What Speed Would an object need to leave the earth at to reach L1? [closed]

Let's say the Earth is an airless sphere. What speed would an object weighing 1 kg need to leave the surface at in order to get to and be motionless at L1, where the Moon's gravity becomes stronger than the Earth's gravity?

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closed as too localized by Waffle's Crazy Peanut, Michael Brown, Manishearth♦Apr 5 '13 at 12:41

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Note that the homework tag "Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright." – Michael Brown Apr 5 '13 at 2:50
At its present form, this question looks much like a "closable" homework because the way it asks, simply declines our policy. @Michael: Recalling the definition. Eh..? ;-) – Waffle's Crazy Peanut Apr 5 '13 at 6:10
Escape velocity (and surely orbital capture velocity as well, as these would only be two sides of the same coin) is independent of object mass. The energy requirement for reaching escape velocity is another matter entirely. Quoting the Wikipedia article on the subject: $v_e = \sqrt{\frac{2GM}{r}}$ – Michael Kjörling Apr 5 '13 at 7:37
@MichaelKjörling Escape velocity is an upper bound since the OP only wants to stop at L1, not to pass by it. – Michael Brown Apr 5 '13 at 8:54
@MichaelBrown Good point there. However, that upper bound is also independent of the object's mass, which was the point I was trying to make. So the mass of the object is irrelevant to the calculation of the speed required when leaving the Earth. – Michael Kjörling Apr 5 '13 at 9:19

First, let's point out that the lagrangian point is not where "Moon's gravity becomes stronger than the Earth's gravity" but rather a point where, in the reference frame where both the Earth and the Moon are at rest, you could stay motionless. The difference is that this reference frame is not inertial and you should add the "centrifugal force" to the Moon's and the Earth's gravity.

If you prefer to stay in an inertial reference frame, let's choose the Moon-Earth barycentric reference frame where the Earth and the Moon are spinning around each other with a period $T$ of around 27 days. In order to stay on the line Moon-Earth, your satellite have to spin at the same period around the barycenter $B$.

Let's pose some notations:

• $E$ the Earth ;
• $M$ the Moon ;
• $L_1$ the lagrangian point ;
• $B$ the barycenter ;
• $M_E$, $M_M$ and $m$ the masses of the Earth, the Moon and the satellite to launch ;
• $r=BL_1$, $r_M=BM$ and $r_E=BE$ the distances to the barycenter of each mass.

The mechanical energy the satellite have to have in order to stay on his lagrangian orbit mixed up gravitationnal energy from the Earth ($-\frac{GM_Em}{r+r_E}$), from the Moon ($-\frac{GM_Mm}{r_M-r}$) and its kinetic energy $\frac12mv^2 = \frac12 m \left(\frac{2\pi r}{T}\right)^2$. You only have to compare this total energy to the one at the surface of the Earth ($-\frac{GM_Em}{R_E}$ where $R_E$ is the Earth radius) to compute the minimal kinetic energy you need to ensure you arrive at $L_1$ with the right speed (in norm).

Note that you won't necessarily be on the right path (id est your velocity will not have the required direction in order to stay on $L_1$), it depends on how you achieve the energy transfer, but that's the minimal energy you have to give to be on the lagrangian orbit.

Note also that the mass $m$ will always simplify itself from the calculation so that the needed speed won't depend on it (by the energy will).

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It wasn't actually a homework assignment or anything, I was just curious.. – Jaren Fleischman Apr 5 '13 at 16:54