Name: Gravitational Potential Energy
Uploaded: 2019-07-19
Duration: 9 min 10 s
Description: The energy possessed by a body because of its position in a gravitational field is known as Gravitational Potential Energy. Let’s learn about it in detail in...

Question

Discuss the variation of acceleration due to gravity with altitude and depth.

Accepted Answer

The gravitational force of attraction at the surface is given by $F_{g} = \frac{G M m}{R^{2}}$ , where $G$ Is the gravitational constant, $M$ Is the mass of earth and $R$ Is the radius of earth.

Therefore, $g = \frac{G M}{R^{2}}$ $. . . . . . (1)$

At heigh $h$ , $R$ Changes to $R'$ , where $R' = R + h$ .

now $g_{h} = \frac{G M}{{(R + h)}^{2}}$ $. . . . . . (2)$

$\frac{(2)}{(1)} = \frac{g_{h}}{g} = \frac{\frac{G M m}{{(R + h)}^{2}}}{\frac{G M m}{R^{2}}} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{R^{2}}{R^{2} (1 + \frac{h}{R})^{2}} = {(1 + \frac{h}{R})}^{- 2}$

Since $h < < R$ .

Therefore, $\frac{g_{h}}{g} = 1 - \frac{2 h}{R} \Rightarrow g_{h} = g (1 - \frac{2 h}{R})$

Let density of earth= $ρ$ , mass = $M$ , radius= $R$

Therefore $M = \frac{4}{3} π r^{3} \times ρ$

$g = \frac{G M}{R^{2}} = \frac{G \frac{4}{3} π R^{3} \times ρ}{R^{2}} = G \frac{4}{3} π R \times ρ . . . . . . (3)$

As depth $d$ Increases, the distance from centre decreases

$R' = R - d$

Therefore, $g_{d} = G \frac{4}{3} π (R - d) \times ρ . . . . . (4)$

$\frac{(4)}{(3)} = \frac{g_{d}}{g} = \frac{G \frac{4}{3} π (R - d) \times ρ}{G \frac{4}{3} π R \times ρ} = \frac{(R - d)}{R} = 1 - \frac{d}{R} \Rightarrow g_{d} = g (1 - \frac{d}{R})$

Discuss the variation of acceleration due to gravity with altitude and depth.

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