# Minimum velocity of the moon

1. Oct 17, 2005

### Orikon

I'm not sure if this has been asked before, I searched the forum but didn't find anything. Anyway, I was wondering what minimum velocity the moon can have without crashing into the earth due to gravity (basically the opposite of the escape velocity).

2. Oct 18, 2005

### whozum

The moon is bound to earth (simplified model) by a centripetal force proportioanl to
$$F = \frac{mv^2}{r}$$ with m being the mass of the moon, v it's velocity, and r its radius. That force is provided by gravity, which is proportional to
$$F_{grav} = \frac{GM_{e}m_{m}}{r^2}$$ with big M being earth's mass. Equating the two gives,
$$\frac{GM_{e}m_{m}}{r^2} = \frac{m_{m}v^2}{r}$$. Cancelling and solving for v gives

$$v = \sqrt{\frac{GM}{r}}$$ which is the exact velocity needed to obtain a circular orbit. Anything different than this would result in a parabolic/elliptic/hyperbolic orbit.

3. Oct 18, 2005

### Tide

The question is highly ambiguous. The minimum speed a moon could have without crashing into the Earth would be zero - if it were situated very far away (infinity!).

4. Oct 18, 2005

### SpaceTiger

Staff Emeritus
Agreed. I think what we want here instead is a minimum angular momentum. The only orbits that will crash into the earth will either be very low energy or very low angular momentum (or both). The former requirement can be approximately derived (assuming a circular orbit) from whozum's equation. For the latter case, I think you can derive the answer by considering:

$$r_p=a(1-e)=R_{earth}$$

where $r_p$ is the moon's distance at closest approach (perigee), a is the semimajor axis of the orbit, and e is the eccentricity of the orbit. We set this equal to the radius of the earth because this is the maximum angular momentum orbit that will bring the moon into collision with the earth.

Now, we just need:

$$l^2 = GM_{earth}a(1-e^2)$$

where l is the angular momentum per unit mass and G is the gravitational constant. This equation can easily be solved for the eccentricity:

$$e=\sqrt{1-\frac{l^2}{GMa}}$$

We know that e~1, since the moon's orbit will have to be extremely eccentric in order for it to crash into earth. Thus, let's just use the binomial expansion:

$$e\simeq(1-\frac{l^2}{2GMa})$$

Pluggiing this into the first equation, we get, finally:

$$l<\sqrt{2GM_{earth}R_{earth}}$$

Note that this is independent of the energy of the orbit in the first-order approximation. To answer the question exactly, one would have to specify both energy and angular momentum.

Last edited: Oct 18, 2005
5. Oct 18, 2005

### Tide

ST,

Nice! But I will resist the temptation to mention the tidal force! :)

6. Oct 18, 2005

### SpaceTiger

Staff Emeritus
Too late, you just did!

What aspect of it were you thinking of? It will certainly cause the orbit to change over time, but the direction of this change will depend on the orbital and rotation periods. I would expect that whatever changes were induced by the tidal forces, the collision would still occur when the moon's orbital parameters coincided with those mentioned above.

7. Oct 18, 2005

### Tide

You'll need to use a somewhat larger value for the moon's perigee if you include tidal forces. At those near distances the moon would be torn apart.

8. Oct 18, 2005

### SpaceTiger

Staff Emeritus
Well, that would be another reason to view the calculation as OOM, but I don't think it makes sense to include it explicitly. For one thing, the moon getting torn apart is not that same as it colliding with earth...and I don't think there's much hope of calculating a priori the orbits of the individual pieces that remain after it's broken. Presumably their average angular momentum per unit mass will be comparable to the original value, so they may well continue on their orbits if the moon's original angular momentum exceeds the above value.

Secondly, it would be a small correction to a calculation that was already OOM. Consider the Roche Radius for a rigid body:

$$d \simeq 1.26R_E(\frac{\rho_E}{\rho_M})^{1/3}$$

which comes out to $\sim1.5R_E$. Not a big correction, but eh, if you wish to include disruption, you can consider the angular momentum limit to be a factor of ~1.2 larger.

9. Oct 18, 2005

### DaveC426913

Quite true. But let's suppose he meant "at its current distance (apogee)".

Now, the minimum it needs to maintain its current orbit is the velocity it is doing now. But if, at apogee, we slowed it down a touch, it would change to an elliptical orbit. Say its perigee were to graze the atmo, (or more practically, the Roche Limit) what is the minimum velocity *at apogee* that it would need to be moving?

I have no idea whether this is what the OP was asking or not, but I think it's an intriguing question.

10. Oct 18, 2005

### SpaceTiger

Staff Emeritus
As do I.

All we have to do is divide the limit for the angular momentum per unit mass by the current distance to the moon:

$$v_a < \frac{l}{d_M} \simeq 0.2~km/s$$