Satellite in Orbit (Quiz#4, 26)

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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 don't explain why

**Please do not use Keplen laws cause I haven't 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
 
on Phys.org
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".
 
the Fc is caused by gravity
 
it is GMn/R^2

but what should I do with it if I am looking for the TIME
 
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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 traveling 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.​
 
sorry i am lost

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

T = d/s
 
We're getting there! What is d for one revolution in a circular orbit of radius R?

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

BUT, they don't want me to use Kepler's law
 
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=...##!

@ehild: what about this 1 hour period infeasibility ?
 
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