A community for students.
Here's the question you clicked on:
 0 viewing
mayankdevnani
 4 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
mayankdevnani
 4 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

This Question is Closed

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0it is absolutely wrong sis? @ShikhaDessai

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm not sure........never solved this kind of a problem before

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kepler's third law bud. \[\Large \frac{T_1^{2}}{T_2^{2}}=\frac{r_1^{3}}{r_2^{3}}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0umm because the universe obeys the law of gravity?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let height be h first and whoose time period is 5hrs h>4h by keplers third lawdw:1354629739334:dwdw:1354629866140:dw

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0but there is no word is used to denote gravity @rajathsbhat

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0C = 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
 4 years ago
Best ResponseYou've already chosen the best response.0it's gravity that keeps the satellite in orbit you know...

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0@Aperogalics where does 64 comes?

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

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 you take very highlevel, i did'nt understand it!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whaddya mean amistere?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 you can do that it will be same :)

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0@rajathsbhat i tihnk there is no gravity in space

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0note 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
 4 years ago
Best ResponseYou've already chosen the best response.0p.s. gravity exists everywhere. in space away from mass, gravity will be very very close to zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mayank, it's very common to think that, believe it or not!! but lemme show you a video...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but i still think its 20h

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0*video from veritasium coming up*

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0@Shadowys formula is correct or @Aperogalics formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.youtube.com/watch?v=d57C2drB_wc&feature=edu&list=PL772556F1EFC4D01C

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0didn't we both use the same thing? i just derived Apero's formula.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mayankdevnani it is true that is actually keplers third law:)

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0why should we use kepler's third law?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0also watch this http://www.youtube.com/watch?v=iQOHRKKNNLQ&feature=edu&list=PL772556F1EFC4D01C

mayankdevnani
 4 years ago
Best ResponseYou've already chosen the best response.0thnx... @shadowys @Aperogalics @rajathsbhat thnx...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.