DLS
  • DLS
If the radius of the earth shrinks by 5%,mass remaining the same,then how much value of acc. due to gravity change?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
radius shrinks by 5% means.. the radius becomes 95/100(Re - current earths radius).. So can you solve now?
DLS
  • DLS
i know that much i did that too didnt get the answer though
anonymous
  • anonymous
wait... so you tell me now.. once you substitue this new value of radius.. what is the ratio of the new value of g to the present value of g?

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More answers

DLS
  • DLS
i have to give the answer is "%" change
anonymous
  • anonymous
Ok we ll work towards it don't worry.. !
anonymous
  • anonymous
answer my question first :P
DLS
  • DLS
6400/6080 ?
anonymous
  • anonymous
:O.. no.. how did you get that? :-/
anonymous
  • anonymous
listen .. use the equation of g..
DLS
  • DLS
\[\frac{95}{100} \times 6400\]
DLS
  • DLS
New one^^
anonymous
  • anonymous
yes.. but i want the value of g.. not the radius!
DLS
  • DLS
\[g= \frac{GM}{R ^{2}}\]
DLS
  • DLS
R'=95/100R
anonymous
  • anonymous
yes exactly.. now put the new value of R .. and get the ratio!!
anonymous
  • anonymous
correct.. now once you put new value.. you ll get a new g.. lets call that g prime... so find out what is the ratio of gprime to g?
DLS
  • DLS
20GM/19R is the new one? :S
anonymous
  • anonymous
if you plug in the values.. you should get g (prime)/ g = 1.108
DLS
  • DLS
yeah..okay
DLS
  • DLS
so 11%?
anonymous
  • anonymous
10.8 percent..
DLS
  • DLS
11.08%?
anonymous
  • anonymous
no.. see.. the increase is .108 ( 1.108-1) ... hence percent is 10.8!
DLS
  • DLS
okayy
DLS
  • DLS
answer is 10% though!
anonymous
  • anonymous
wait . lemme see if i did some calc. mistake!
DLS
  • DLS
the solution is solved using differentiation n stuff
anonymous
  • anonymous
!? :-/ why differentiation? :O :O
DLS
  • DLS
DK
DLS
  • DLS
\[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{-2}) \]
DLS
  • DLS
Solving this we got, \[\frac{-2g}{r}\]
anonymous
  • anonymous
ohh.. like that.. ! thats not really required!!
DLS
  • DLS
dg/g=-2dR/R
DLS
  • DLS
:O
DLS
  • DLS
at last we get 0.1*100=10%!!
anonymous
  • anonymous
i mean it can be done in the normal method i told :P.. but this method holds too.. i dunno why we getting answer difference :D.. maybe something is wrong!! lemme recheck and get back
anonymous
  • anonymous
i find no flaw in calc.. i dunno really what went wrong :-/
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
There is a difference, because the differential method gives only a fist-order approximation of the result. The real answer is the one in which you work out the new value of g. If the shrinking had been 20%, the differential method would have lead to a very wrong value. On the other hand, if the shrinking were 0.1%, then your calculator would probably have problems coping with two big numbers almost identical to subtract, whereas the differential method would give a reliable answer.
anonymous
  • anonymous
ow.. thaks vincent for clearing that up! But why does the differential method give only an approximation!?... dg/dr, means change in the value of g for a change in value of r.. so why is it an approximation?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
|dw:1353660718894:dw| My dr should have been negative to match your problem. If dr is too big, the first order differential can lead you (by following the tangent and not the curve) to a value far from the real one.
anonymous
  • anonymous
aha i get it now.. thanks .. ! :)
anonymous
  • anonymous
hence differentiation is used for instantaneous change right?!...
anonymous
  • anonymous
the new value og g will be 110.80% og the original g and hence the difference is 10.8%. \[g =GM divR ^{2}\]. The effective R= .95 R original. All the other values remain the same. Just divide by .95sq.

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