Why are v and a decreasing as the oribit radius increases?

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  • #1
simphys
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Homework Statement:
The planets Venus, Earth, and Mars all move in approximately
circular orbits around the sun. Use the data in the table to find (a) the
speed of each planet in its orbit and (b) the centripetal acceleration of each
planet. (c) As the size of a planet’s orbit increases, does the speed increase
or decrease? Does the centripetal acceleration increase or decrease?

Planet Orbital radius (m): Venus 1.08 * 1011 Orbital period (days): 225
Planet Orbital radius (m): Earth 1.50 * 1011 Orbital period (days): 365
Planet Orbital radius (m): Mars 2.28 * 1011 Orbital period (days): 687
Relevant Equations:
a = v^2 / R
I don't understand part (c) the answer is decreasing for both. I would have it decreasing for the centripetal acceleration and then increasing for the velocity, why is that not correct?

Thanks in advance!
 

Answers and Replies

  • #2
BvU
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You have been given ##R## and ##T## so you can calculate v -- if you have the right relevant equation (not ##a = v^2 / R##). You see the ##v## are decreasing with ##R##.
I would have it decreasing for the centripetal acceleration and then increasing for the velocity, why is that not correct?
Well, perhaps you have the wrong relevant equation ? What does ##a## have to do with anything ?

##\ ##
 
  • #3
simphys
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You have been given ##R## and ##T## so you can calculate v -- if you have the right relevant equation (not ##a = v^2 / R##). You see the ##v## are decreasing with ##R##.

Well, perhaps you have the wrong relevant equation ? What does ##a## have to do with anything ?

##\ ##
not the right one?? So are we talking Kepler's law that or something like that??

That is strange.. this chapter is about 2d-3d motion, I don't know orbital motions yet.
 
  • #4
PeroK
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Homework Statement:: The planets Venus, Earth, and Mars all move in approximately
circular orbits around the sun. Use the data in the table to find (a) the
speed of each planet in its orbit and (b) the centripetal acceleration of each
planet. (c) As the size of a planet’s orbit increases, does the speed increase
or decrease? Does the centripetal acceleration increase or decrease?

Planet Orbital radius (m): Venus 1.08 * 1011 Orbital period (days): 225
Planet Orbital radius (m): Earth 1.50 * 1011 Orbital period (days): 365
Planet Orbital radius (m): Mars 2.28 * 1011 Orbital period (days): 687
Relevant Equations:: a = v^2 / R

I don't understand part (c) the answer is decreasing for both. I would have it decreasing for the centripetal acceleration and then increasing for the velocity, why is that not correct?

Thanks in advance!
The planets are held in orbit by the Sun's gravity, which obeys an inverse square law. Everything follows from that.
 
  • #5
BvU
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For a circular orbit, the ##a## is constant and provided by the force of gravity -- for which you can think up a relevant equation of the form ##a=\ldots \ ##.

The two ##a## drop out and there remains something of the form ##v\propto \ldots \ ##.

##\ ##
 
  • #6
simphys
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For a circular orbit, the ##a## is constant and provided by the force of gravity -- for which you can think up a relevant equation of the form ##a=\ldots \ ##.

The two ##a## drop out and there remains something of the form ##v\propto \ldots \ ##.

##\ ##
The planets are held in orbit by the Sun's gravity, which obeys an inverse square law. Everything follows from that.
is this law a = 1/d^2 by any chance, if not it is okay. if so it'd be decreasing.
 
  • #8
Orodruin
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Well, perhaps you have the wrong relevant equation ? What does a have to do with anything ?
I disagree. It is the only relevant equation to compute acceleration from the given data apart from the equation for finding the velocity. That this is caused by Newton’s law of gravitation is a separate issue.
 
  • #9
PeroK
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is this law a = 1/d^2 by any chance, if not it is okay. if so it'd be decreasing.
The gravitational force decreases with distance, hence so must the centripetal acceleration. It would be better if you equated these formulas to obtain expressions for ##a## and ##v##.
 
  • #10
Orodruin
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The gravitational force decreases with distance, hence so must the centripetal acceleration. It would be better if you equated these formulas to obtain expressions for ##a## and ##v##.
Again, I do not believe that that is the purpose of this assignment. The assignment gives radius and orbital period and together with the kinematics of circular motion this is sufficient. The reference to gravity as the centripetal force is superfluous. In fact, the given data leads us to conclude the inverse square law for gravitational acceleration rather than assuming it as input.
 
  • #11
PeroK
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Again, I do not believe that that is the purpose of this assignment. The assignment gives radius and orbital period and together with the kinematics of circular motion this is sufficient. The reference to gravity as the centripetal force is superfluous. In fact, the given data leads us to conclude the inverse square law for gravitational acceleration rather than assuming it as input.
You're right. I didn't read the question!
 
  • #12
BvU
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I downplayed ##a## and at the same time put ##a=a## at center stage. Appears inconsistent, so oro objected. I meant to steer away from the value to the two proportionalities, one as1/r, the other as 1/r^2.
 
  • #13
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I downplayed ##a## and at the same time put ##a=a## at center stage. Appears inconsistent, so oro objected. I meant to steer away from the value to the two proportionalities, one as1/r, the other as 1/r^2.
Solving the problem requires the relationship between the radius and circumference of a circle. The importance of elementary geometry! :smile:
 
  • #14
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Solving the problem requires the relationship between the radius and circumference of a circle. The importance of elementary geometry! :smile:
Technically, for (c), you just need to know that there is a linear relationship 😉
 
  • #15
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Technically, for (c), you just need to know that there is a linear relationship 😉
That's what I meant! :smile:
 
  • #16
simphys
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That's what I meant! :smile:
Technically, for (c), you just need to know that there is a linear relationship 😉
yeah, yeah of course guys, but.. a = 4pi^2R / T^2 gives us an increase no decrease? That's what I don't get
 
  • #17
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yeah, yeah of course guys, but.. a = 4pi^2R / T^2 gives us an increase no decrease? That's what I don't get
I think you are just supposed to calculate and compare six numbers (in whatever units you prefer): centripetal acceleration of Venus, Earth and Mars; and, orbital speed of Venus, Earth and Mars.
 
  • #18
simphys
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I think you are just supposed to calculate and compare six numbers (in whatever units you prefer): centripetal acceleration of Venus, Earth and Mars; and, orbital speed of Venus, Earth and Mars.
What do you mean?
It's about part c right? where there needs to be answered with increase/decrease of centripetal acceleration and velocity as the radius is increasing
 
  • #19
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yeah, yeah of course guys, but.. a = 4pi^2R / T^2 gives us an increase no decrease? That's what I don't get
No, it does not because T and R are not independent. You need to actually compute the velocity and acceleration and see what comes out.
 
  • #20
simphys
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No, it does not because T and R are not independent. You need to actually compute the velocity and acceleration and see what comes out.
Got it, thanks guys!
 

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