# 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