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mayankdevnani

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
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    20h i guess

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

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

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

    • one year ago
  5. rajathsbhat
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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
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    ok...then

    • one year ago
  7. mayankdevnani
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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
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    umm because the universe obeys the law of gravity?

    • one year ago
  9. Aperogalics
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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
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    but there is no word is used to denote gravity @rajathsbhat

    • one year ago
  11. amistre64
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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
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    it's gravity that keeps the satellite in orbit you know...

    • one year ago
  13. mayankdevnani
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    ok

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

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

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

    • one year ago
  18. rajathsbhat
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    whaddya mean amistere?

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

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

    • one year ago
  21. Shadowys
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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
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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
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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
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    who is lemme?

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

    • one year ago
  26. mayankdevnani
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    where is video?

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

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

    • one year ago
  29. mayankdevnani
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    ohkk

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

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

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

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

    • one year ago
  34. mayankdevnani
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    ok

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

    • one year ago
  36. Aperogalics
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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
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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
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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
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    also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C

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

    • one year ago
  41. Aperogalics
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    ur welcome:)

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

    • one year ago
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