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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
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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ShikhaDessai Group TitleBest ResponseYou've already chosen the best response.0
20h i guess
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
it is absolutely wrong sis? @ShikhaDessai
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 @Callisto @pratu043 @inkyvoyd @eyust707 @rajathsbhat @Aperogalics @sirm3d @Shadowys plz help me guys:)
 one year ago

ShikhaDessai Group TitleBest ResponseYou've already chosen the best response.0
i'm not sure........never solved this kind of a problem before
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
kepler's third law bud. \[\Large \frac{T_1^{2}}{T_2^{2}}=\frac{r_1^{3}}{r_2^{3}}\]
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
ok...then
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
why should we use kepler's 3 law, why should'nt we use 5*4=20 hrs
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
umm because the universe obeys the law of gravity?
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
let height be h first and whoose time period is 5hrs h>4h by keplers third lawdw:1354629739334:dwdw:1354629866140:dw
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
but there is no word is used to denote gravity @rajathsbhat
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
it's gravity that keeps the satellite in orbit you know...
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@Aperogalics where does 64 comes?
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
4^3=4*4*4=64:)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
but, if we are using celestial laws :) isnt the area swept out the same for each condition?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 you take very highlevel, i did'nt understand it!
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
whaddya mean amistere?
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
@amistre64 you can do that it will be same :)
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@rajathsbhat i tihnk there is no gravity in space
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
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

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
p.s. gravity exists everywhere. in space away from mass, gravity will be very very close to zero.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
mayank, it's very common to think that, believe it or not!! but lemme show you a video...
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
who is lemme?
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
lemme=let me.
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
where is video?
 one year ago

ShikhaDessai Group TitleBest ResponseYou've already chosen the best response.0
but i still think its 20h
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
*video from veritasium coming up*
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
ohkk
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@Shadowys formula is correct or @Aperogalics formula
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
http://www.youtube.com/watch?v=d57C2drB_wc&feature=edu&list=PL772556F1EFC4D01C
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
didn't we both use the same thing? i just derived Apero's formula.
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
@mayankdevnani it is true that is actually keplers third law:)
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
why should we use kepler's third law?
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
http://www.tutorvista.com/content/physics/physicsiii/gravitation/satellites.php#periodofrevolutionortimeperiodofasatellite Read this it will help u a lot .................as i think ur basic concepts are very week
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
@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

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
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

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
thnx... @shadowys @Aperogalics @rajathsbhat thnx...
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.1
ur welcome:)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
you're welcome :)
 one year ago
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