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mayankdevnani

  • 2 years ago

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

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  1. ShikhaDessai
    • 2 years ago
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    20h i guess

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

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

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

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

  6. mayankdevnani
    • 2 years ago
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    ok...then

  7. mayankdevnani
    • 2 years ago
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    why should we use kepler's 3 law, why should'nt we use 5*4=20 hrs

  8. rajathsbhat
    • 2 years ago
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    umm because the universe obeys the law of gravity?

  9. Aperogalics
    • 2 years ago
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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|

  10. mayankdevnani
    • 2 years ago
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    but there is no word is used to denote gravity @rajathsbhat

  11. amistre64
    • 2 years ago
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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 :)

  12. rajathsbhat
    • 2 years ago
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    it's gravity that keeps the satellite in orbit you know...

  13. mayankdevnani
    • 2 years ago
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    ok

  14. mayankdevnani
    • 2 years ago
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    @Aperogalics where does 64 comes?

  15. Aperogalics
    • 2 years ago
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    4^3=4*4*4=64:)

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

  17. mayankdevnani
    • 2 years ago
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    @amistre64 you take very high-level, i did'nt understand it!

  18. rajathsbhat
    • 2 years ago
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    whaddya mean amistere?

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

  20. mayankdevnani
    • 2 years ago
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    @rajathsbhat i tihnk there is no gravity in space

  21. Shadowys
    • 2 years ago
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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.

  22. Shadowys
    • 2 years ago
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    p.s. gravity exists everywhere. in space away from mass, gravity will be very very close to zero.

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

  24. mayankdevnani
    • 2 years ago
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    who is lemme?

  25. rajathsbhat
    • 2 years ago
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    lemme=let me.

  26. mayankdevnani
    • 2 years ago
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    where is video?

  27. ShikhaDessai
    • 2 years ago
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    but i still think its 20h

  28. rajathsbhat
    • 2 years ago
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    *video from veritasium coming up*

  29. mayankdevnani
    • 2 years ago
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    ohkk

  30. mayankdevnani
    • 2 years ago
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    @Shadowys formula is correct or @Aperogalics formula

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

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

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

  34. mayankdevnani
    • 2 years ago
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    ok

  35. mayankdevnani
    • 2 years ago
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    why should we use kepler's third law?

  36. Aperogalics
    • 2 years ago
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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

  37. Aperogalics
    • 2 years ago
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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;)

  38. Shadowys
    • 2 years ago
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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.

  39. rajathsbhat
    • 2 years ago
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    also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C

  40. mayankdevnani
    • 2 years ago
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    thnx... @shadowys @Aperogalics @rajathsbhat thnx...

  41. Aperogalics
    • 2 years ago
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    ur welcome:)

  42. Shadowys
    • 2 years ago
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    you're welcome :)

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