If the radius of earth shrinks by 1.5 % (mass | vedclass

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Question

(c)

$g=\frac{G M}{R^2}$

$g^{\prime}=\frac{G M}{(0.985 R)^2}$

$g^{\prime}=(1.0306) \frac{G M}{R^2}$

$\Rightarrow g^{\prime}=1.0306 g$

$\

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c)

$g=\frac{G M}{R^2}$

$g^{\prime}=\frac{G M}{(0.985 R)^2}$

$g^{\prime}=(1.0306) \frac{G M}{R^2}$

$\Rightarrow g^{\prime}=1.0306 g$

$\Rightarrow$ Acceleration changes by

$\frac{\Delta g}{g} 	imes 100=+3 \%$

Alternate method:

$g^{\prime}=\frac{G M}{(R+\Delta R)^2}$

$g^{\prime}=G M(R+\Delta R)^{-2}$

$g^{\prime}=\frac{G M}{R^2}\left(1+\frac{\Delta R}{R}ight)^{-2}$

for $\frac{\Delta R}{R} \ll 1$, we can use binomial and approximately,

$g^{\prime}=\frac{G M}{R^2}\left(1-\frac{2 \Delta R}{R}ight)$

$\Rightarrow g^{\prime}=g-g \frac{2 \Delta R}{R}$

$\Rightarrow \frac{\Delta g}{g}=\frac{-2 \Delta R}{R}=-2 	imes\left(\frac{-1.5}{100}ight)=\frac{+3}{100}=3 \% \quad\left[g^{\prime}-g=\Delta gight]$

If the radius of earth shrinks by $1.5 \%$ (mass remaining same), then the value of gravitational acceleration changes by ......... $\%$

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