Calculating Earth's Effective Spring Constant for its Orbit Around the Sun

[Edel Crine]

Homework Statement

Consider the nearly circular orbit of Earth around the Sun as
seen by a distant observer standing in the plane of the orbit.
What is the effective “spring constant” of this simple harmonic
motion?

Relevant Equations

T=2π√m/k

My first attempt was using the period equation of a spring system.
I've changed it into k=((2π)^2*m)/T^2, then put Earth's mass into "m" (5.972*10^24), then put the time required for one revolution of Earth around the Sun, 365 days into seconds, 31536000 sec, to "T"
So I got (2.371*10^11 kg/s^2) as the answer, but not sure whether is right or not...
I'll appreciate every single help from you!
Thank you!

 

[haruspex]

I'm not sure what the question intends. Another interpretation would be to consider the work required to increase the orbital radius by x and write W=kx²/2. Maybe that gives the same.

 

[jbriggs444]

Consider the nearly circular orbit of Earth around the Sun as
seen by a distant observer standing in the plane of the orbit.
What is the effective “spring constant” of this simple harmonic
motion?

With respect to the question of interpretation...

The idea that I have is to adopt a rotating frame of reference that keeps pace with the orbitting Earth. In this frame, you have a radial distance which is a function of time and should be a decent approximation to simple harmonic motion. Tangential position is transformed into irrelevance.

Now for strategy...

If we continue to use this frame, we can concentrate on the radial forces acting on the Earth. If we add up those forces and write down a formula for force in terms of radius, it should give a result of zero at the equilibrium point (mean orbital radius of Earth).

What radial forces exist?

One wants to end up with a formula that only depends on r. It may be useful to remember that angular momentum and energy (both calculated in the inertial frame, of course) are constant throughout the orbit as r varies.

If one has a formula for force versus radial distance then the first derivative of this formula gives the effective spring constant.

 

[haruspex]

With respect to the question of interpretation...

The idea that I have is to adopt a rotating frame of reference that keeps pace with the orbitting Earth. In this frame, you have a radial distance which is a function of time and should be a decent approximation to simple harmonic motion. Tangential position is transformed into irrelevance.

Now for strategy...

If we continue to use this frame, we can concentrate on the radial forces acting on the Earth. If we add up those forces and write down a formula for force in terms of radius, it should give a result of zero at the equilibrium point (mean orbital radius of Earth).

What radial forces exist?

One wants to end up with a formula that only depends on r. It may be useful to remember that angular momentum and energy (both calculated in the inertial frame, of course) are constant throughout the orbit as r varies.

If one has a formula for force versus radial distance then the first derivative of this formula gives the effective spring constant.

Isn't that equivalent to the interpretation in post #2?

 

[jbriggs444]

Isn't that equivalent to the interpretation in post #2?

Might be. I've not calculated it out.

I have trouble wrapping an intuition around the work approach. Conservation of energy applies in the inertial frame. [In the scheme that I have in mind, work (or potential energy) enters into a calculation of the centrifugal force].

 

[haruspex]

Might be. I've not calculated it out.

I have trouble wrapping an intuition around the work approach. Conservation of energy applies in the inertial frame. [In the scheme that I have in mind, work (or potential energy) enters into a calculation of the centrifugal force].

Seems both are considering an oscillation resulting from a perturbation, so should give the same answer.
But as we know, the new orbit would be an ellipse on pretty much the same period as in the circular orbit, so I suspect it gives the same as the OP's idea.

 

[Lnewqban]

... Consider the nearly circular orbit of Earth around the Sun as
seen by a distant observer standing in the plane of the orbit.
What is the effective “spring constant” of this simple harmonic
motion?
...

Could this be what the question means (disregarding the difference in distances between perihelion and aphelion)?:

https://en.m.wikipedia.org/wiki/Earths_orbit

 

[Chestermiller]

Could this be what the question means (disregarding the difference in distances between perihelion and aphelion)?:

https://en.m.wikipedia.org/wiki/Earths_orbit

View attachment 268509

Yes. I agree with this interpretation. The far-away observer is looking at the orbital plane on edge, and, to this observer, the Earth seems to be oscillating horizontally in simple harmonic motion caused by an invisible spring. The oscillation is about the "equilibrium position" at which the Earth and sun are aligned with his line of sight, and the amplitude of the oscillation is the radius of the orbit.

 

[haruspex]

Could this be what the question means (disregarding the difference in distances between perihelion and aphelion)?:

https://en.m.wikipedia.org/wiki/Earths_orbit

View attachment 268509

Again, I think all interpretations so far lead to essentially this.