Question

Hey !!


If the radius of the earth shrinks by 2.0%,
mass remaining constant, then how would the value of
acceleration due to gravity change ?

Provide elaborated explanation please!

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Answer 1

Answer:

Here's your answer in above attachment.

hope it's helpful

Answer 2

Answer:

The value of acceleration due to gravity change by 4%.

Explanation:

=> Things we know

$g \: = G \times \frac{M}{ {R}^{2} } ........(1)$

Also , to find the rate of change we use differentiation method,

Given , Radius shrinks by 2 %

so We can say

$= > \frac{dR}{R} \times 100\%= - 2\% \\ \\ = > \frac{dR}{R} = - 0.02 \: \: \: \: \: \:....(2) \\ \\ negative \: because \: of \: reduction \: in \: radius.$

To find the change in acceleration due to gravity :

We need to find ,

$\frac{dg}{g} \times 100\%$

So by differentiating equation (1) with respect to R ,we get

$= > dg = GM( - \frac{2}{ {R}^{3} } dR) \: \: \: \\ \\ = > dg = \frac{GM}{ {R}^{2} } \times ( - 2\frac{dR}{R}) \\ \\ = > dg = g \times ( - 2 \times \frac{dR}{R} )....(3)$

Now diving both Sides by g we get ,

$= > \frac{dg}{g} = - 2 \times ( - 0.02 )= 0.04$

since we know ,

$= > \frac{dR}{R} = - 0.02$

So the change in acceleration due to gravity will be

$= > \frac{dg}{g} \times 100\% = 0.04 \times 100\% \\ \\ = > \frac{dg}{g} \times 100\% = 4\% \: \: \: ...(ans)$