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mayankdevnani Group Title

The time period of a satellite is 5 hours. If the separation between the earth and the satellite is increased to 4 times what will be the new time period? a) 20 hrs b) 10 hrs c) 80 hrs d) 40 hrs

  • one year ago
  • one year ago

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  1. ShikhaDessai Group Title
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    20h i guess

    • one year ago
  2. mayankdevnani Group Title
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    it is absolutely wrong sis? @ShikhaDessai

    • one year ago
  3. mayankdevnani Group Title
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    @amistre64 @Callisto @pratu043 @inkyvoyd @eyust707 @rajathsbhat @Aperogalics @sirm3d @Shadowys plz help me guys:)

    • one year ago
  4. ShikhaDessai Group Title
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    i'm not sure........never solved this kind of a problem before

    • one year ago
  5. rajathsbhat Group Title
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    kepler's third law bud. \[\Large \frac{T_1^{2}}{T_2^{2}}=\frac{r_1^{3}}{r_2^{3}}\]

    • one year ago
  6. mayankdevnani Group Title
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    ok...then

    • one year ago
  7. mayankdevnani Group Title
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    why should we use kepler's 3 law, why should'nt we use 5*4=20 hrs

    • one year ago
  8. rajathsbhat Group Title
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    umm because the universe obeys the law of gravity?

    • one year ago
  9. Aperogalics Group Title
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    let height be h first and whoose time period is 5hrs h---->4h by keplers third law|dw:1354629739334:dw||dw:1354629866140:dw|

    • one year ago
  10. mayankdevnani Group Title
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    but there is no word is used to denote gravity @rajathsbhat

    • one year ago
  11. amistre64 Group Title
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    C = 2 pi (Er+d) C/2pi = Er+d C/2pi - Er = d 4C/2pi - 4Er = 4d 4C = 2pi 4(Er +d) when the distance between E and satillite is 4times as much, given that i havent completely gotten a botched idea in me head :)

    • one year ago
  12. rajathsbhat Group Title
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    it's gravity that keeps the satellite in orbit you know...

    • one year ago
  13. mayankdevnani Group Title
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    ok

    • one year ago
  14. mayankdevnani Group Title
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    @Aperogalics where does 64 comes?

    • one year ago
  15. Aperogalics Group Title
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    4^3=4*4*4=64:)

    • one year ago
  16. amistre64 Group Title
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    but, if we are using celestial laws :) isnt the area swept out the same for each condition?

    • one year ago
  17. mayankdevnani Group Title
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    @amistre64 you take very high-level, i did'nt understand it!

    • one year ago
  18. rajathsbhat Group Title
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    whaddya mean amistere?

    • one year ago
  19. Aperogalics Group Title
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    @amistre64 you can do that it will be same :)

    • one year ago
  20. mayankdevnani Group Title
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    @rajathsbhat i tihnk there is no gravity in space

    • one year ago
  21. Shadowys Group Title
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    note that \(F_g = \frac{GMm}{r^2}\) and \(g=\frac{GM}{r^2}\), \(v=\frac{2 \pi}{T}\) you can also use this: \(a=g=\frac{v^2}{r}=\frac{4\pi^2}{T^2 r}\) rearranging, and substituing \(g=\frac{GM}{r^2}\), \(\frac{GM}{r^2}=\frac{4\pi^2}{T^2 r}\) \(\frac{T^2}{r^3}=\frac{4\pi^2}{GM}= constant\) so you have to use this formula.

    • one year ago
  22. Shadowys Group Title
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    p.s. gravity exists everywhere. in space away from mass, gravity will be very very close to zero.

    • one year ago
  23. rajathsbhat Group Title
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    mayank, it's very common to think that, believe it or not!! but lemme show you a video...

    • one year ago
  24. mayankdevnani Group Title
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    who is lemme?

    • one year ago
  25. rajathsbhat Group Title
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    lemme=let me.

    • one year ago
  26. mayankdevnani Group Title
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    where is video?

    • one year ago
  27. ShikhaDessai Group Title
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    but i still think its 20h

    • one year ago
  28. rajathsbhat Group Title
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    *video from veritasium coming up*

    • one year ago
  29. mayankdevnani Group Title
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    ohkk

    • one year ago
  30. mayankdevnani Group Title
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    @Shadowys formula is correct or @Aperogalics formula

    • one year ago
  31. rajathsbhat Group Title
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    http://www.youtube.com/watch?v=d57C2drB_wc&feature=edu&list=PL772556F1EFC4D01C

    • one year ago
  32. Shadowys Group Title
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    didn't we both use the same thing? i just derived Apero's formula.

    • one year ago
  33. Aperogalics Group Title
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    @mayankdevnani it is true that is actually keplers third law:)

    • one year ago
  34. mayankdevnani Group Title
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    ok

    • one year ago
  35. mayankdevnani Group Title
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    why should we use kepler's third law?

    • one year ago
  36. Aperogalics Group Title
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    http://www.tutorvista.com/content/physics/physics-iii/gravitation/satellites.php#period-of-revolution-or-time-period-of-a-satellite Read this it will help u a lot .................as i think ur basic concepts are very week

    • one year ago
  37. Aperogalics Group Title
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    @mayankdevnani well it is same to ask y 2+2=4 :) well kepler derives these laws if you dont use them then you still use them;)

    • one year ago
  38. Shadowys Group Title
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    The satellite is in orbit, so there must be a force acting on it preventing it from flying away. the only force available is gravity, which exists everywhere. And Kepler's third law follows the law of Newtons (derived above) so they hold valid.

    • one year ago
  39. rajathsbhat Group Title
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    also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C

    • one year ago
  40. mayankdevnani Group Title
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    thnx... @shadowys @Aperogalics @rajathsbhat thnx...

    • one year ago
  41. Aperogalics Group Title
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    ur welcome:)

    • one year ago
  42. Shadowys Group Title
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    you're welcome :)

    • one year ago
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