## DLS 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

1. Mashy

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

2. DLS

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

3. Mashy

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?

4. DLS

i have to give the answer is "%" change

5. Mashy

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

6. Mashy

7. DLS

6400/6080 ?

8. Mashy

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

9. Mashy

listen .. use the equation of g..

10. DLS

$\frac{95}{100} \times 6400$

11. DLS

New one^^

12. Mashy

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

13. DLS

$g= \frac{GM}{R ^{2}}$

14. DLS

R'=95/100R

15. Mashy

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

16. Mashy

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?

17. DLS

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

18. Mashy

if you plug in the values.. you should get g (prime)/ g = 1.108

19. DLS

yeah..okay

20. DLS

so 11%?

21. Mashy

10.8 percent..

22. DLS

11.08%?

23. Mashy

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

24. DLS

okayy

25. DLS

26. Mashy

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

27. DLS

the solution is solved using differentiation n stuff

28. Mashy

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

29. DLS

DK

30. DLS

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

31. DLS

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

32. Mashy

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

33. DLS

dg/g=-2dR/R

34. DLS

:O

35. DLS

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

36. Mashy

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

37. Mashy

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

38. 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.

39. Mashy

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?

40. 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.

41. Mashy

aha i get it now.. thanks .. ! :)

42. Mashy

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.