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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0radius shrinks by 5% means.. the radius becomes 95/100(Re  current earths radius).. So can you solve now?

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0i know that much i did that too didnt get the answer though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait... 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?

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0i have to give the answer is "%" change

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok we ll work towards it don't worry.. !

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer my question first :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:O.. no.. how did you get that? :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0listen .. use the equation of g..

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{95}{100} \times 6400\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes.. but i want the value of g.. not the radius!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes exactly.. now put the new value of R .. and get the ratio!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0correct.. 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
 4 years ago
Best ResponseYou've already chosen the best response.020GM/19R is the new one? :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you plug in the values.. you should get g (prime)/ g = 1.108

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no.. see.. the increase is .108 ( 1.1081) ... hence percent is 10.8!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait . lemme see if i did some calc. mistake!

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0the solution is solved using differentiation n stuff

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0!? :/ why differentiation? :O :O

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{2}) \]

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0Solving this we got, \[\frac{2g}{r}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh.. like that.. ! thats not really required!!

DLS
 4 years ago
Best ResponseYou've already chosen the best response.0at last we get 0.1*100=10%!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0i find no flaw in calc.. i dunno really what went wrong :/

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.0There 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ow.. 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?

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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
 4 years ago
Best ResponseYou've already chosen the best response.0aha i get it now.. thanks .. ! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hence differentiation is used for instantaneous change right?!...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 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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