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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

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

    • one year ago
  2. DLS Group Title
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    i know that much i did that too didnt get the answer though

    • one year ago
  3. Mashy Group Title
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    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?

    • one year ago
  4. DLS Group Title
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    i have to give the answer is "%" change

    • one year ago
  5. Mashy Group Title
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    Ok we ll work towards it don't worry.. !

    • one year ago
  6. Mashy Group Title
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    answer my question first :P

    • one year ago
  7. DLS Group Title
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    6400/6080 ?

    • one year ago
  8. Mashy Group Title
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    :O.. no.. how did you get that? :-/

    • one year ago
  9. Mashy Group Title
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    listen .. use the equation of g..

    • one year ago
  10. DLS Group Title
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    \[\frac{95}{100} \times 6400\]

    • one year ago
  11. DLS Group Title
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    New one^^

    • one year ago
  12. Mashy Group Title
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    yes.. but i want the value of g.. not the radius!

    • one year ago
  13. DLS Group Title
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    \[g= \frac{GM}{R ^{2}}\]

    • one year ago
  14. DLS Group Title
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    R'=95/100R

    • one year ago
  15. Mashy Group Title
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    yes exactly.. now put the new value of R .. and get the ratio!!

    • one year ago
  16. Mashy Group Title
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    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?

    • one year ago
  17. DLS Group Title
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    20GM/19R is the new one? :S

    • one year ago
  18. Mashy Group Title
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    if you plug in the values.. you should get g (prime)/ g = 1.108

    • one year ago
  19. DLS Group Title
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    yeah..okay

    • one year ago
  20. DLS Group Title
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    so 11%?

    • one year ago
  21. Mashy Group Title
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    10.8 percent..

    • one year ago
  22. DLS Group Title
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    11.08%?

    • one year ago
  23. Mashy Group Title
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    no.. see.. the increase is .108 ( 1.108-1) ... hence percent is 10.8!

    • one year ago
  24. DLS Group Title
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    okayy

    • one year ago
  25. DLS Group Title
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    answer is 10% though!

    • one year ago
  26. Mashy Group Title
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    wait . lemme see if i did some calc. mistake!

    • one year ago
  27. DLS Group Title
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    the solution is solved using differentiation n stuff

    • one year ago
  28. Mashy Group Title
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    !? :-/ why differentiation? :O :O

    • one year ago
  29. DLS Group Title
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    DK

    • one year ago
  30. DLS Group Title
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    \[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{-2}) \]

    • one year ago
  31. DLS Group Title
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    Solving this we got, \[\frac{-2g}{r}\]

    • one year ago
  32. Mashy Group Title
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    ohh.. like that.. ! thats not really required!!

    • one year ago
  33. DLS Group Title
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    dg/g=-2dR/R

    • one year ago
  34. DLS Group Title
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    :O

    • one year ago
  35. DLS Group Title
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    at last we get 0.1*100=10%!!

    • one year ago
  36. Mashy Group Title
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    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

    • one year ago
  37. Mashy Group Title
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    i find no flaw in calc.. i dunno really what went wrong :-/

    • one year ago
  38. Vincent-Lyon.Fr Group Title
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    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.

    • one year ago
  39. Mashy Group Title
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    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?

    • one year ago
  40. Vincent-Lyon.Fr Group Title
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    |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.

    • one year ago
  41. Mashy Group Title
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    aha i get it now.. thanks .. ! :)

    • one year ago
  42. Mashy Group Title
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    hence differentiation is used for instantaneous change right?!...

    • one year ago
  43. pkhawadiya Group Title
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    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.

    • one year ago
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