DLS Group Title 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 Group Title

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

2. DLS Group Title

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

3. Mashy Group Title

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 Group Title

i have to give the answer is "%" change

5. Mashy Group Title

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

6. Mashy Group Title

7. DLS Group Title

6400/6080 ?

8. Mashy Group Title

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

9. Mashy Group Title

listen .. use the equation of g..

10. DLS Group Title

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

11. DLS Group Title

New one^^

12. Mashy Group Title

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

13. DLS Group Title

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

14. DLS Group Title

R'=95/100R

15. Mashy Group Title

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

16. Mashy Group Title

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 Group Title

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

18. Mashy Group Title

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

19. DLS Group Title

yeah..okay

20. DLS Group Title

so 11%?

21. Mashy Group Title

10.8 percent..

22. DLS Group Title

11.08%?

23. Mashy Group Title

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

24. DLS Group Title

okayy

25. DLS Group Title

26. Mashy Group Title

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

27. DLS Group Title

the solution is solved using differentiation n stuff

28. Mashy Group Title

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

29. DLS Group Title

DK

30. DLS Group Title

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

31. DLS Group Title

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

32. Mashy Group Title

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

33. DLS Group Title

dg/g=-2dR/R

34. DLS Group Title

:O

35. DLS Group Title

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

36. Mashy Group Title

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 Group Title

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

38. Vincent-Lyon.Fr Group Title

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 Group Title

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 Group Title

|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 Group Title

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

42. Mashy Group Title

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.