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

20h i guess

2. mayankdevnani Group Title

it is absolutely wrong sis? @ShikhaDessai

3. mayankdevnani Group Title

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

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

5. rajathsbhat Group Title

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

6. mayankdevnani Group Title

ok...then

7. mayankdevnani Group Title

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

8. rajathsbhat Group Title

umm because the universe obeys the law of gravity?

9. Aperogalics Group Title

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 Group Title

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

11. amistre64 Group Title

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 Group Title

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

13. mayankdevnani Group Title

ok

14. mayankdevnani Group Title

@Aperogalics where does 64 comes?

15. Aperogalics Group Title

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

16. amistre64 Group Title

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

17. mayankdevnani Group Title

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

18. rajathsbhat Group Title

19. Aperogalics Group Title

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

20. mayankdevnani Group Title

@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.

23. rajathsbhat Group Title

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

24. mayankdevnani Group Title

who is lemme?

25. rajathsbhat Group Title

lemme=let me.

26. mayankdevnani Group Title

where is video?

but i still think its 20h

28. rajathsbhat Group Title

*video from veritasium coming up*

29. mayankdevnani Group Title

ohkk

30. mayankdevnani Group Title

@Shadowys formula is correct or @Aperogalics formula

31. rajathsbhat Group Title

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

33. Aperogalics Group Title

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

34. mayankdevnani Group Title

ok

35. mayankdevnani Group Title

why should we use kepler's third law?

36. Aperogalics Group Title

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 Group Title

@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.

39. rajathsbhat Group Title
40. mayankdevnani Group Title