \(h\) above the surface of the earth.
\(
\begin{aligned}
&g=\frac{G M}{r^2} \\
&g_h=\frac{G M}{(r+h)^2}
\end{aligned}
\)
Dividing \(g_h\) by \(g\)
\(
\begin{aligned}
\frac{g_h}{g} &=\frac{r^2}{(r+h)^2} \\
g_h &=\frac{g}{\left(1+\frac{h}{r}\right)^2}
\end{aligned}
\)
This equation shows that as \(h\) increases, the value of \(g_h\) decreases. At a considerable height, \(h>>r\), acceleration due to gravity becomes negligibly small. The earth’s gravity does not influence objects there.

\(
g_h=\frac{F_h}{m}=\frac{G M}{\left(r+h\right)^2} .
\)
This is clearly less than the value of \(g\) on the surface of earth : \(g=\frac{G M}{r^2}\). For \(h \ll r\), we can expand the RHS of the equation:
\(
g_h=\frac{G M}{r^2\left(1+h / r\right)^2}=g\left(1+h / r\right)^{-2}
\)
For \(\frac{h}{r}<<1\), using binomial expression, \(g_h \cong g\left(1-\frac{2 h}{r}\right)\).
The equation above thus tells us that for small heights \(h\), the value of \(g\) decreases by a factor \(\left(1-2 h / r\right)\)

Variation in \(g\) Due to Depth

Let \(g_d\) be the acceleration due to gravity at a depth, \(d\) below the surface of the earth. Since it is below the surface, we must consider the acceleration in terms of density.
\(
\begin{aligned}
&g=\frac{4 \pi}{3} G \rho r \\
&g_d=\frac{4 \pi}{3} G \rho(r-d)
\end{aligned}
\)
Dividing \(g_d\) by \(g we get,
[latex]
g_d=g\left(1-\frac{d}{r}\right)
\)
At surface, \(d=0, g_d=g\)
At the centre of the earth, \(d=r\). So, \(g_d=0\). Acceleration due to gravity at the centre of the earth is zero.

Difference Between Mass and Weight

Mass and weight sound similar but they represent different physical quantity. Following are the differences between them.

Since weight is a force due to gravity, it has the formula, Weight, \(W=m g\)
Where \(g\) is the acceleration due to gravity at that place.

Comparison of Weight on Earth and on Moon

The mass of the earth is approximately 100 times that of the moon. The radius of the earth is about 4 times that of the moon.

Acceleration due to gravity for the earth:
\(
g=\frac{G M}{r^2}
\)
Acceleration due to gravity for the moon:
\(
g_o=\frac{G M_o}{r_o{ }^2}
\)
Given,
\(
M=100 M_o
\)
Dividing,
\(
\begin{aligned}
&r=4 r_o \\
&\frac{g}{g_o}=\frac{M}{M_o} \times\left(\frac{r_o}{2}\right)^2=100 \times\left(\frac{1}{4}\right)^2=6.25
\end{aligned}
\)
Since \(m\) remains the same both on the Earth and the Moon,
\(
\frac{W}{W_o}=6.25
\)
Weight on an object is approximately 6 times that on the moon, or weight on the moon is \(\left(\frac{1}{6}\right)^{\text {th }}\) of that on earth.

Example 1: The mass of the earth is 10 times that of another planet. Radius of the earth is 2 times its radius. Compare the weight of an object on the planet with that on the earth.

Solution:

Given,
\(
\begin{aligned}
&M=10 M_p \\
&r=2 r_p
\end{aligned}
\)
Dividing,
\(
\frac{g}{g_p}=\frac{M}{M_p} \times\left(\frac{r_p}{r}\right)^2=10 \times\left(\frac{1}{2}\right)^2=2.25
\)
Since weight \(W=m g\), and \(\mathrm{m}\) remains the same everywhere, \(\frac{W}{W_p}=2.25\)
Weight on the planet is approximately half of that on earth.