Will slinky length be different on the moon?

If we hold one end of a slinky and leave other end free, the earth's gravity applies force on the slinky and it expands. If we do the same on the moon with the same slinky, will the acquired height of the slinky be different?

• Interestingly, the time it takes the bottom to start falling if the top is dropped doesn't depend on gravitational intensity: insidescience.org/news/secrets-levitating-slinky – MatrixManAtYrService Nov 18 '20 at 1:29
• @uhoh The time it takes the bottom of the extended spring to fall, when the top is let go, is not zero. It will remain stationary for a while until the top has come closer, since the spring force that is holding up the bottom part is still present for a while. – Steeven Nov 18 '20 at 10:57
• @Steeven yes you are right! There's this about that, a whole world of "slinky science" I hadn't known about but now can't stop thinking about. – uhoh Nov 18 '20 at 11:01
• @uhoh It is indeed quite fascinating. Just have a look at the Gif here: physics.stackexchange.com/questions/56833/… – Steeven Nov 18 '20 at 11:14

Think about taking two extreme cases :

How much does a slinky extend in a gravity-free space? None at all

How much would it extend if it was on perhaps Jupiter or even a black hole ?It should extend by a large amount.

Gravity does play a role.

• I checked slinky video in space, and after that I was thinking the same. I needed confirmation from experts. Thanks for reply. – Tarun Mishra Nov 17 '20 at 17:10

For a typical slinky (wiki) and comparing the gravity of earth and moon, this will be the case as outlined by the other answers.

However, for a general spring (wiki) or weaker gravity fields, the spring may be in a state where it is fully contracted and the weight of its own mass is insufficient to stretch it at all. In that regime, gravity does not play a role.

• So long as there is any gravity I do not believe that state exists. It might be too small to notice, but there will be some stretching. – Loren Pechtel Nov 19 '20 at 5:07
• @LorenPechtel Check out the pictures in the second link. – Wolpertinger Nov 19 '20 at 10:57

If a slinky is hanging vertically in a gravitational field, the amount of stretch in any short section depends on the weight of the coil hanging below that section. Less gravity will produce less stretch.

Does the solution to:

$$\frac{dU}{dz} = \frac d {dz}[U_{spring}(z)+U_{gravity}(z)] =\frac d {dz}[\frac 1 2 k z^2 - \frac 1 2 mgz]=0$$

depend on $$g$$?

• I'm intrigued. As written, the $U_{spring}(z)$ term assumes uniform spreading out of the slinky so that both the mass density and strain profiles are flat. Is that what happens? It's been so long that I've held one that I don't remember, but I thought that the spreading is markedly non-uniform. – uhoh Nov 18 '20 at 6:22
• Since your Stack Exchange answer relies solely on the validity of the equation and nothing else, this is important. – uhoh Nov 18 '20 at 6:25
• @uhoh I'll go with non-uniform. I base my claim on this video: youtube.com/watch?v=uiyMuHuCFo4 – Theraot Nov 18 '20 at 6:36
• @Theraot oh lovely! So this equation is not even close to correct. – uhoh Nov 18 '20 at 10:31
• This is the equilibrium condition for a mass connected to an ideal spring (mass less spring), which the slinky is obviously not. The slinky is instead approximated by chopping that spring into many tiny pieces, and connecting them in series with tiny masses in between. – Luo Zeyuan Nov 18 '20 at 11:30