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If the radius of the earth shrinks by 5%,mass remaining the same,then how much value of acc. due to gravity change?
 one year ago
 one year ago
If the radius of the earth shrinks by 5%,mass remaining the same,then how much value of acc. due to gravity change?
 one year ago
 one year ago

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MashyBest ResponseYou've already chosen the best response.1
radius shrinks by 5% means.. the radius becomes 95/100(Re  current earths radius).. So can you solve now?
 one year ago

DLSBest ResponseYou've already chosen the best response.0
i know that much i did that too didnt get the answer though
 one year ago

MashyBest ResponseYou've already chosen the best response.1
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?
 one year ago

DLSBest ResponseYou've already chosen the best response.0
i have to give the answer is "%" change
 one year ago

MashyBest ResponseYou've already chosen the best response.1
Ok we ll work towards it don't worry.. !
 one year ago

MashyBest ResponseYou've already chosen the best response.1
answer my question first :P
 one year ago

MashyBest ResponseYou've already chosen the best response.1
:O.. no.. how did you get that? :/
 one year ago

MashyBest ResponseYou've already chosen the best response.1
listen .. use the equation of g..
 one year ago

DLSBest ResponseYou've already chosen the best response.0
\[\frac{95}{100} \times 6400\]
 one year ago

MashyBest ResponseYou've already chosen the best response.1
yes.. but i want the value of g.. not the radius!
 one year ago

MashyBest ResponseYou've already chosen the best response.1
yes exactly.. now put the new value of R .. and get the ratio!!
 one year ago

MashyBest ResponseYou've already chosen the best response.1
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?
 one year ago

DLSBest ResponseYou've already chosen the best response.0
20GM/19R is the new one? :S
 one year ago

MashyBest ResponseYou've already chosen the best response.1
if you plug in the values.. you should get g (prime)/ g = 1.108
 one year ago

MashyBest ResponseYou've already chosen the best response.1
no.. see.. the increase is .108 ( 1.1081) ... hence percent is 10.8!
 one year ago

MashyBest ResponseYou've already chosen the best response.1
wait . lemme see if i did some calc. mistake!
 one year ago

DLSBest ResponseYou've already chosen the best response.0
the solution is solved using differentiation n stuff
 one year ago

MashyBest ResponseYou've already chosen the best response.1
!? :/ why differentiation? :O :O
 one year ago

DLSBest ResponseYou've already chosen the best response.0
\[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{2}) \]
 one year ago

DLSBest ResponseYou've already chosen the best response.0
Solving this we got, \[\frac{2g}{r}\]
 one year ago

MashyBest ResponseYou've already chosen the best response.1
ohh.. like that.. ! thats not really required!!
 one year ago

DLSBest ResponseYou've already chosen the best response.0
at last we get 0.1*100=10%!!
 one year ago

MashyBest ResponseYou've already chosen the best response.1
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
 one year ago

MashyBest ResponseYou've already chosen the best response.1
i find no flaw in calc.. i dunno really what went wrong :/
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
There is a difference, because the differential method gives only a fistorder 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.
 one year ago

MashyBest ResponseYou've already chosen the best response.1
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?
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
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.
 one year ago

MashyBest ResponseYou've already chosen the best response.1
aha i get it now.. thanks .. ! :)
 one year ago

MashyBest ResponseYou've already chosen the best response.1
hence differentiation is used for instantaneous change right?!...
 one year ago

pkhawadiyaBest ResponseYou've already chosen the best response.0
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.
 one year ago
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