# Conceptual problem with gravity

1. Jun 24, 2009

### zooboodoo

This came from a practice MCAT test and I thought about it for a while and was still confused, I have the correct answer along with the explanation the text book gave and was wondering if anyone had any light to shed on this. I didn't want to post this in the homework section because I'm not looking for an answer per say, more of a discussion; in either case, if this is in the wrong section my apologies. The question, my thoughts, and the answer in the following order:
A rocket is launched from earth to explore our solar system and beyond. As the rocket moves out of the earth's atmosphere and into deep space, the gravitational constant. g. decreases and approaches zero, and the gravitational potential energy of the rocket:
A. also decreases and approaches zero.
b. continually increases.
c. remains constant.
d. increases at first and then decreases and approaches zero.

I thought at first, the relationship between g and height would be the important factor so i thought that since Force of gravity goes by the inverse square, as you increase height, or distance of the rocket from earth in this case the r^2 value effectively decreases the value for g from 9.81 m/s^2 --> something much smaller as you get far away to other solar systems as the question implies, but as g gets smaller, h, from Ug=MGH gets bigger, but since H=R (maybe this is where my assumption was wrong) the rate at which g gets smaller and h gets bigger based on the distances, it seemed to me like at some point the gravitational potential energy would stop increasing (i thought a pretty large distance away from the center of the earth when this would happen cuz i think of it like 2 functions ones linear and ones exponential (mgh v. gmm/r^2) and they would eventually intersect and then i thought of how that could apply to the work energy theorem and i started getting confused) but either wayso i thought about it for a while and i picked letter D. the answer from the book was........ (spoiler bellow)

B., the explanation given was:
Energy is required to separate attracting bodies. The rocket is attracted to earth by gravity. Gravity is a conservative force so the added energy goes into potential energy.
Doesn't make sense to me.

2. Jun 24, 2009

### zooboodoo

Sorry, for some more clarity what I thought was that: g is getting smaller faster than h is getting bigger in the equation for Gravitational potential energy (GPE=mgh). This is why the math / concept didn't make sense to m; like when you put height from GPE=mgh into the equation F=GMm/r^2, h goes in for r, r is related to g through inverse square (or at least this is what I thought), so as r doubles, and h doubles, g, as calculated from F=GMm/r^2 decreases by a factor of 4, h increases by a factor of two .:. GPE, I feel like, should be decreasing as height increases?

3. Jun 24, 2009

Staff Emeritus
Step 1. Don't try to shoehorn their explanation into your previous conceptions. In particular, you are "fishing for the right equation". This is not going to help you.

Step 2. Read the explanation, and think of it entirely on its own terms. Now, what part doesn't make sense? Is it:

1. Energy is required to separate attracting bodies.
2. The rocket is attracted to earth by gravity.
3. Gravity is a conservative force so the added energy goes into potential energy.

Which part don't you understand?

4. Jun 24, 2009

### zooboodoo

The individual parts make sense, on their own. I was thinking that at a point, some distance, r/h gravitational potential energy would actually stop increasing and be decreasing, concepts of the spaceship being at a very large distance and having an incredibly high gravitational potential energy was making me think about the equations, and that's when I thought something was probably wrong in my relationship between the two, but that's what I dont understand, I know that everything isn't intuitive, but it doesn't feel right that an object an infinite distance away will have the greatest gravitational potential energy

5. Jun 24, 2009

### zooboodoo

And another part of the problem I had is why gravity is necessarily/how it was defined as a conservative force; that part I will start looking up soon.

6. Jun 24, 2009

### uart

I'm not sure if this will help, but a better answer would be the following :

the gravitational potential energy of the rocket:

e) Continually increases, asymptotically approaching zero from below.

7. Jun 24, 2009

### Mapes

All energy values are relative; replace the meaningless "zero" with "a constant" (or briefer, "Continually increases to an asymptote."), and you have an even better answer.

