# Hover above the moon

1. Feb 19, 2015

### shlosmem

Suppose a rocket take off from the moon surface and is hovering above the moon at a constant height, does it stay in the same place relative to the surface like an helicopter on earth or because there is no atmosphere there it will shifted eventually because of the moon spin?

2. Feb 19, 2015

### A.T.

You can do both, if you control the thrust vector in the right way.

3. Feb 19, 2015

Staff Emeritus
Well, technically, you can do either.

4. Feb 19, 2015

### shlosmem

Sure, what I meant is what if the thrust vector is pointed toward the surface and only resists the moon gravity. The rocket gets it initial speed in X direction from the moon surface, but because is higher than surface I thought maybe its needs to be faster than the moon in order to stay on the same place.

5. Feb 19, 2015

### phinds

In the case you describe, then just like a rocket that moves out of the Earth's atmosphere and "hovers", it would see the body rotate under it.

6. Feb 19, 2015

### shlosmem

In what speed the body will rotate under it?

7. Feb 19, 2015

### phinds

Well, the Earth will rotate at one day per day. The moon's rotation is something like 1.5 Earth days per Earth day; whatever is required for a tidally locked moon.

8. Feb 19, 2015

### Staff: Mentor

That doesn't sound right. If you hover just above the earth or moon, you'll stay pretty much in the same place because you keep the linear speed from the rotation. The higher you go/longer you hover, the more the coriolis effect will impact it though.

9. Feb 19, 2015

### A.T.

Yes, with radial thrust only, it will lag behind the Moon's rotation. From the rotating frame this is attributed to the Coriolis effect Russ mentioned.

10. Feb 19, 2015

### Bandersnatch

That would be more like once per month, wouldn't it?

That depends on how high the rocket hovers.

You start with a rocket sitting on the lunar equator. Its period of rotation is
$T=2πR/V$
where T is one lunar month (~27 days; sidereal), V is the tangential velocity and R is the lunar radius.
By raising the rocket to height h, we end up with a new period of rotation
$T'=2π(R+h)/V$
(the tangential velocity remains unchanged)

Imagine we draw a line from the rocket at the new height with its new period of rotation towards the centre of the moon. At the intersection with the surface the point of contact (like the 'shadow' of the rocket) would have tangential velocity
$V'=2πR/T'$
Substituting we get
$V'=RV/(R+h)$
R and V are known (check the wikipedia; alternatively V can be extracted from the first equation knowing the period T=1 sidereal month). The only variable left is the height h over the surface.

The difference between V' and V will net the velocity the rocket's 'shadow' would lag behind the surface.

For example, at height h equal to R (so, two radii from the centre), V' equals 1/2V, and the 'shadow' lags behind the surface by V'- V = - 1/2V, or about - 2.3 m/s

11. Feb 19, 2015

### phinds

Hm ... one a day, once a month ... I just make this stuff up as I go (which is why I added "
whatever is required for a tidally locked moon" )

12. Feb 20, 2015

### Imager

Definitely not scale. Just wondered if this picture captures the concept correctly?

13. Feb 20, 2015

### jbriggs444

For a rising rocket, Coriolis acceleration is anti-spinward. So if you are trying to represent the trajectory from the viewpoint of the rotating frame, the curvature is in the wrong direction.

If you are trying to represent the trajectory from the viewpoint of an inertial frame, the initial direction is wrong. A slowly rising rocket will start out nearly tangent to the moon's surface since its initial velocity due to rotation will be larger than its vertical velocity.

14. Feb 20, 2015

### HallsofIvy

Note that, initially, when the rocket is sitting on the moon, it already has the same motion as the moon. When it takes of it will, in addition to the upward velocity due to the rocket, it will have a motion tangent to the moons surfact. That will increase its height but also keep it above the point from which it took off.

15. Feb 20, 2015

### Staff: Mentor

Looks like it's drawn in the inertial frame and is correct.

16. Feb 21, 2015

### willem2

In the Inertial frame the rocket starts out going 4.62 m/s to the left. (length moon equator/siderial period)
The acceleration of the rocket will be upwards, curving the path to the right and not to the left. Since 4.62 m/s is so slow compared to the likely speed a rocket will get, the path will really look like a straight line at this scale.

17. Feb 22, 2015

### Staff: Mentor

Ahh, you're right -- I read the "curvature in the wrong direction" direction of the vector, not the direction of the change in vector and didn't pay attention to that. It's still pointing to the left, but you guys are correct that the velocity vector should be rotating to to the right.

18. Feb 23, 2015

### Imager

If one is on the surface of the moon, I can see how it would be to the right. I intended for the diagram to be from a point of view that is looking down at the moon/rocket (my bad, that I didn't state this in first place). From above the moon's pole, wouldn't the rocket appear to be travelling to the left?

I thought the rocket would start to fall behind (to the right) of the launch point. Using Willem2's 4.62 m/s on the surface, wouldn't the rocket start to fall behind as its height and the radius increase. I'm thinking it speed remains the same but the distance it needs to travel is greater.

19. Feb 23, 2015

### A.T.

Yes, right at the start, not after curving to the left as you have drawn.

20. Feb 23, 2015

### Imager

Is this closer?