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?
Newtonian Mechanics – Gravitational Potential Energy of Mass Between Two Planets
newtonian-gravitynewtonian-mechanicspotential energy
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There are many fundamental concepts that both you and Trimok have misunderstood.
First of all, you can completely ignore the mass energy. The kinetic energy in no way "compensates" for the loss in mass energy because no mass energy is lost in the example that you gave. Sure, the mass of the rocket decreased as it ejected propellant to move forward, but that mass hasn't disappeared. The propellant is still out there, floating around in space. You can't just ignore its mass energy because it's not in the rocket anymore, it still exists. In other words your total $E_m$ term is always the same throughout this situation, at no point is matter converted into another form of energy, like kinetic energy as you seem to suggest.
Second, this is conceptually wrong :
The gravitational attraction between me and the earth will significantly decrease and after a certain distance it will cease to act as the inertia of my rocket would be bigger than the gravitational force between my rocket and the earth.
You can't compare the inertia of a system with a force (they're not even measured in the same unit). One can't be "greater" than another. Regardless of the mass of an object, however small a force you apply on it it will still have an impact. There's no cutoff. This doesn't matter much because indeed, you can get to a point where this is small enough that you consider yourself to be in a situation where the potentially energy is now maximal (you seem to think that because this is 0 the potential energy has now disappeared, but this is not the case, it is sometimes chosen to be 0 at infinity but keep in mind that this is still an increase because in such conventions this potential energy is negative at lift off). Potential energy increases as you get further from the earth. This is because, intuitively enough, now that you are further from the earth you now have the potential to gain more kinetic energy by falling back towards the earth through a longer distance, gaining more speed in the process.
Finally, let's work out the actual energy distribution throughout all this. Like I said we can ignore mass energy because the total mass is conserved. Instead what we actually need to consider is :
$E_k$ : the kinetic energy of the rocket.
$E_c$ : the chemical potential energy stored in the propellant. It is this energy that will propel the rocket, not mass energy as you seem to think.
$E_g$ : the gravitational potential energy.
In the beginning, the rocket is sitting on the surface of the earth, and $E_c$ is maximal, $E_g$ is minimal, and $E_k$ is just $0$. As you start burning propellant, you release the energy stored in your propellant and $E_c$ begins to decrease as $E_k$ increases. To completely compute $E_k$ you actually have to take into account both the kinetic energy of the rocket and the kinetic energy of the ejected propellant. If you do, you will notice that this is actually not enough to make up for the loss in $E_c$. Indeed, the difference between the two corresponds to the gain in $E_g$ as the rocket goes further and further away from the earth, and indeed we will always have conservation of $E_k + E_c + E_g$.
The formula $U=mgh$ is simplified and assumes that you are near a planet's surface ($g$ itself is only valid near the Earth's surface). This sounds kind of like a homework problem, so I'll give you some hints. First, you need to pick an origin (usually the star at the center of the system is a good place). Now you need to figure out the forces acting on the planets. Since the problem is asking for gravitational potential energy, you know that you only need to consider gravitational forces. Identify the gravitational forces and then integrate the force along the paths in question to calculate the potential energy. Be sure to maintain the same coordinate system throughout!
Making an assumption about your situation: you might be able to neglect the planet to planet gravitational force if the central force is much larger (which it should be, but I don't know what your problem statement was) and just calculate the potential energy due to the interaction between the planet and the star.
EDIT: I feel i can be more specific --
Gravity, when modeled as a central force (ie, Newtonian gravity) has a value of $F=-G\frac{Mm}{r^2}$, where the different m's are the masses of the two separate objects and r is the distance between them. To find the potential energy due to gravity, you essentially want to calculate the negative of the work done in order to "put" the planet where it is in orbit. Work is just the integral of force from where the potential is defined as 0. In this case, you would need to bring the planet in from $\infty$ where the force vanishes (approaches 0).
So, in math terms, $U=-\int_{\infty}^{r} -G\frac{Mm}{r^{\prime2}} dr^{\prime}=-G\frac{Mm}{r}$. This result is for one planet in relation to the sun, calculated assuming one dimension of force (which is valid for only 2 bodies). If you have more bodies that you want to account for (like other planets), then you need to complicate that expression to higher dimensions and add all the terms from the gravitational forces due to every object with the proper vectors in tact. Luckily, you can use superposition of forces for that so it's not all that messy, but would be a little much here (I think). In practical terms, the gravitational force from the star at the center of a solar system far outweighs that of the individual planets, that you can neglect those terms and come up with a very good approximation of the total gravitational potential.
Best Answer
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.