Suppose I want to launch a rocket from earth to some point $O$ between the center of earth and the center of moon (on a straight line connecting their centers), where the gravitational force of the moon 'cancels out' the gravitational force of the earth (this point is located at $\approx 54 R_E$ from the center of earth where $R_E$ is the radius of earth). I want to know how much energy I should spend in order for the rocket to get there (neglecting the atmosphere and the rotation of the earth around its axis). So, I know that this is basically the difference between the potential energy at the start point and at the end point of the destination. However, $O$ is located not only in the gravitational field of the earth, but also in the gravitational field of the moon. And it seems that I cannot neglect the gravitational potential energy of the body at the moon's gravitational field. So my question is - how can I combine these two? How can I calculate the total GPE of the body in two (or even more) intersecting gravitational fields?

  • $\begingroup$ You are looking for terms "Delta-V required to get to the Earth-Moon Lagrange point no.1". $\endgroup$ May 1, 2013 at 18:55
  • $\begingroup$ @DeerHunter - sorry, but I'm not familiar with Lagrange mechanics. I'm just starting with physics and Newtonian mechanics. $\endgroup$
    – grjj3
    May 1, 2013 at 21:08
  • $\begingroup$ @grjj3 He's not talking about Lagrangian mechanics but about the Lagrange points. Same guy, different ideas. $\endgroup$ May 2, 2013 at 18:37
  • $\begingroup$ @dmckee - Thanks for the remark. But for me - still no idea of what it is. Hopefully, one day I'll learn that too. $\endgroup$
    – grjj3
    May 2, 2013 at 18:46

1 Answer 1


Gravitational Potential is a scalar quantity so can be added algebraically directly for both(or more) bodies.

Also GPE is just Gravitational potential times mass. $$E=\underbrace{\big(\sum P\big)}_{\text{due to all bodies in vicinity}}\times m$$

Now , rest of your aproach is allright ! Continue using this.

  • $\begingroup$ So, the gravitational potential energy of a single body is just the mass of body multiplied by the sum of the quantity $- G \frac{M}{R}$ for each planet that its gravitational field influences the body? $\endgroup$
    – grjj3
    May 1, 2013 at 15:46
  • $\begingroup$ @grjj3 Potential due to a single body is $-GM/R$ at a distance $R$ from it's center if it's a sphere. Mass times this quantity is GPE of that body and M(planet) and not just GP.I don't understand what u mean by " that its gravitational field influences the body?" $\endgroup$
    – ABC
    May 1, 2013 at 15:49
  • $\begingroup$ I know. I just ask about the sum. I know that $-GMm/R$ is GPE. I'm asking how GPE will look for a single body in multiple gravitational fields. Is it $(-GM_E/R_{E,O} - GM_{m}/R_{m, O}) m$ where $M_E$, $R_{E,O}$ are mass of earth and distance from center of earth to $O$ and $M_m$, $R_{m,O}$ are mass of moon and distance from center of moon to $O$? $\endgroup$
    – grjj3
    May 1, 2013 at 15:54
  • $\begingroup$ Yes , you are correct . They are scalars and can be added algebraically . $\endgroup$
    – ABC
    May 1, 2013 at 15:56
  • 1
    $\begingroup$ young man then :) I myself student too. $\endgroup$
    – grjj3
    May 1, 2013 at 16:08

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