Question

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

Answer 1

radius shrinks by 5% means.. the radius becomes 95/100(Re - current earths radius).. So can you solve now?

Answer 2

i know that much i did that too didnt get the answer though

Answer 3

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?

Answer 4

i have to give the answer is "%" change

Answer 5

Ok we ll work towards it don't worry.. !

Answer 6

answer my question first :P

Answer 7

6400/6080 ?

Answer 8

:O.. no.. how did you get that? :-/

Answer 9

listen .. use the equation of g..

Answer 10

\[\frac{95}{100} \times 6400\]

Answer 11

New one^^

Answer 12

yes.. but i want the value of g.. not the radius!

Answer 13

\[g= \frac{GM}{R ^{2}}\]

Answer 14

R'=95/100R

Answer 15

yes exactly.. now put the new value of R .. and get the ratio!!

Answer 16

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?

Answer 17

20GM/19R is the new one? :S

Answer 18

if you plug in the values.. you should get

g (prime)/ g = 1.108

Answer 19

yeah..okay

Answer 20

so 11%?

Answer 21

10.8 percent..

Answer 22

11.08%?

Answer 23

no.. see.. the increase is .108 ( 1.108-1) ... hence percent is 10.8!

Answer 24

okayy

Answer 25

answer is 10% though!

Answer 26

wait . lemme see if i did some calc. mistake!

Answer 27

the solution is solved using differentiation n stuff

Answer 28

!? :-/ why differentiation? :O :O

Answer 29

DK

Answer 30

\[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{-2}) \]

Answer 31

Solving this we got,
\[\frac{-2g}{r}\]

Answer 32

ohh.. like that.. ! thats not really required!!

Answer 33

dg/g=-2dR/R

Answer 34

:O

Answer 35

at last we get 0.1*100=10%!!

Answer 36

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

Answer 37

i find no flaw in calc.. i dunno really what went wrong :-/

Answer 38

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.

Answer 39

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?

Answer 40

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.