How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?

I'm looking to calculate parameters around launching, say, a model rocket STRAIGHT to the moon. This does not mean through orbital insertion into a stable ~17,000 MPH relative to earth's surface and then Hohmann Transfer maneuvers.

What I mean is reaching an altitude that would correspond to the Earth-Moon L1 Lagrangian Point https://en.wikipedia.org/wiki/Lagrangian_points, with a minimal amount of velocity left to fall toward the Moon. For the purposes of the calculation, this could mean getting there with zero velocity (though it would need just a slight bit more, and have to be timed perfectly with the Moon's rotation).

Tentative Values:

1. Where is Earth-Moon L1, on average? 345,000km from Earth's center?
2. Earth's Radius at the Equator? 6,384km?
3. What altitude is the Earth-Moon L1? 338,616km?
4. Without air resistance, how much energy would be required to get a given mass to this altitude? (straight up, no orbital insertion)
5. What does the formula for this look like (getting to a specified altitude of a given astronomical body)?
6. If you were to launch from, say, 100km up with a few weather balloons, how would the formula need to be adjusted?
7. What type and how many rocket engines would be required to get a tiny model rocket to this altitude, regardless of feasibility?
8. Other than the issues of landing safely, would this be a more efficient way to transport supplies to the moon if they could survive the crash landing?
• Your question misses the point that the L1 point is rotating around the earth at the same angular velocity as the moon. Your image is to get to L1 with zero velocity in inertial space, which is a well-defined operation and DavePhD has shown how to calculate the energy requirement for that. However, it is incorrect to think you will just "fall to the moon" from there. The moon is running away from you with its orbital velocity. You have a lot of sideways velocity relative to the moon, so are likely to miss. – Ross Millikan Mar 27 '14 at 16:42
• If launched from earth's equator, that would give 459 m/s of angular velocity. The moon rotates with 1024 m/s of velocity, so yes the rocket would need to 'add' 566 m/s in the rotational direction of the moon to simply 'Fall In'. However, I'm looking for the rest of the component, and I suspect that one would not need all 566 m/s if some fancy maneuvers were added. – Ehryk Apr 22 '14 at 9:51

2 Answers

4.Without air resistance, how much energy would be required to get a given mass to this altitude?

Energy Needed ($E$) = Potential Energy at L1 ($V_{L1}$) - Potential Energy at Earth's Surface ($V_e$)

$$V_e = -Gm(\frac{M_e}{r_e} +\frac{M_l}{LD - r_e})$$

$$V_{L1} = -Gm(\frac{M_e}{d_{L1}} +\frac{M_l}{LD - d_{L1}})$$

where $m$ is the transported mass, $M_e$ is Earth's mass, $M_l$ is the Moon's mass, $LD$ is the center to center Earth-Moon distance, $r_e$ is Earth's radius, and $d_{L1}$ is the distance from the center of the Earth to L1.

5.What does the formula for this look like (getting to a specified altitude of a given astronomical body)?

The formula above it for a two body system, along the center line of such a system. You could replace $d_{L1}$ with a smaller distance if you want to go less than all the way to L1 along this line.

6.If you were to launch from, say, 100km up with a few weather balloons, how would the formula need to be adjusted?

In the formula for $V_e$ substitute $r_e + 100km$ for $r_e$

7.What type and how many rocket engines would be required to get a tiny model rocket to this altitude, regardless of feasibility?

This subquestion doesn't have a specific answer. Even if a zero mass payload is assumed, the propellant and structure holding the propellant have mass. The chemical nature and mass of the propellant, mass of the inert sturcture, and arrangement of stages would be major factors. Multiple stages are advantageous to reduce the mass during flight by eliminating no longer needed inert sturcture.

8.Other than the issues of landing safely, would this be a more efficient way to transport supplies to the moon if they could survive the crash landing?

If fuel is not used to make a soft landing on the Moon, this would definitely increase efficiency.

Addtional considerations:

None of the above considers the gravitational potential of the Sun. If you don't mind crashing into the Moon, launching when the Moon is between the Earth and the Sun would minimize the amount of energy needed. This would add a third term involving the Sun's mass and distances to the Sun to each of the potential energy equations.

As previously suggested by user "I like Serena", since Earth is rotating about its own axis, a rocket launched from Earth will have an initial velocity component due to this rotation. It is optimal to launch from near the equator to take advantage of the maximum velocity due to rotation, as well as greater Earth diameter/less gravity. Launch should be timed such that the rotational velocity component is directed to the Moon as much as possible, at which time the direction of the Moon will be generally eastward.

