Gravitational Energy of Earth<Moon (what distance)

[Mikemorgan12]

Homework Statement

Not give any information just the equation Fg= G m1m2/r^2
Supposed to find the distance at which the moons gravity has a greater force on a spacecraft than the Earth does.

Homework Equations

My prof told me to put G m1m2/r^2= G m1m2/r^2 and substitute one r in for another i.e. (r-r₂)^2

The Attempt at a Solution

I put the two equations equal to each other and got rid of what I could

mE/r^2= mM/(r-r₂)²
I try to plug numbers in and get a quadratic but I keep getting a square root of a negative number or i just get the actual distance from the Earth to the moon 3.85x10⁸ m
Help would be greatly appreciated have been working on this for a few days now.

 

[PeroK]

Homework Statement

Not give any information just the equation Fg= G m1m2/r^2
Supposed to find the distance at which the moons gravity has a greater force on a spacecraft than the Earth does.

Homework Equations

My prof told me to put G m1m2/r^2= G m1m2/r^2 and substitute one r in for another i.e. (r-r₂)^2

The Attempt at a Solution

I put the two equations equal to each other and got rid of what I could

mE/r^2= mM/(r-r₂)²
I try to plug numbers in and get a quadratic but I keep getting a square root of a negative number or i just get the actual distance from the Earth to the moon 3.85x10⁸ m
Help would be greatly appreciated have been working on this for a few days now.

You need to decide what these different variable radii are. I would tend to use $R$ for the distance between the Earth and the Moon, as this is fixed, and then $r$ for the variable distance the spacecraft is from Earth.

With that, what does your equation become?

 

[Mikemorgan12]

You need to decide what these different variable radii are. I would tend to use $R$ for the distance between the Earth and the Moon, as this is fixed, and then $r$ for the variable distance the spacecraft is from Earth.

With that, what does your equation become?

Okay, so I would get something similar to what I had?
mE/ R² = mM/ (R-r)²
Then plug in R and solve for r? I think my problem is more with the algebra involved but maybe I am plugging in the wrong numbers...

 

[PeroK]

Okay, so I would get something similar to what I had?
mE/ R² = mM/ (R-r)²
Then plug in R and solve for r? I think my problem is more with the algebra involved but maybe I am plugging in the wrong numbers...

What does $m_E/R^2$ represent? Remember $R$ is the fixed distance from the Earth to the Moon.

 

[Mikemorgan12]

What does $m_E/R^2$ represent? Remember $R$ is the fixed distance from the Earth to the Moon.

I guess it would represent Fg/ G mM. Am I going about this in the wrong way? I am having trouble understanding.

 

[Mikemorgan12]

I obtained mE/R² by cancelling things from each side of the equation

 

[PeroK]

I guess it would represent Fg/ G mM. Am I going about this in the wrong way? I am having trouble understanding.

$m_E/R^2$

That represents the gravitational force of the Earth on something the same distance as the moon. So, that's the force on the spacecraft only when the spacecraft has reached the moon.

 

[ZapperZ]

First of all, I am not sure if you even understood what and why you are doing this. You stated that you are doing what your professor told you, but do you even know why?

Secondly, in ANY physics problems at this level, you must show a sketch! Otherwise, we are dealing with all these symbols that you never defined and become rather vague.

So instead, *I* spent time in presenting a sketch that is relevant to this problem.

I can explain what each of the symbols mean, but I am guessing that the sketch should be very clear in what they are. In addition, I also drew the FBD on the mass "m" which is the object of interest. Those "F's" in the sketch represents the forces acting on it.

At some distance, the NET force acting on "m" will be zero. This means that if "m" gets any closer to the Moon, the force acting on "m" by the moon will be larger than the force acting on "m" by the earth. So this is the location that you want, where the net force is zero.

So solve for F_E = F_M

but with the added info that r_E + r_M = R.

Zz.

 

[PeroK]

... just to align @ZapperZ's diagram with what I had suggested: he uses $r_E$ where I used $r$.

 

[Mikemorgan12]

First of all, I am not sure if you even understood what and why you are doing this. You stated that you are doing what your professor told you, but do you even know why?

Secondly, in ANY physics problems at this level, you must show a sketch! Otherwise, we are dealing with all these symbols that you never defined and become rather vague.

So instead, *I* spent time in presenting a sketch that is relevant to this problem.

View attachment 233042
I can explain what each of the symbols mean, but I am guessing that the sketch should be very clear in what they are. In addition, I also drew the FBD on the mass "m" which is the object of interest. Those "F's" in the sketch represents the forces acting on it.

At some distance, the NET force acting on "m" will be zero. This means that if "m" gets any closer to the Moon, the force acting on "m" by the moon will be larger than the force acting on "m" by the earth. So this is the location that you want, where the net force is zero.

So solve for F_E = F_M

but with the added info that r_E + r_M = R.

Zz.

Thanks and this is a question on an assignment I am doing.

 

[ZapperZ]

Thanks and this is a question on an assignment I am doing.

This doesn't tell me anything.

Almost every question posted in the HW forum is "... a question on an assignment...". You have not stated whether you actually understood how to set up the the solution and if you actually found what you want.

Zz.

 

[PeroK]

Thanks and this is a question on an assignment I am doing.

First, you should look carefully at @ZapperZ 's diagram. Do you understand it? Second, you should have had a diagram like that. Third, you need to relate the equations (in particular $F = \frac{GMm}{r^2}$) to the diagram. That's the point at which you move from the physical set up of the problem to the (in this case algebraic) solution.

The good thing about the diagram and the algebra is that it generalises the problem. It could be the Earth and the Moon; or the Earth and the Sun; or the Sun and Jupiter, for example.

The final step is, of course, to plug in the values for the Earth-Moon problem to get a specific answer.