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This is the maximum reduction possible due to rotation. Indeed, we can neglect this variation for all practical purposes except where very high accuracy is required.
In this section, we shall discuss the effect of the vertical position of the point of measurement. For this, we shall consider Earth as a perfect sphere of radius “R” and uniform density, “ρ”. Further, we shall first consider a point at a vertical height “h” from the surface and then a point at a vertical depth “d” from the surface.
Gravitational acceleration due to Earth on its surface is equal to gravitational force per unit mass and is given by :
$${g}_{0}=\frac{F}{m}=\frac{GM}{{R}^{2}}$$
where “M” and “R” are the mass and radius of Earth. It is clear that gravitational acceleration will decrease if measured at a height “h” from the Earth’s surface. The mass of Earth remains constant, but the linear distance between particle and the center of Earth increases. The net result is that gravitational acceleration decreases to a value “g’” as given by the equation,
$$\Rightarrow g\prime =\frac{F\prime}{m}=\frac{GM}{{\left(R+h\right)}^{2}}$$
We can simplify this equation as,
$$\Rightarrow g\prime =\frac{GM}{{R}^{2}{\left(1+\frac{h}{R}\right)}^{2}}$$
Substituting for the gravitational acceleration at the surface, we have :
$$\Rightarrow g\prime =\frac{{g}_{0}}{{\left(1+\frac{h}{R}\right)}^{2}}$$
This relation represents the effect of height on gravitational acceleration. We can approximate the expression for situation where h<<R.
$$\Rightarrow g\prime =\frac{{g}_{0}}{{\left(1+\frac{h}{R}\right)}^{2}}={g}_{0}{\left(1+\frac{h}{R}\right)}^{-2}$$
As h<<R, we can neglect higher powers of “h/R” in the binomial expansion of the power term,
$$\Rightarrow g\prime ={g}_{0}\left(1-\frac{2h}{R}\right)$$
We should always keep in mind that this simplified expression holds for the condition, h<<R. For small vertical altitude, gravitational acceleration decreases linearly with a slope of “-2/R”. If the altitude is large as in the case of a communication satellite, then we should resort to the original expression,
$$\Rightarrow g\prime =\frac{GM}{{\left(R+h\right)}^{2}}$$
If we plot gravitational acceleration .vs. altitude, the plot will be about linear for some distance.
In order to calculate gravitational acceleration at a depth “d”, we consider a concentric sphere of radius “R-d” as shown in the figure. Here, we shall make use of the fact that gravitational force inside a spherical shell is zero. It means that gravitational force due to the spherical shell above the point is zero. On the other hand, gravitational force due to smaller sphere can be calculated by treating it as point mass. As such, net gravitational acceleration at point “P” is :
$$\Rightarrow g\prime =\frac{F\prime}{m}=\frac{GM\prime}{{\left(R-d\right)}^{2}}$$
where “M’” is the mass of the smaller sphere. If we consider Earth as a sphere of uniform density, then :
$$M=V\rho $$
$$\Rightarrow \rho =\frac{M}{\frac{4}{3}\pi {R}^{3}}$$
Hence, mass of smaller sphere is equal to the product :
$$M\prime =\rho V\prime $$
$$\Rightarrow M\prime =\frac{\frac{4}{3}\pi {\left(R-d\right)}^{3}XM}{\frac{4}{3}\pi {R}^{3}}=\frac{{\left(R-d\right)}^{3}XM}{{R}^{3}}$$
Substituting in the expression of gravitational acceleration, we have :
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