The L1 point is calculated considering gravitational potential and centrifugal force of a body in the rotating frame. The point of maximum gravitational potential along a line joining the Earth and Moon would be a somewhat different point. It would be more correct to find the maximum gravitation potential along this line and use that potential energy value, although it should be similar to the potential energy to get to L1.

For more information on low energy transport to the moon, without using Hohmann transfer, and without crash landing, see Low Energy Transfer to the Moon. An alternative low-energy approach was used by the 1991 Japanese Hiten mission.

• Nailed it. Would this be correct for solar gravity inclusion: $V_e = −Gm(\frac {M_e} {r_e} + \frac {M_l} {LD−r_e} + \frac {M_s} {SD-r_e})$ and $V_{L1} = −Gm(\frac {M_e} {d_{L1}} + \frac {M_l} {LD−d_{L1}} + \frac {M_s} {SD-d_{L1}})$ ? – Ehryk Mar 27 '14 at 18:20
• Also, one of the things I'm confused about is atmospheric orbital speed: if the balloon setup sat at 100km altitude for long enough to synchronize with the atmosphere at that height (what little of it there is); would it speed up to the orbital rotation at that height (I.E. track the same geostationary location once settled), or slow down, or ? – Ehryk Mar 27 '14 at 18:25
• And, (finalish question) - if I got a small rocket precisely to L1 and used the surface rotation timing, how much velocity/energy would it take for lunar capture? Or would it just speed past the moon unless slowed down significantly? Would it need to get much closer to the moon than L1 to actually be captured with the KE from earth's rotation? – Ehryk Mar 27 '14 at 18:29
• yes(to the "Would this be correct for solar gravity inclusion" comment), but only at a moment that the three bodies are aligned with the moon in the middle (a solar eclipse) – DavePhD Mar 27 '14 at 18:30
• @Ehryk as to the balloon comment, balloons travel with the atmosphere, which rotates at the same angular velocity as the suface, a somewhat faster linear velocity than the surface – DavePhD Mar 27 '14 at 18:40

To calculate the energy required, use the law of preservation of energy, which dictates that:

Potential energy at earth's surface + kinetic energy at the surface + invested energy = potential energy in L1 + kinetic energy in L1

Potential energy = $-\frac{GM_{earth}m}{r_{earth}} -\frac{GM_{moon}m}{r_{moon}}$

Kinetic energy at earth's surface = $\frac 1 2 m v_{earth}^2$

$v = \frac{2\pi r}{T}$

Btw, note that a rocket would and should not launch straight up, but in the direction of earth's rotation to gain the maximum benefit of that rotation.

• Wouldn't launching off-normal (straight up) 'add velocity' at the expense of some altitude? What angle would maximize altitude from Earth's surface? – Ehryk Mar 20 '14 at 0:28
• I think you're thinking of an orbital insertion; in my case I don't want or need any additional 'horizontal' velocity; unless it can be used to increase the vertical. I just want to reach the highest point I can. – Ehryk Mar 20 '14 at 5:46
• Simplified for only the normal vector, and just reaching the Earth-Moon L1 from rest on earth's surface: $PE_{i} + KE_{i} + E_{rocket} = PE_{L1} + KE_{L1}$, so $0 + 0 + E_{rocket engine} = PE_{L1} + 0$, or $E_{required} = PE_{L1}$. I'm lost on how to calculate the Potential Energy at L1, since $PE = mgh$ doesn't really work (g = 0?). Your equation above doesn't seem to care about the mass of the object, a 100kg object surely has more PE at L1 than a 1kg object. Can you elaborate? – Ehryk Mar 20 '14 at 6:32
• You are quite right. The formula for $PE$ should include the mass of the object. I have edited my post to reflect that. The formula $PE=mgh$ is an approximation that only holds true close to the earth surface. The formula I gave is the general formula where $r_{earth}$ is the distance of the object to the center of the earth. Neglecting friction by air, the highest altitude will be reached when the rocket flies horizontally in the direction of earth's rotation. That way it is maximizing its kinetic energy that will then be converted to potential energy. – Klaas van Aarsen Mar 20 '14 at 15:59
• Can you explain the process by which horizontal kinetic energy is converted to vertical potential energy, without a ramp or some similar structure? I don't think $KE_{horizontal}$ or lack thereof has any effect on the height an object will reach, unless we're leaving Earth's rotating reference frame or something. – Ehryk Mar 20 '14 at 18:51