## mayankdevnani 3 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

1. anonymous

20h i guess

2. mayankdevnani

it is absolutely wrong sis? @ShikhaDessai

3. mayankdevnani

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

4. anonymous

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

5. anonymous

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

6. mayankdevnani

ok...then

7. mayankdevnani

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

8. anonymous

umm because the universe obeys the law of gravity?

9. anonymous

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

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

11. 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 :)

12. anonymous

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

13. mayankdevnani

ok

14. mayankdevnani

@Aperogalics where does 64 comes?

15. anonymous

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

16. amistre64

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

17. mayankdevnani

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

18. anonymous

19. anonymous

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

20. mayankdevnani

@rajathsbhat i tihnk there is no gravity in space

21. 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.

22. anonymous

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

23. anonymous

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

24. mayankdevnani

who is lemme?

25. anonymous

lemme=let me.

26. mayankdevnani

where is video?

27. anonymous

but i still think its 20h

28. anonymous

*video from veritasium coming up*

29. mayankdevnani

ohkk

30. mayankdevnani

@Shadowys formula is correct or @Aperogalics formula

31. anonymous
32. anonymous

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

33. anonymous

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

34. mayankdevnani

ok

35. mayankdevnani

why should we use kepler's third law?

36. 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

37. 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;)

38. 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.

39. anonymous
40. mayankdevnani

41. anonymous

ur welcome:)

42. anonymous

you're welcome :)