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The value of acceleration due to gravity at

earth's surface is 9.8 $ms^{-2}$ . The altitude

above its surface at which the acceleration

due to gravity decreases to 4.9 $ms^{-2}$ , is

close to : (Radiuas of earth = 6.4 x $10^{^{6}}$ m)

Option 1)
2.6 X $10^{6}$ m

Option 2)
6.4 x $10^{6}$ m

Option 3)
9.0 x $10^{6}$ m

Option 4)
1.6 x $10^{^{6}}$ m

Answers (1)

best_answer

Acceleration due to gravity (g) -

Force extended by earth on a body is gravity.

Formula: $g=\frac{GM}{R^{2}},$

$g=\frac{4}{3}\pi \rho \, GR$

$g\rightarrow$ gravity

$\rho \rightarrow$ density of earth

$R \rightarrow$ Radius of earth

- wherein

It's average value is $9.8\: m/s^{2}\; \; or \; \; 981cm/sec^{2}\; or\; 32feet/s^{2}$ on the surface of earth

Variation in 'g' with height -

$g'\rightarrow$ gravity at height $h$ from surface of earth.

$R\rightarrow$ Radius of earth

$h\rightarrow$ height above surface

- wherein

$g'\alpha\, \frac{1}{r^{2}}$

$r=R+h$

$g=\frac{GM}{R^{2}}$

$g'=\frac{g}{2}=\frac{GM}{(R+h)^{2}}$

=> $\frac{1}{2}=\frac{R^{2}}{(R+h)^{2}}$

=> $R+h=\sqrt 2 \: \: R$

=> $R=\frac{h}{\sqrt2-1}$

=> $h=(\sqrt2-1) R$

=> $h=0.41\times 6.4\times 10^{6}m$

=> $h=2.6\times 10^{6}m$

Option 1)

2.6 X $10^{6}$ m

Option 2)

6.4 x $10^{6}$ m

Option 3)

9.0 x $10^{6}$ m

Option 4)

1.6 x $10^{^{6}}$ m

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