Orbital energy of object overcoming Moon's gravity

[salmayoussef]

Homework Statement

If the mass of the Moon is 7.4 x 10²² kg and its radius is 1.74 x 10⁶ m, compute the speed with which an object would have to be fired in order to sail away from it, completely overcoming the Moon’s gravity pull.

This might seem like a stupid question to ask, but I'm not sure what equation to use. That's my only problem.

My classmate used V = √(2GM/r) but I'm not sure how he got it. Any help would be great!

 

[Nathanael]

Let's find the minimum speed necessary.

With the minimum speed, the object will just barely overcome the gravity, and so it will have zero velocity and zero gravitational potential energy.

Zero gravitational potential energy happens at an infinite distance (when you're "out of reach" of the moon's gravity). So the velocity of the object will approach zero as the distance approaches infinity.

If the kinetic energy becomes zero and the GPE becomes zero (you could have nonzero KE, but we're solving for the minimum) then (according to Conservation of Energy) what must the KE be when you're on the surface of the moon (where GPE is not zero)?
P.S.
I'm sorry if this was an awful hint. (I can't think of the best way to explain it, perhaps someone else will do better)

 

[salmayoussef]

Let's find the minimum speed necessary.

With the minimum speed, the object will just barely overcome the gravity, and so it will have zero velocity and zero gravitational potential energy.

Zero gravitational potential energy happens at an infinite distance (when you're "out of reach" of the moon's gravity). So the velocity of the object will approach zero as the distance approaches infinity.

If the kinetic energy becomes zero and the GPE becomes zero (you could have nonzero KE, but we're solving for the minimum) then (according to Conservation of Energy) what must the KE be when you're on the surface of the moon (where GPE is not zero)?

I appreciate the attempt, but I'm pretty clueless! Tried writing out your hint but I'm still not getting it...

 

[Nathanael]

I appreciate the attempt, but I'm pretty clueless! Tried writing out your hint but I'm still not getting it...

Do you know the formula for GPE? (The integral of gravity with respect to distance)

Edit:
Nevermind, I'm probably going to end up confusing you (not because it's a confusing problem, I'm just not a good teacher)Basically the energy of the object (Kinetic plus GPE) must be zero on the surface of the moon (the reason, which I tried to explain in my last post, is basically just that "GPE=0" is what it takes to overcome gravity, and "KE=0" means after it's overcome gravity it has no KE left (and so that would be the minimum velocity needed to escape gravity). And the energy is the same the whole time (on the surface and once it's escaped) so you can figure out the velocity that way.)

I'm still making it complicated hahaha I just really can't think of how to say it :\

 

[salmayoussef]

Isn't it Egp = mgh ?

 

[Nathanael]

Isn't it Egp = mgh ?

Yes (well it's actually negative mgh)

What would "h" be in this situation?

And what would "g" be?

P.S. I edited my last post

 

[salmayoussef]

What would "h" be in this situation?

I think I understand what you're trying to say. Wouldn't h be the radius?

 

[Nathanael]

I think I understand what you're trying to say. Wouldn't h be the radius?

Yes, and what would "g" be?

 

[salmayoussef]

what would "g" be?

Fg = Gm1m2/r², no?

 

[Nathanael]

Fg = Gm1m2/r², no?

Yes, but g is not the force, it's just the acceleration. So it would only involve the mass of the moon.

Then you multiply it by h (r) and m and you get GPE=-Gm1m2/r

Then you set GPE+KE = to zero and therefore KE = |GPE| and then solve for V and you get the formula your classmate told you (sorry for being so direct but the explanation (of why GPE+KE=0) is difficult for me to articulate, I can't do much better than my last posts)

Edit:
There is logic behind it, though!

 

[salmayoussef]

The explanation (of why GPE+KE=0) is difficult for me to articulate, I can't do much better than my last posts)

No, no, no! That made lots of sense actually! Had to read it over a few times but I got it! Thanks so much for your help. I appreciate it. :)

 

[Nathanael]

No, no, no! That made lots of sense actually! Had to read it over a few times but I got it! Thanks so much for your help. I appreciate it. :)

No problem, I'm glad you got something out of it!

 

[ehild]

Isn't it Egp = mgh ?

The formula GPE=mgh is applied only when the gravitational force is constant. It is true enough near the surface of the Earth where the force of gravity is mg, with g about 9.8 m/s².

At longer distances you need Newton's Law for Universal Gravitation.

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:

$$F = G \frac{m_1 m_2}{r^2}\$$,

where:

F is the force between the masses,
G is the gravitational constant,
m1 is the first mass,
m2 is the second mass, and
r is the distance between the centers of the masses.

The gravitational potential energy at distance r between two objects is $$GPE = - G \frac{m_1 m_2}{r}\$$

It is negative, and becomes zero when the objects are infinite far from each other.
As Nathanael explained, the total energy of the object must be at least zero to be able to escape from the Moon's gravity: E=KE+PE=0. On the surface of the Moon that means

$$0.5 m v^2-G\frac{mM}{R}=0$$
where M is the mass of the Moon, and R is its radius.

ehild