Satellite in Orbit (Quiz#4, 26)

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A satellite in orbit around Earth has a period of one hour, while a second satellite is placed in an orbit with a radius nine times larger. The discussion revolves around determining the period of the second satellite without using Kepler's laws. Participants explore the relationship between gravitational force, centripetal acceleration, and the time period of circular orbits, emphasizing the need to derive the period using basic physics principles. The conclusion drawn is that the period of the second satellite is 27 hours, based on the relationship between radius and orbital period. The conversation highlights the challenges faced by those unfamiliar with the underlying physics concepts.
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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
 
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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".
 
It's Kepler's Laws, not Keplen.
 
the Fc is caused by gravity
 
And what is the expression for the force of gravity?

ehild
 
it is GMn/R^2

but what should I do with it if I am looking for the TIME
 
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How is time related to speed and distance traveled ?

What distance travels a satellite along its whole orbit?
 
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
 
  • #10
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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  • #12
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
 
  • #13
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 ?
 
  • #14
Read the last paragraph. It solves the problem with circular orbits.
 
  • #15
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
  • #17
ill read that
 
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