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

- mayankdevnani

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- anonymous

20h i guess

- mayankdevnani

it is absolutely wrong sis? @ShikhaDessai

- mayankdevnani

@amistre64 @Callisto @pratu043 @inkyvoyd @eyust707 @rajathsbhat @Aperogalics @sirm3d @Shadowys plz help me guys:)

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## More answers

- anonymous

i'm not sure........never solved this kind of a problem before

- anonymous

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

- mayankdevnani

ok...then

- mayankdevnani

why should we use kepler's 3 law, why should'nt we use 5*4=20 hrs

- anonymous

umm because the universe obeys the law of gravity?

- anonymous

let height be h first and whoose time period is 5hrs
h---->4h
by keplers third law|dw:1354629739334:dw||dw:1354629866140:dw|

- mayankdevnani

but there is no word is used to denote gravity @rajathsbhat

- amistre64

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 :)

- anonymous

it's gravity that keeps the satellite in orbit you know...

- mayankdevnani

ok

- mayankdevnani

@Aperogalics where does 64 comes?

- anonymous

4^3=4*4*4=64:)

- amistre64

but, if we are using celestial laws :) isnt the area swept out the same for each condition?

- mayankdevnani

@amistre64 you take very high-level, i did'nt understand it!

- anonymous

whaddya mean amistere?

- anonymous

@amistre64 you can do that it will be same :)

- mayankdevnani

@rajathsbhat i tihnk there is no gravity in space

- anonymous

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.

- anonymous

p.s. gravity exists everywhere. in space away from mass, gravity will be very very close to zero.

- anonymous

mayank, it's very common to think that, believe it or not!! but lemme show you a video...

- mayankdevnani

who is lemme?

- anonymous

lemme=let me.

- mayankdevnani

where is video?

- anonymous

but i still think its 20h

- anonymous

*video from veritasium coming up*

- mayankdevnani

ohkk

- mayankdevnani

@Shadowys formula is correct or @Aperogalics formula

- anonymous

http://www.youtube.com/watch?v=d57C2drB_wc&feature=edu&list=PL772556F1EFC4D01C

- anonymous

didn't we both use the same thing? i just derived Apero's formula.

- anonymous

@mayankdevnani it is true that is actually keplers third law:)

- mayankdevnani

ok

- mayankdevnani

why should we use kepler's third law?

- anonymous

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

- anonymous

@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;)

- anonymous

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.

- anonymous

also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C

- mayankdevnani

thnx... @shadowys @Aperogalics @rajathsbhat thnx...

- anonymous

ur welcome:)

- anonymous

you're welcome :)

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