Last edited: Jun 24, 2009
8. Jun 24, 2009

### zooboodoo

if its increasing asymptotically towards 0 would that say that GPE is always negative, and relatively decreasing? Can anyone shed some light on the relationship of the equations F=GMm/r^2 and if / how it can be related to GPE=mgh in regards to the r and h values if they are infact related at all

9. Jun 24, 2009

### uart

Good point. I was just taking the most common reference point of infinity for the zero potential energy in the hope of alerting the OP to how PE is usually referenced for "space" as opposed to "terrestrial" type problems. (and trying not to confuse him too much in the process ).

10. Jun 24, 2009

### uart

Yes, this is the question I hoped you would ask.

With terrestrial type problems (close to Earth) the gravitational force stays relatively constant so we get a simple linear function for PE (PE = mgh). With such a simple linear function there is no real problem with taking any arbitrary reference point for zero energy, so we usually just choose ground level as zero potential.

In a "space" type problem however the gravitation force falls off as 1/r^2 and it turns out that this implies that the PE falls off as -1/r (negative reciprocal of distance from center of body). In particular (for the simple two body case) :

$$PE = const - \frac{m M G}{r}$$

In this case it's not quite so clear exactly what point we should take as the zero potential "reference point". As Mapes pointed out above, we still can choose any arbitrary point for our zero reference (eg the surface of the Earth if we wish), as the "free" constant ("const") in the above equation can be adjusted accordingly. If however we do go this route then we have the rather awkward situation that our formula will only work for objects orbiting (or escaping from) Earth, to use the equation for other planets or other bodies would generally require a different value of "const" each time.

The awkwardness is greatly relieved if we simply choose the zero PE point to be at r = infinity, making the constant "const"=0, and resulting in negative values of PE for all finite distances.

Last edited: Jun 24, 2009
11. Jun 24, 2009

### Bob S

Both the Earth's gravitational potential and the Sun's gravitational potential increase as the rocket leaves the solar system. But some of the solar potential is included in the orbital angular momentum of the Earth circling the Sun (2.95 miles per sec). But even then, mgh is not a conservative potential for the following reason. The rocket ship must use (lose) mass to get thrust. So the mass of rocket ship is less when it leaves the solar system than when it left Earth. Since the mass is less, then mgh is less. So although gravity is a conservative force, mgh is not a conservative potential.

12. Jun 24, 2009

### PatPwnt

Yes, the GPE is always negative with respect to the GPE at r=infinity and no, the the GPE is increasing when you go farther out into space. When you take off in a rocket you have to do work the whole time you are overcoming the gravitational force, even as the gravity is getting weaker.

The gravitational force is conservative because if I fixed two points A and B around my planet, no matter what path I take to go from one to the other, the energy required will be the same for all of the paths.

Think of the gravitational potential energy at a distance r from a planet as the energy you gain by traveling from that distance r to a point at infinity. When you do this you gain negative energy which is just a fancy way of saying you lost energy or did work. You have to do work because the gravitational force is acting against you.

13. Jun 24, 2009

### Bob S

Yes, but if you use 1/4 the total rocket mass as thrust to get to infinity, your potential energy is not m earthinfg(r) dr, but closer to 3/4mearthinfg(r) dr.

14. Jun 25, 2009

Staff Emeritus
True. And completely irrelevant for the purposes of the poster.

The guy is trying to study for the MCAT. Tossing in a zillion complications is not helpful.

15. Jun 25, 2009

Staff Emeritus
Why not?

The higher you drop something from, the harder it hits right? This is just the limiting case of that statement.

16. Jun 25, 2009

### 337

2 cent :

Look at the basics : the rocket takes off, it is using energy to overcome earth's gravity (ignoring atmospheric effects), which is gradually weakening as the rocket is moving away from the source of the gravitational force.

Potential energy can be "cashed in" and turned into kinetic energy (and then impact) only for as long as the force exerted by earth's gravity is strong enough to pull the rocket back (in sum with other masses in space).

At an infinite distance the force of earth's gravity becomes infinitely small.