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DLS

  • 3 years ago

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

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

  2. DLS
    • 3 years ago
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    i know that much i did that too didnt get the answer though

  3. Mashy
    • 3 years ago
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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?

  4. DLS
    • 3 years ago
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    i have to give the answer is "%" change

  5. Mashy
    • 3 years ago
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    Ok we ll work towards it don't worry.. !

  6. Mashy
    • 3 years ago
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    answer my question first :P

  7. DLS
    • 3 years ago
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    6400/6080 ?

  8. Mashy
    • 3 years ago
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    :O.. no.. how did you get that? :-/

  9. Mashy
    • 3 years ago
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    listen .. use the equation of g..

  10. DLS
    • 3 years ago
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    \[\frac{95}{100} \times 6400\]

  11. DLS
    • 3 years ago
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    New one^^

  12. Mashy
    • 3 years ago
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    yes.. but i want the value of g.. not the radius!

  13. DLS
    • 3 years ago
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    \[g= \frac{GM}{R ^{2}}\]

  14. DLS
    • 3 years ago
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    R'=95/100R

  15. Mashy
    • 3 years ago
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    yes exactly.. now put the new value of R .. and get the ratio!!

  16. Mashy
    • 3 years ago
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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?

  17. DLS
    • 3 years ago
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    20GM/19R is the new one? :S

  18. Mashy
    • 3 years ago
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    if you plug in the values.. you should get g (prime)/ g = 1.108

  19. DLS
    • 3 years ago
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    yeah..okay

  20. DLS
    • 3 years ago
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    so 11%?

  21. Mashy
    • 3 years ago
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    10.8 percent..

  22. DLS
    • 3 years ago
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    11.08%?

  23. Mashy
    • 3 years ago
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    no.. see.. the increase is .108 ( 1.108-1) ... hence percent is 10.8!

  24. DLS
    • 3 years ago
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    okayy

  25. DLS
    • 3 years ago
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    answer is 10% though!

  26. Mashy
    • 3 years ago
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    wait . lemme see if i did some calc. mistake!

  27. DLS
    • 3 years ago
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    the solution is solved using differentiation n stuff

  28. Mashy
    • 3 years ago
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    !? :-/ why differentiation? :O :O

  29. DLS
    • 3 years ago
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    DK

  30. DLS
    • 3 years ago
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    \[\frac{dg}{dR} = \frac{d}{dR} ( \frac {GM}{R ^{2}} ) = GM \frac{d}{dR} (R^{-2}) \]

  31. DLS
    • 3 years ago
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    Solving this we got, \[\frac{-2g}{r}\]

  32. Mashy
    • 3 years ago
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    ohh.. like that.. ! thats not really required!!

  33. DLS
    • 3 years ago
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    dg/g=-2dR/R

  34. DLS
    • 3 years ago
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    :O

  35. DLS
    • 3 years ago
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    at last we get 0.1*100=10%!!

  36. Mashy
    • 3 years ago
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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

  37. Mashy
    • 3 years ago
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    i find no flaw in calc.. i dunno really what went wrong :-/

  38. Vincent-Lyon.Fr
    • 3 years ago
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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.

  39. Mashy
    • 3 years ago
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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?

  40. Vincent-Lyon.Fr
    • 3 years ago
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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.

  41. Mashy
    • 3 years ago
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    aha i get it now.. thanks .. ! :)

  42. Mashy
    • 3 years ago
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    hence differentiation is used for instantaneous change right?!...

  43. pkhawadiya
    • 3 years ago
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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.

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