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Q.
When the radius of earth is reduced by $1\%$ without changing the mass, then change in the acceleration due to gravity will be:
AFMCAFMC 2002
Solution:
From law of gravitation.
$F=G \frac{M_{e} m}{R_{e}^{2}}$ ...(i)
Also from Newton's law
$F=m g$ ...(ii)
Equating Eqs. (i) and (ii), we get
$g =\frac{G M_{e}}{R_{e}^{2}}$
$\frac{g_{1}}{g_{2}} =\left(\frac{R_{2}}{R_{1}}\right)^{2}$
$=\left(\frac{0.99 R}{R}\right)^{2}$
$\frac{g_{1}}{g_{2}} =0.98$
$g_{2} =\frac{g_{1}}{0.98}=1.02\, g_{1}$
Hence, change in acceleration due to gravity
$=g_{2}-g_{1}=1.02\, g_{1}-g_{1}=0.02\, g_{1}=2 \% g_{1}$
Since, sign is positive, acceleration due to gravity will increase by $2 \%$