Satellite in Orbit (Quiz#4, 26)

1. Oct 2, 2014

gcombina

• Warning: Infraction points awarded to gcombina for not using the homework template.
A satellite in orbit around the earth has a period of one hour. An identical satellite is placed in an orbit having a radius that is nine times larger than that of the first satellite. What is the period of the second satellite?

(a) 27 h (c) 4 h (e) 0.04 h

(b) 3 h (d) 9 h

** I truly don't know where to start.
I have seen (online) people using equation for force of gravity, and centripetal acceleration but they dont explain why

**Please do not use Keplen laws cause I havent studied that

My attempt:
The period is the distance of the orbit which is 2piR
The first satellite moves because it has centripetal force which is Fc= mv^2/R
The first satellite also has acceleration which is ac = v^2/R

2. Oct 2, 2014

BvU

What is causing the Fc ? You need an expression for that, preferably one that also has R in it.
Use the template. Read the guidelines. Don't write "I don't know where to start".

3. Oct 2, 2014

SteamKing

Staff Emeritus
It's Kepler's Laws, not Keplen.

4. Oct 3, 2014

gcombina

the Fc is caused by gravity

5. Oct 3, 2014

ehild

And what is the expression for the force of gravity?

ehild

6. Oct 3, 2014

gcombina

it is GMn/R^2

but what should I do with it if I am looking for the TIME

Last edited: Oct 3, 2014
7. Oct 3, 2014

ehild

How is time related to speed and distance travelled ?

What distance travels a satellite along its whole orbit?

8. Oct 3, 2014

BvU

Ah, good to see you now have an expression for the force that functions as centripetal force. And it has an R in it too! Splendid! You can work it around to get something for v. And travelling a circular orbit of radius R brings you back to the same point in a time ....

By the way, doesn't anyone of the helpers worry about the 1 hour period of the first satellite ? I got something of an underground orbit... ? Fortunately, the exercise can be continued nevertheless: IF the R satellite circles in 1 hour, THEN the 9R satellite does so in ... hours.​

9. Oct 5, 2014

gcombina

sorry i am lost

you asked what is the relation between STD? speed time and distance?

T = d/s

10. Oct 5, 2014

BvU

We're getting there! What is d for one revolution in a circular orbit of radius R?

 Ah, I see you answered that already in post # 1: 'the distance of the orbit which is 2piR'

@ehild: can't we assume circular orbits? It's already difficult enough for GC I would guess...

Last edited: Oct 5, 2014
11. Oct 5, 2014

ehild

12. Oct 5, 2014

gcombina

yeah i read Kepler which is more simple
t^2/R^3 = T^2/(9R)^3
T = 27

BUT, they dont want me to use Kepler's law

13. Oct 5, 2014

BvU

Goodness. Post 1 said no Kepler. And it has R and 9R, so I take it they want you to derive this for circular orbits. You have $GMm/R^2=mv^2/R$ , you have $T = d/v$ and you have $d = 2\pi R$. Play around until you have $T=...$!

14. Oct 5, 2014

ehild

Read the last paragraph. It solves the problem with circular orbits.

15. Oct 5, 2014

gcombina

ok sorry, i suck at physics but need to pass this only course which is a prereq for my major which has nothing to do with physics!. Please be patient

16. Oct 5, 2014

BvU

17. Oct 5, 2014