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Question: If the radius of the earth shrinks by 1%, its mass remaining the same, what is the change in acceler...

If the radius of the earth shrinks by 1%, its mass remaining the same, what is the change in acceleration due to gravity on the surface of the earth?
(A) Decrease by 2%
(B) Decrease by 0.5%
(C) Increase by 2%
(D) Increase by 0.5%

Explanation

Solution

Percent error formula is the absolute value of the difference of the measured value and the actual value divided by the actual value and multiplied by 100 i.e.
%Error\text{ }in\text{ g; }\Delta \text{g}=\dfrac{g'-g}{g}\times 100
Where g’ is the new value and g is the actual value of acceleration due to gravity.

Complete step by step solution
Here the value of radius R shrinks by 1% so,
ΔR=−1\Delta R=-1
R′−RR×100=−1 R′−R=−R100 R′=R−R100 R′=0.99R \begin{aligned} & \dfrac{R'-R}{R}\times 100=-1 \\\ & R'-R=\dfrac{-R}{100} \\\ & R'=R-\dfrac{R}{100} \\\ & R'=0.99R \\\ \end{aligned}
Now the new value of g i.e. g’ becomes:
g′=GMR′2 g′=GM(0.99R)2 g′=g(0.99)2 \begin{aligned} & g'=\dfrac{GM}{R{{'}^{2}}} \\\ & g'=\dfrac{GM}{{{\left( 0.99R \right)}^{2}}} \\\ & g'=\dfrac{g}{{{\left( 0.99 \right)}^{2}}} \\\ \end{aligned}
Percentage change in g is
Δg=g′−gg×100 =g(0.99)2−gg×100 =g(1(0.99)2−1)g×100 =1−0.98010.9801×100 =0.01990.9801×100 =0.0203×100 \begin{aligned} & \Delta g=\dfrac{g'-g}{g}\times 100 \\\ & =\dfrac{\dfrac{g}{{{\left( 0.99 \right)}^{2}}}-g}{g}\times 100 \\\ & =\dfrac{g\left( \dfrac{1}{{{(0.99)}^{2}}}-1 \right)}{g}\times 100 \\\ & =\dfrac{1-0.9801}{0.9801}\times 100 \\\ & =\dfrac{0.0199}{0.9801}\times 100 \\\ & =0.0203\times 100 \\\ \end{aligned}
Here the positive sign indicates that the value of g is decreasing.
Therefore the value of acceleration due to gravity is increased by 2%.

Note
Alternate method:
Percentage error in any term Z having formula Z=ApBqCrZ=\dfrac{{{A}^{p}}{{B}^{q}}}{{{C}^{r}}} is given by:
ΔZZ×100=(pΔAA+qΔBB−rΔCC)×100\dfrac{\Delta Z}{Z}\times 100=\left( p\dfrac{\Delta A}{A}+q\dfrac{\Delta B}{B}-r\dfrac{\Delta C}{C} \right)\times 100
Here
g=GMR2 g α 1R2 Δgg×100=−2ΔRR×100 \begin{aligned} & g=\dfrac{GM}{{{R}^{2}}} \\\ & g\text{ }\alpha \text{ }\dfrac{1}{{{R}^{2}}} \\\ & \dfrac{\Delta g}{g}\times 100=-2\dfrac{\Delta R}{R}\times 100 \\\ \end{aligned}
As ΔRR×100=−1  \begin{aligned} & \dfrac{\Delta R}{R}\times 100=-1 \\\ & \\\ \end{aligned}
So,
Δgg×100=−2×(−1) \begin{aligned} & \dfrac{\Delta g}{g}\times 100=-2\times \left( -1 \right) \\\ \end{aligned}
Therefore the value of g increases by 2%.