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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?
radius shrinks by 5% means.. the radius becomes 95/100(Re - current earths radius).. So can you solve now?
i know that much i did that too didnt get the answer though
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?
i have to give the answer is "%" change
Ok we ll work towards it don't worry.. !
answer my question first :P
:O.. no.. how did you get that? :-/
listen .. use the equation of g..
\[\frac{95}{100} \times 6400\]
yes.. but i want the value of g.. not the radius!
yes exactly.. now put the new value of R .. and get the ratio!!
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?
20GM/19R is the new one? :S
if you plug in the values.. you should get g (prime)/ g = 1.108
no.. see.. the increase is .108 ( 1.108-1) ... hence percent is 10.8!
wait . lemme see if i did some calc. mistake!
the solution is solved using differentiation n stuff
!? :-/ why differentiation? :O :O
\[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{-2}) \]
Solving this we got, \[\frac{-2g}{r}\]
ohh.. like that.. ! thats not really required!!
at last we get 0.1*100=10%!!
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
i find no flaw in calc.. i dunno really what went wrong :-/
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.
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?
|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.
aha i get it now.. thanks .. ! :)
hence differentiation is used for instantaneous change right?!...
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.