The acceleration due to gravity at height above the earth, if (radius of earth), is given by [2023]

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Q.
The acceleration due to gravity at height $h$ above the earth, if $h ≪ R$ (radius of earth), is given by [2023]

1 $g' = g (1 - \frac{2 h}{R})$

2 $g' = g (1 - \frac{h}{2 R})$

3 $g' = g (1 - \frac{2 h^{2}}{R^{2}})$

4 $g' = g (1 - \frac{h^{2}}{2 R^{2}})$

Ans.
(1)

$For a point outside the surface of Earth, g = \frac{G M}{r^{2}}$

$r = distance from centre of Earth$

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

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

$If h ≪ R, {(1 + \frac{h}{R})}^{- 2} \approx 1 - \frac{2 h}{R}$

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

$\Rightarrow g (h) = g_{surface} (1 - \frac{2 h}{R}), \frac{G M}{R^{2}} = g_{surface}$

Solution

Q.
The acceleration due to gravity at height $h$ above the earth, if $h ≪ R$ (radius of earth), is given by [2023]

1 $g' = g (1 - \frac{2 h}{R})$

2 $g' = g (1 - \frac{h}{2 R})$

3 $g' = g (1 - \frac{2 h^{2}}{R^{2}})$

4 $g' = g (1 - \frac{h^{2}}{2 R^{2}})$

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Solution

Q. The acceleration due to gravity at height h above the earth, if h≪R (radius of earth), is given by [2023]

1 g'=g(1-2hR)

2 g'=g(1-h2R)

3 g'=g(1-2h2R2)

4 g'=g(1-h22R2)

Ans. (1) For a point outside the surface of Earth, g=GMr2 r=distance from centre of Earth ⇒ g(h)=GM(R+h)2=GMR2(1+hR)2 ⇒ g(h)=GMR2(1+hR)-2 If h≪R,(1+hR)-2≈1-2hR ⇒ g(h)=GMR2(1-2hR) ⇒ g(h)=gsurface(1-2hR),GMR2=gsurface

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Q.
The acceleration due to gravity at height $h$ above the earth, if $h ≪ R$ (radius of earth), is given by [2023]

1 $g' = g (1 - \frac{2 h}{R})$

2 $g' = g (1 - \frac{h}{2 R})$

3 $g' = g (1 - \frac{2 h^{2}}{R^{2}})$

4 $g' = g (1 - \frac{h^{2}}{2 R^{2}})$