Potential energy in gravity

1. Sep 16, 2009

Villhelm

1. The problem statement, all variables and given/known data
What is the minimum (net) energy needed to take a 1kg object from the surface of the earth to the surface of the moon. Take gm = 0.16ge, where g= GM/R^2 is the grav. force per unit mass at the surface.

2. Relevant equations
Potential energy?

3. The attempt at a solution
I calculated the work done by the test mass against earths gravity minus that done on the test mass by the moons gravity across the distance earth surface -> moon surface just using the potential energy equation for an isolated gravitational field (GMm/R). I got 60.4MJ using the constants I fed in for the masses of the earth, moon and the surface radii and distance between them - the answer in the textbook was 60MJ, so I'm happy that this is at least correct.

My problem is that I didn't do the question the way that it's almost certainly meant to be done (I never used the ratio given in the question at all) and was wondering if anyone could help me with the insight that I think I've clearly missed somewhere along the line. I've spent a good while trying to figure out what, but to no avail.

edit: This is Harris Benson Ch.8 ex.53 pt.a

2. Sep 16, 2009

ideasrule

gm=GMm/rm2 and ge=GMe/re2, so GMm/rm=gmrm and GMe/rm=gere. Using gmrm and gere is probably how the book wanted you to do the problem.

3. Sep 16, 2009

Villhelm

Maybe I'm being dense, but I can't see how to make use of those either ... everytime I try, I just end up doing the same thing which just ignores the ratios

4. Sep 17, 2009

saunderson

Your Problem dealt with 3 bodies
• body 1 : earth
• body 2 : moon
• body 3 : object
and you are only interested in the body i called 'body 3' with the mass

$$m_3=1 \rm{kg}$$​

So the force exerts on body 3 is
$$\vec F_3(\vec r_3) = \vec F_{31} + \vec F_{32}$$​
(F_{ij} denotes the force on the ith body due to the jth body!)

If 'body 3' takes the direct way along a line, we only have to consider one component of the force, so the expression above reduces to
$$F_3(x_3) = F_{31} + F_{32} = - \gamma \, m_3 \cdot \left( {\frac{M_1}{(x_3 - x_1)^2} - {\frac{M_2}{(x_3 - x_2)^2} \right)$$​
(See the negative sign for F_{32} !!!) I've tried to visualize the above equation in the image below!

http://go-krang.de/physicsforums.com[/URL] [Broken] - thread - 337947.png[/PLAIN] [Broken]​

So if we try to think about your problem again, there must be a point on the line, where the resulting force is zero, so we have to solve the equation

$$0 = F_3(x_3) ~~~ \rm{for} ~~~ x_3 = R_z$$​

I hope you know, that every Force, that only depends on x (in this case x_3) have a potential, defined as

$$- \frac{dV(x_3)}{dx_3} = F_3(x_3)$$​

In our case we find

$$V(x_3) = - \gamma \, m_3 \cdot \left( {\frac{M_1}{(x_3 - x_1)} - {\frac{M_2}{(x_3 - x_2)} \right)$$​

So the energy, with that 'body 3' reachs the point R_z where

$$F_{31} = F_{32}$$​

and therefore can reach the moon (cause then 'body 3' only falls on the moon) is

$$E = V(R_z) - V(R_e)$$​

hope my text is not too complicated and i could help

with best regards...

Last edited by a moderator: May 4, 2017
5. Sep 17, 2009

Villhelm

Saunderson, you've worked out the energy required to move from the earth surface to the unstable equilibrium point between the earth & moon and not the net energy. The result your equations seem to give (unless I've made a calculation error) is 63.5MJ, which differs significantly to the textbooks answer of 60MJ and my answer of 60.4MJ, which falls within the range of uncertainty in the textbooks 2 significant figure answer.

Where my problem lay was in the question having quoted the ratio of surface gravitation gm=0.16ge for the earth and moon and in what way that would have been useful in calculating the net energy change. Typically the questions in this textbook have not given useless relationships with regards to questions. Neither my solution nor yours made use of the ratio given, which is why I am wondering what insight(s) I may have missed that would allow the ratio to have been useful, or whether this is simply the first major bug in a question of the textbook.

