Earth-Moon System Homework: Net Force, Potential Energy & Escape Velocity

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In summary: MAt x=d_M, potential energy is minimum and kinetic energy is maximum. At x=0, potential energy is zero and kinetic energy is maximum. At x=∞, potential energy is zero and kinetic energy is zero.In summary, the given conversation discusses the Earth-Moon system and its various parameters such as mass, radius, and distance. It also introduces the universal gravitational constant and considers an object placed on the Earth-Moon line. The conversation then poses questions about zero points for gravitational force and potential energy, as well as the kinetic energy of an object released from these points. Finally, it asks about the escape velocity of an object launched towards the moon, taking into
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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.
 
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brindha-rocks said:
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.

---------------------------------------------------------------

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##
 

1. What is net force and how does it relate to the Earth-Moon system?

Net force is the overall force acting on an object, taking into account all individual forces acting upon it. In the Earth-Moon system, the net force is what keeps the Moon in orbit around the Earth. It is the combined gravitational force between the two bodies.

2. How is potential energy involved in the Earth-Moon system?

Potential energy is the energy an object has due to its position or configuration. In the Earth-Moon system, the potential energy is involved in the gravitational interaction between the two bodies. As the Moon moves closer or further away from the Earth, its potential energy changes. This potential energy is what keeps the Moon in orbit around the Earth.

3. How is escape velocity related to the Earth-Moon system?

Escape velocity is the minimum velocity required for an object to escape the gravitational pull of a celestial body. In the Earth-Moon system, the escape velocity is the minimum velocity required for an object to escape the gravitational pull of the Earth or the Moon. This can be calculated using the mass and radius of the body, as well as the universal gravitational constant.

4. How does the distance between the Earth and Moon affect the net force between them?

The distance between the Earth and Moon affects the net force between them because the force of gravity decreases as the distance between two objects increases. As the Moon moves further away from the Earth, the net force between them decreases, which can result in changes in the Moon's orbit.

5. How do the concepts of net force, potential energy, and escape velocity apply to other celestial bodies?

The concepts of net force, potential energy, and escape velocity apply to all celestial bodies, as they are all subject to the force of gravity. These concepts help us understand how objects move in relation to each other and how they can escape the gravitational pull of a larger body. They can be applied to planets, moons, stars, and other objects in the universe.

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