# Earth-Moon System

brindha-rocks

## Homework Statement

Consider the Earth-Moon system. The mass of the Moon is 7.36 × 1022 kg . The radius of
the Moon is 1.74×106 m. The mass of the Earth is 5.97×1024 kg. The radius of the
Earth is 6.37×106 m. The universal gravitation constant is 6.67×10−11 N⋅m2 ⋅kg−2 . The
average distance from the Earth to the Moon is 3.84 × 108 m . Suppose an object of mass m is placed on a line passing between the centers of the Earth and Moon (Earth-Moon
line).

a) Where on the Earth-Moon line is the net gravitational force zero?

b) Choose the zero point for the potential energy to be at infinity. Is the net potential energy zero anywhere else on the earth-moon line?

c) Suppose an object of mass m = 1.00 kg is released from the point at which the net
gravitational force is zero. The object is given a very small initial velocity toward the earth. What is the kinetic energy of the object when it just arrives at the Earth’s surface?

d) Suppose the object is released from a point the same distance from the center of the Earth as in part c), but from the side of the Earth away from the Moon. What is the kinetic energy of the object when the object just reaches the Earth’s surface?

e) What is the escape velocity of an object at the surface of the Earth along the when it is launched directly towards the moon when (i) you only take into account the gravitational force of the Earth, (ii) you take into account the gravitational force of both the Earth and the Moon? You may neglect the effect of the Sun’s gravitational force and any effect due to the rotation of the Earth.

## Homework Equations

Conservation of kinetic and potential energy.

## The Attempt at a Solution

I'm having some difficulty with part b). I would say the net potential energy is zero at the surfaces of the moon and the earth. But what about the point where the net gravitational force is zero?

And with part c), I can't seem to understand what's happening exactly. I'm visualizing a graph of potential energy vs distance from the earth. And the graph starts at (0,0) (the earth's position), slopes down to a minimum and rises up to (x,0) where the net gravitational force is 0, (or should it be at the moon's position?) like an upside down bell. Could you please guide me, I think the only way to solve for c), is if I can truly understand the changes in potential energy with distance in the system.

## Answers and Replies

NihalSh
I'm having some difficulty with part b). I would say the net potential energy is zero at the surfaces of the moon and the earth. But what about the point where the net gravitational force is zero?

And with part c), I can't seem to understand what's happening exactly. I'm visualizing a graph of potential energy vs distance from the earth. And the graph starts at (0,0) (the earth's position), slopes down to a minimum and rises up to (x,0) where the net gravitational force is 0, (or should it be at the moon's position?) like an upside down bell. Could you please guide me, I think the only way to solve for c), is if I can truly understand the changes in potential energy with distance in the system.

For part (b), you can solve for potential energy of the system using just the variables. When you do that it will become clear that it won't become zero anywhere except infinity, not even at the surfaces of earth and moon and the same applies to the point where net gravitational force is zero.

You need to calculate potential energy as a function of distance. Due to energy conservation, change in potential energy appears as kinetic energy.

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P.S. Total Potential Energy (Earth-Moon-Particle system):

$$U(x)=-\frac{GM_{E}M_{M}}{d_{M}}-\frac{GM_{E}M_{P}}{x}-\frac{GM_{M}M_{P}}{d_{M}-x}$$

##x## is center-to-center distance from earth to the particle. ##d_{M}## is the center-to-center distance of moon from earth

##ΔU(x)=ΔK##