6. Sep 17, 2009

saunderson

Yes! The kinetic energy at the start moment must be at least

$$T = E = V(R_z) - V(R_e)$$​

Only in this case the object can reach the point of unstable equilibrium and gets in the space where the attraction of the moon is greater then the attraction of the earth ! If the object is behind the point of unstable equilibrium the gravitational field of the moon expends the work on the object! I think if the result that my equation yields is different to yours, maybe it's designated to consider the angular velocity of the earth!?

I'm pretty sure that my approach is basically right

with best regards

7. Sep 17, 2009

Villhelm

"I'm pretty sure that my approach is basically right."
Not with respect to what the question is actually asking, which is for the net energy change.

"maybe it's designated to consider the angular velocity of the earth"
Nope. I wrote the question word for word in my original post.

Even if so, to take into account the angular velocity of the earth, then this would only contribute 0.11MJ (equatorial tangential velocity is ~470m/s).

8. Sep 17, 2009

saunderson

Hi Villhelm,

the problem statement is:

So if you shoot the object from the earth with a kinetic energy of

$$T < V(r_{_{st.eq}}) - V(r_{_{earth}})$$​

the object never reaches the moon, it falls back to the earth again (For the notation see the image below!).

So if you calculate

$$V(r_{_{st.eq}}) - V(r_{_{earth}}) = 63.5MJ$$​

and with your approach you get

$$\approx 60 MJ$$​

then in your case the object never reaches the moon (assumed that you've taken the same values for the constants)!

http://www.go-krang.de/physicsforums.com[/URL] [Broken] - thread - 337947_2.png[/PLAIN] [Broken]​

Maybe a third can give a comment to that!?

Last edited by a moderator: May 4, 2017
9. Sep 17, 2009

Villhelm

This is pointless, you're repeatedly ignoring probably the most important word in the question definition and just inventing your own question that isn't the one in the textbook for which I am asking about. You've also ignored what I've said several times to be what is confusing me, namely the ratio gm=0.16ge and it's significance to the question in the textbook.

Scans of the question and answer:

Question:-
http://villhelm.webs.com/Q.jpg [Broken]
http://villhelm.webs.com/A.jpg [Broken]

You'll notice the question isn't asking for the initial kinetic energy, it's asking for the net energy change.

Last edited by a moderator: May 4, 2017
10. Sep 18, 2009

saunderson

Hi Villhelm,

but the problem statement is:

so if the book says that 60MJ are needed to take a 1kg object from the surface of the earth to the surface of the moon it's wrong, because then the object can't pass the point of unstable equilibrium and falls back to earth!

Of course it's theoretical possible to regain the energy of 3.5MJ, but even then 63.5MJ are initially needed to take the object to the surface of the moon!
The object can't borrow energy during the flight to the moon!

Hope you understand!?

No offense!

11. Sep 18, 2009

Villhelm

"energy needed"
vs
"(net) energy needed"

Can you explain why you seem to think there's no difference between these two items of text? Do you understand what net energy change actually means? Also, why is it that me, the textbook authors and all the proof readers who checked it and, tacitly, the user ideasrule (for not spotting my "mistake") are all wrong - while you who chops words seemingly at will from the question statement MUST be correct and yet your result, out of the three stated, is the outlier?

Are you aware of the difference that glossing over, changing, or simply removing parts can make to a sentence?

I shot "..." the queen.

I shot "a picture of" the queen.

The difference is obvious. One is an act of high treason, while the other is an act of tourism. One carries the maximum penalty applicable in UK law, the others only conceivable penalty is a tacky memento. Moral: details matter.

Similarly, the question statement I originally provided:

vs, something which I didn't write, nor exists in the textbook:
If the question asked the latter of the two, then yes indeed the value being sought would be the energy required to reach and just surpass the equilibrium point between the earth and the moon. However, the question is actually worded as the former, which further idealises the system into one in which the energy of the object falling into the moon from the equilibrium point is regained to the somewhat mystical reservoir of usable energy that many of such similar questions tap into. It might be "magic" to idealise so, but consider that exercise questions are almost always incredibly simplified and idealised in such manners.

How does the 1kg mass actually gain the energy it requires? A rocket? Rockets are inefficient - how much energy went into moving the 1kg mass and the rocket and all it's unspent fuel until the point at which the 1kg mass had enough energy on it's own to reach the moon? How much atmospheric drag, what effect did the other planets, the sun, the asteroid belt ... the motion of the various bodies themselves during the actual motion have on the energy needed? What were the exact initial conditions of the solar system when the 1kg mass is sent to the moon? Was the moon at it's minimum, maximum or somewhere inbetween distance to the earth during it's orbit? What about the solar wind, comets, radiation pressure, relativistic effects? What if the 1kg mass was radioactive and lost mass/energy via nuclear decay? What if it was 1kg of plutonium just on the verge of critical density and one stray particle from space just happens to hit it?

