# 4.1 Gravity  (Page 6/6)

 Page 6 / 6

$⇒g\prime =\frac{G{\left(R-d\right)}^{3}XM}{{\left(R-d\right)}^{2}{R}^{3}}$

Inserting gravitational acceleration at the surface ( ${g}_{0}=GM/{R}^{2}$ ), we have :

$⇒g\prime =\frac{{g}_{0}{\left(R-d\right)}^{3}}{{\left(R-d\right)}^{2}R}=\frac{{g}_{0}\left(R-d\right)}{R}$

$g\prime ={g}_{0}\left(1-\frac{d}{R}\right)$

This is also a linear equation. We should note that this expression, unlike earlier case of a point above the surface, makes no approximation . The gravitational acceleration decreases linearly with distance as we go down towards the center of Earth. Conversely, the gravitational acceleration increases linearly with distance as we move from the center of Earth towards the surface.

The plot above combines the effect of altitude and the effect of depth along a straight line, starting from the center of Earth.

## Gravitational acceleration .vs. measured acceleration

We have made distinction between these two quantities. Here, we shall discuss the differences once again as their references and uses in problem situations can be confusing.

1: For all theoretical discussion and formulations, the idealized gravitational acceleration ( ${g}_{0}$ ) is considered as a good approximation of actual gravitational acceleration on the surface of Earth, unless otherwise told. The effect of rotation is indeed a small value and hence can be neglected for all practical purposes, unless we deal with situation, requiring higher accuracy.

2: We should emphasize that both these quantities ( ${g}_{0}$ and g) are referred to the surface of Earth. For points above or below, we use symbol (g’) for effective gravitational acceleration.

3: If context requires, we should distinguish between “g0” and “g”. The symbol “ ${g}_{0}$ ” denotes idealized gravitational acceleration on the surface, considering Earth (i) uniform (ii) spherical and (iii) stationary. On the other hand, “g” denotes actual measurement. We should, however, be careful to note that measured value is also not the actual measurement of gravitational acceleration. This will be clear from the point below.

4: The nature of impact of “rotation” on gravitational acceleration is different than due to other factors. We observed in our discussion in this module that “constitution of Earth” impacts the value of gravitational acceleration for a point below Earth’s surface. Similarly, shape and vertical positions of measurements affect gravitational acceleration in different ways. However, these factors only account for the “actual” change in gravitational acceleration. Particularly, they do not modify the gravitational acceleration itself. For example, shape of Earth accounts for actual change in the gravitational acceleration as polar radius is actually smaller than equatorial radius.

Now, think about the change due to rotation. What does it do? It conceals a part of actual gravitational acceleration itself. A part of gravitational force is used to provide for the centripetal acceleration. We measure a different gravitational acceleration than the actual one at that point. We should keep this difference in mind while interpreting acceleration. In the nutshell, rotation alone affects measurement of actual gravitational acceleration, whereas other factors reflect actual change in gravitational acceleration.

5: What is actual gravitational acceleration anyway? From the discussion as above, it is clear that actual gravitational acceleration on the surface of Earth needs to account for the part of the gravitational force, which provides centripetal force. Hence, actual gravitational acceleration is :

${g}_{\mathrm{actual}}=g+{\omega }^{2}R\mathrm{cos}\phi$

Note that we have made correction for centripetal force in the measured value (g) – not in the idealized value ( ${g}_{0}$ ). It is so because measured value accounts actual impacts due to all factors. Hence, if we correct for rotation – which alone affects measurement of actual gravitational acceleration, then we get the actual gravitational acceleration at a point on the surface of the Earth.

#### Questions & Answers

List the application of projectile
Luther Reply
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Mercy Reply
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show that 1w= 10^7ergs^-1
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A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
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Amar
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isaac Reply
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Riya
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Mustapha
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x=5.8-3.22 x=2.58
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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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