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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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!
No spam please... !
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Answer:
The value of acceleration due to gravity change by 4%.
Explanation:
=> Things we know
[tex]g \: = G \times \frac{M}{ {R}^{2} } ........(1)[/tex]
Also , to find the rate of change we use differentiation method,
Given , Radius shrinks by 2 %
so We can say
[tex] = > \frac{dR}{R} \times 100\%= - 2\% \\ \\ = > \frac{dR}{R} = - 0.02 \: \: \: \: \: \:....(2) \\ \\ negative \: because \: of \: reduction \: in \: radius.[/tex]
To find the change in acceleration due to gravity :
We need to find ,
[tex] \frac{dg}{g} \times 100\%[/tex]
So by differentiating equation (1) with respect to R ,we get
[tex] = > 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)[/tex]
Now diving both Sides by g we get ,
[tex] = > \frac{dg}{g} = - 2 \times ( - 0.02 )= 0.04[/tex]
since we know ,
[tex] = > \frac{dR}{R} = - 0.02[/tex]
So the change in acceleration due to gravity will be
[tex] = > \frac{dg}{g} \times 100\% = 0.04 \times 100\% \\ \\ = > \frac{dg}{g} \times 100\% = 4\% \: \: \: ...(ans)[/tex]