Sign in to follow this  
Followers 0
NomenNescio

Momentum and Rotating Atmospheres

4 posts in this topic

I've been wondering for a while, if you send up a small rocket, but the rocket never leaves the atmosphere, am I right in saying the planet still rotates beneath the rocket even though the momentum from the planet, and the atmosphere spinning, should nudge the rocket along and keep it more or less above the launching point? Do you see what I mean? It's like how when you're in plane the planet shouldn't rotate relative to you, but it seems to me that in KSP it does. This is really just a question of how elaborate the simulation is, and I'm really just curious to see if I'm right.

Share this post


Link to post
Share on other sites

First, let's assume there was no atmosphere, as this makes things a bit easier to understand. The oceans of Kerbin rotate with a given speed. So do the mountains of Kerbin. We know from observations, that Kerbin's mountains do not change their positions. (Same goes for any other planet and moon out there). Now the peak of a mountain and the sea level rotate 360 degrees in the same amount of time - they do have the same angular velocity  (360degrees/6h). However, since the peak of the mountain is higher up, it needs to go a longer way around the center of the planet, then the sea level needs to go. That means the peak covers a longer way in the same amount of time. It has a faster tangential velocity. If you sent up a rocket from the sea level to the height of the mountain the planet would start to rotate beneath you, as long as you do not cover the difference of the tangential velocities between the sea level and the mountain peak.

Now let's switch on the atmosphere and presume it had the same density everywhere. Why does the atmosphere rotate at all? There is friction between the surface of the planet and the lowest layer of air. That lowest layer gets 'pulled' with the planet so both have the same tangential velocities. Now the lowest layer of air interacts with the layer above and so on. Finally your atmosphere would rotate bottom to top but there is nothing pulling it faster than the surface. So the upper layers of air would be rotating with lower angular velocities than the planet.

If we consider the exponentially decreasing density of the atmosphere, then there are less air molecules per cubic meter in one layer of air than there are in the layer beneath. So the friction between two layers becomes less and less the higher you go. Likewise, your rocket experiences less drag higher up which is why you do a gravity turn or build a space plane.

So what's the conclusion? The atmosphere rotates with the surface speed or lower speeds if you are higher up but does not increase that speed in higher layers. This is why a rocket will not be pulled with the planet.

In reality, of course, there are such things as pressure, energy transfer from the sun and wind which will change the simple model presented above to "some extend". If you're really in to that, I propose to read some meteorological books or refer to Wikipedia as a start. Well, read wikipedia or the pages of the NAOO or whatever. There's so much stuff out there for free...

1 person likes this

Share this post


Link to post
Share on other sites

Posted (edited)

Yes this is a thing. That is why there are 2 modes on the navball speed indicator: surface and orbital. Surface shows your velocity relative to the surface of the planet and orbital shows your speed relative to the (imaginative) surface off the planet that is not rotating at all. So if you are landed on a planet surface speed says your speed is 0m/s (obviously since you are on the surface) and orbital speed says your speed is the speed at which the surface of the planet is rotating.

 

But there is still a reason (IRL reason) why your craft wont land back in the same exact spot if you just launch it straight up. It is quite hard for me to explain it without a picture but I'll try:

1. First let's assume that the rotational speed of a given point on planets surface is 100m/s. And R is the distance from center of the planet to the surface of the planet.

2. Let's then say you launch a ship straight up so that you fly high up in to the atmosphere and hit the ground again after 100 seconds.

During this 100s the spot on the surface you started from has moved 100s*100m/s = 10km, which in other words means that the planet has rotated 10km/R radians.

Simultaneously during this time your ship is has also traveled 10km BUT since your craft is above the ground at some given height (let's call it h). And this in turn means that it will have rotated less radians around the axis of the planet than the surface has since the distance from the planets center has been R+h (which is obviously larger than R) at any given time --> 10km/(R+h) < 10km/R.

And because of this difference in rotational speeds the craft will not land back to where it started.

EDIT: plus what @somethingsaid above about the higher atmosphere and friction between different atmospheric "layers".

Edited by tseitsei89
1 person likes this

Share this post


Link to post
Share on other sites

It's all a practical joke.  Your 'friends' pull the planet away while you aren't looking.

1 person likes this

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now
Sign in to follow this  
Followers 0