All of that and more is ignored for the purposes of providing sandbox examples to test the understanding of the very basic concepts without muddying the waters. Part of this specific sandbox is that we are only concerned with the net energy change. Not how such a thing could occur, only what it would be. Perhaps there is a large, ideal pneumatic cylinder that is used to losslessly store the incoming KE for later use by some perfectly efficient machinery on the moon itself which is later used to perhaps shoot some mass back at the earth which enters a similar system on earth that stores the KE from the fall losslessly for use in shooting mass at the moon ... and so on.

Does it matter what that setup is? No. The point of the question and many others like it is simply to use the physical concepts and mathematics to work out a (sometimes arbitrary) target value, but nevertheless exercising those concepts in an attempt to embed and associate them in memory. Nothing more. I've been quite amused if I be honest, that despite the number of times I've said it, you still seem to think I don't understand the difference between the energy required to reach and surpass the equilibrium point from earth and the net energy change associated with the motion to the moons surface. But I'm also unsurprised that this has devolved into a pissing contest, given the condescending tone of your initial response I can see you being unable to admit you may have simply misread the question, having already invested so much of your ego into it. Upon correcting your inadequate, unchecked suggestion that the difference arises from the KE associated with launch at earths maximum tangential velocity - did the thought even occur to check for any other unfounded assumptions regarding the question?

12. Sep 19, 2009

Hello Villhelm,I think that the author of the book should have expressed the question more clearly because it is easy to see an apparent ambiguity there.That aside have you solved the problem yet?In order to help I think that people here need to know what data the question allows you to use.I am assuming that you are given the radius of the earth,the radius of the moon and perhaps the distance between the earth and the moon(although you can get a reasonably accurate answer without the latter information)

13. Sep 19, 2009

saunderson

Hi Vilhelm,

I only want to help!

So, back to this "strange" problem:

The potential Energy of the object in the gravitational field of the earth is (reference point center of the earth):

$$V_e(r) = - \gamma \frac{M_e m}{r^3}$$​

and in the gravitational field of the moon (reference point center of the moon):

$$V_m(r) = - \gamma \frac{M_m m}{r^3}$$​

The gravitational force on the earth is

$$m g = \gamma \frac{G M_e m}{R_e^2}$$​

so

$$V_e(R_e) = -R_e \cdot g_e \, m$$​

and

$$V_m(R_m) = -R_m \cdot g_m \, m = -R_m \cdot 0.16 \cdot g_e \, m$$​

So if we neglect that the gravitational field of the earth still interacts with the object on the moon, we get

$$V_m(R_m) - V_e(R_e) = g_e m \left( R_e - 0.16 \cdot R_m\right) \approx 59,8 MJ$$​

maybe this is the way the "honorable" author of the book wants to solve the problem (but mind: the relative deviation applied on the result of the author is still 0,33444816054%)!?

Strange!

with best regards...

14. Sep 19, 2009

Villhelm

Thank you!

I don't quite understand the forms of your first set of equations, especially what $$\gamma$$ is, but I redone the question using more familiar forms for V and the insight it seems I was lacking, to take the potential from each surface individually.

It makes sense given the potential energy diagram you posted earlier:

The chapter from which the question comes covers those, but I hadn't realised it seems to be valid to work out the isolated surface potentials like so and take the energy level (which I think is what's being done?). Is that the ambiguity you refer to Dadface?

Using the values for the constants in the book, the result comes to 59.71 MJ, which seems agreeably close, the answer is only given to 2 significant figures as I doubt 60MJ is an exact value given the nature of the problem. The calculated value rounds to it nicely though and it has made use of the given ratio - so I'm happy, you've cracked it and I've been able to learn something worthwhile from that It's definitely been one of the more arbitrary exercises I've come across so far though!

I'm sorry I got a bit angry, but it was frustrating to be spending the time here to make no progress.

ge=9.807ms-2
Re=6.37 x 106m
Rm=1.67 x 106m

Last edited by a moderator: May 4, 2017
15. Sep 19, 2009