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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

Physics
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20h i guess
it is absolutely wrong sis? @ShikhaDessai

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Other answers:

i'm not sure........never solved this kind of a problem before
kepler's third law bud. \[\Large \frac{T_1^{2}}{T_2^{2}}=\frac{r_1^{3}}{r_2^{3}}\]
ok...then
why should we use kepler's 3 law, why should'nt we use 5*4=20 hrs
umm because the universe obeys the law of gravity?
let height be h first and whoose time period is 5hrs h---->4h by keplers third law|dw:1354629739334:dw||dw:1354629866140:dw|
but there is no word is used to denote gravity @rajathsbhat
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 :)
it's gravity that keeps the satellite in orbit you know...
ok
@Aperogalics where does 64 comes?
4^3=4*4*4=64:)
but, if we are using celestial laws :) isnt the area swept out the same for each condition?
@amistre64 you take very high-level, i did'nt understand it!
whaddya mean amistere?
@amistre64 you can do that it will be same :)
@rajathsbhat i tihnk there is no gravity in space
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.
p.s. gravity exists everywhere. in space away from mass, gravity will be very very close to zero.
mayank, it's very common to think that, believe it or not!! but lemme show you a video...
who is lemme?
lemme=let me.
where is video?
but i still think its 20h
*video from veritasium coming up*
ohkk
@Shadowys formula is correct or @Aperogalics formula
http://www.youtube.com/watch?v=d57C2drB_wc&feature=edu&list=PL772556F1EFC4D01C
didn't we both use the same thing? i just derived Apero's formula.
@mayankdevnani it is true that is actually keplers third law:)
ok
why should we use kepler's third law?
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
@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;)
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.
also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C
ur welcome:)
you're welcome :)

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