Escape Velocity from Earth-Moon System

[acfryman]

Homework Statement

A projectile is fired straight away from the moon from a base on the far side of the moon, away from the earth. What is the projectile's escape speed from the earth-moon system?

m_earth= 5.97x10^24kg
m_moon= 7.348x10^22kg
radius_earth= 6.371x10^6m

Homework Equations

V_esc=√(2GM/R)
KE=.5mv^2
PE=mgh

The Attempt at a Solution

My thought was that you would use the escape velocity equation and take into account the mass of both the Earth and moon, and for the radius, sum the radius of Earth, the distance between the Earth and moon, and the diameter of the moon. That wasn't right.

I then considered the kinetic energy required to escape both systems separately, and then went to sum those and solve for the velocity using KE=.5mv^2, but without a mass for the body that won't work. I also know that escape velocity isn't dependent on mass of the object escaping, so I also disregarded this method.

 

[rude man]

Homework Statement

A projectile is fired straight away from the moon from a base on the far side of the moon, away from the earth. What is the projectile's escape speed from the earth-moon system?

m_earth= 5.97x10^24kg
m_moon= 7.348x10^22kg
radius_earth= 6.371x10^6m

Homework Equations

V_esc=√(2GM/R
KE=.5mv^2
PE=mgh

The Attempt at a Solution

I then considered the kinetic energy required to escape both systems separately, and then went to sum those and solve for the velocity using KE=.5mv^2, but without a mass for the body that won't work.

m appears to the 1st power in potential as well as kinetic energy so it cancels.

Take a mass m at infinity. How much work does it take to bring it to the Moon's far surface if the Earth were not there?

Same deal, how much work to bring it to within the distance between the Earth's c.m. and the far side of the Moon if the Moon were not there?

 

[acfryman]

m appears to the 1st power in potential as well as kinetic energy so it cancels.

Take a mass m at infinity. How much work does it take to bring it to the Moon's far surface if the Earth were not there?

Same deal, how much work to bring it to within the distance between the Earth's c.m. and the far side of the Moon if the Moon were not there?

I'm assuming that you're setting KE and PE equal to one another to cancel out mass, as you mentioned, and then using that to solve for the escape velocities, and that makes sense. I have the values for the escape velocities. To escape Earth's gravitational pull, the velocity is 11.2 km/s and to escape the moon, you need a velocity of 2.375 km/s.

I'm confused about the second part though. How do you find the work value for each of those scenarios?

 

[rude man]

The escape velocity from Earth does not enter the picture.

Look at it a bit differently from what I said before. m located a certain distance from Earth (from its c.m. to be precise) on the far side of the Moon. Pretend the Moon is not there. How much work would you have to apply to m to move it to infinity? Call it W1.

Now, same question, but the Earth is not there & m is on the far side of the Moon. How much work W2 to move m to infinity?

Then, what is the relationship between W1, W2 and the escape velocity from the Moon's surface, heading to infinity, where the velocity finally reaches zero?

Do not try to use formulas for escape velocity. Reason the problem out, keeping conservation of energy in mind.

 

[Nickie]

Finally, I looked at the potential energy, and I'm not sure if I can use this to solve for escape velocity. However, I know that the potential energy at infinity is 0, so I set the potential energy of the projectile at the moon's surface equal to the potential energy at infinity.

PE=mgh
0=mgh
√(2GM/R)=√(2GM/R)

This doesn't give me a numerical answer, but it does show that the escape velocity from the Earth-moon system is equal to the escape velocity from the moon's surface, which makes sense since the projectile is already on the moon's surface. Additionally, this equation does take into account the mass of both the Earth and moon, as well as the distance between them. However, this approach assumes that the projectile is starting from rest on the moon's surface, which may not be the case in the given scenario. To accurately calculate the escape velocity, more information about the initial conditions of the projectile would be needed.