Does a Rocket Hover Above the Moon?

[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?

 

[A.T.]

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?

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

 

[Vanadium 50]

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

Well, technically, you can do either.

 

[shlosmem]

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

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.

 

[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.

 

[shlosmem]

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.

In what speed the body will rotate under it?

 

[phinds]

In what speed the body will rotate under it?

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.

 

[russ_watters]

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.

 

[A.T.]

I thought maybe its needs to be faster than the moon in order to stay on the same place.

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.

 

[Bandersnatch]

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.

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

In what speed the body will rotate under 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

 

[phinds]

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

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" )

 

[Imager]

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

 

[jbriggs444]

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

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.

 

[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.

 

[russ_watters]

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

 

[willem2]

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

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.

 

[russ_watters]

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.

 

[Imager]

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.

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 traveling to the left?

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.

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.

 

[A.T.]

From above the moon's pole, wouldn't the rocket appear to be traveling to the left?

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

 

[Imager]

Is this closer?