Dot Physics

xkcd and Gravity Wells

Wow. In xkcd 681 comic, there is an impressive illustration of the common term “gravity well”. Here is a small part of that large image:

i-ef55d5b21410dcf4300673dd012e834b-2009-12-29_xkcd.jpg

I can’t resist. I must talk about this awesome illustration. My goal for this post is to help someone understand that comic (although the comic itself does a pretty good job).

Energy

Energy is the key here. Here, I will talk about two types of energy – kinetic energy and field energy. In this case, kinetic energy is basically just the energy associated with something moving. Field energy is the energy stored in the gravitational field. You could also think of the field energy as the gravitational potential energy stored in the configuration of a system. I know that I didn’t talk about particle energy (you know, the E = mc2 stuff because it doesn’t matter here)

In a closed system, energy is conserved. This means that I can write:

i-524dba7678bffdabbb85a206bdb0ed6d-2009-12-30_la_te_xi_t_1_6.jpg

Just to say – a closed system is one that has no work done on it. Maybe the best way to explain a closed system is with an example. If I drop a ball and let it fall to the Earth, the ball by itself would be an open system. The ball plus the Earth would be a closed system. I really don’t want to go too much into work-energy principles, just enough to get to where I want to go (explaining xkcd).

So, back to the energy equation above. For this situation, I can write the kinetic energy (K) and the gravitational potential (Ug) as:

i-e877965c1ff528380f665454c4ccd589-2009-12-30_la_te_xi_t_1_7.jpg

I guess I should say that G is the gravitational constant (the big G, not the little g). ME is the mass of the Earth (change this if you are on a different planet) and the little m is the mass of the object you are looking at. Why is the gravitational potential negative? How about I just say that it is for now. How about a plot of Ug/m for an object somewhere around the Earth? (starting from r = Radius of Earth)

i-ee73792b4a8197fe650d9882922c29ed-2010-01-03_pot_6.jpg

I plotted the distance in units of “radius of the Earth”. Also, I included a “zoomed-in” portion of the graph. This zoomed in part is a plot of the same thing except from r = radius of the Earth to 10,000 meters higher. You will notice in this part, it looks pretty darn linear. In fact, I could even fit a linear function to that part of the data. Here is that function (where r is now in units of meters and measured from the center of the Earth)

i-c547fce1dec088ff91dfe6bbdb45288a-2010-01-03_la_te_xi_t_1.jpg

See anything familiar? I know you see “g” in there. Yes, that is the same g you know. This is where you get that function in the textbooks:

i-4dfe1499ee9b0e0761a381ce8fd6e0f7-2010-01-03_la_te_xi_t_1_1.jpg

The y-intercept is left off because only the change in potential matters. Ok, now for an example. Suppose I throw a ball up from the ground. If I consider the time after the ball left my hand AND I consider the system to be the ball and the Earth, then there is no work done on the system and the energy is constant. I can write:

i-312e22b75880d2ea5d8db01492be108b-2010-01-03_la_te_xi_t_1_2.jpg

Notice that both K and Ug have an m term in it. So, the mass doesn’t really matter. Now let me represent this as a sketch of a graph.

i-871321e1c2832bb7a5365cfb44e7c822-2010-01-03_untitled_2.jpg

The green line represents the total energy. This means that for any possible height, the difference between E and U is the kinetic energy. Notice that there is a maximum height for this given energy. If the ball were to exist in this energy plot to the right of that line, the kinetic energy would have to be negative. This is a problem in that it would imply an imaginary velocity. Also notice that this plot does not show you the trajectory of the thrown object. It just shows you what the speed will be for a given position.

Now back to the real potential energy plot. Here is the same thing as the above diagram for a ball that is thrown faster (ignoring the work done by air resistance). For this plot, I am going to pretend I throw a ball straight up with a speed of 10 km/s (yes, that is fast). Note that for this plot, the vertical axis is energy/mass.

i-9117a27ec65df93a70e9c2a7f27d0490-2010-01-03_untitled_3.jpg

In this case, the ball (or whatever it is) will get about 5 Earth-radii away from the surface before it starts to fall back down. But there is one big difference with this real potential function and the linear one from above. The linear function keeps increasing. If that was the potential, you could never get an infinite distance from the planet. However, with the real potential you can get an infinite distance away. If the total energy is

i-1d158b64dd156a602526ab7c6c8d7cf8-2010-01-03_la_te_xi_t_1_3.jpg

Since Ug goes to zero as r goes to infinity, then an object CAN escape. If the total energy is zero, then I can solve for the velocity needed to escape:

i-fda684132a7f05cdb03bb1d688da1b46-2010-01-03_la_te_xi_t_1_4.jpg

You can think of this velocity need tot escape as “the escape velocity”. Really, you should think about the “escape energy” which is the energy needed to get away from the planet and never return. The escape velocity assumes it is a free falling object. The problem is that it could be a combination of several things like the rotational motion of the object on the rotating planet or extra rockets or whatever.

How about a plot of Earth’s gravity well?

i-bd74c72e91d7fa6ed1ba0b9bcad7930c-2010-01-03_untitled_4.jpg

I added the Earth in there just to make it pretty.

The xkcd version

My well looks different than Randall’s (the xkcd author). He writes that the planets are not space out to scale so I guess he just artistically drew the wells (to look like wells). Also, he writes:

“Each well is scaled such that the rising out of a physical well of that depth – in constant Earth surface gravity – would take the same energy as escaping from that planet’s gravity in reality”

Let me check and see if this works. First, I will need to take some measurements. Sure, you could use photoshop or gimp or something to measure, but I will use Tracker Video Analysis. It’s free and does images also. Now, which planet should I look at? How about Uranus, because it is fun to say.

Step one – use the radius of the Earth to scale the picture.

i-49cecc9a41ec0db8fca846723d760094-2010-01-03_tracker.jpg

Now to measure the “height” of the Uranus gravity well. Using the same technique, I get that the well is about 3.8 Earth radii. So, what is the gravitational potential for the surface of Uranus? According to google, the mass of Uranus is 8.68 x 1025 kg and its radius is 2.55 x 107 m. This gives a gravitational potential per mass of:

i-013c2724f0769e437cc8a93ad5ebc66c-2010-01-03_la_te_xi_t_1_5.jpg

Now, how high would a “well” on Earth have to be to have the same change in potential per kg? (yes, this assumes the slope of the potential stays constant). Remember from before, on the surface of the Earth:

i-02b649708ba22d9c60eb85255f2d8f47-2010-01-03_la_te_xi_t_1_10.jpg

The real change in potential for Uranus is also positive since the final potential is zero. So, setting the Ug/m to the value for Uranus and solving for h:

i-57d85c038093a5fb943371058ca8c5f0-2010-01-03_la_te_xi_t_1_11.jpg

Wow. It worked. So, you can see where Randall gets the general expression for the height of the well in his drawing. He sets the real potential to mass equal to the gh Earth potential and gets:

i-2a6d5bdebf3b74a6fa2566278bd4686b-2010-01-03_la_te_xi_t_1_12.jpg

I love this drawing (or comic – not sure what to call it other than AWESOME).

The rest of this image could be left alone and be part of Dan Meyer’s What Can You Do With This series. But I can’t contain myself. Here are some suggested homework problems.

  • How big of a piece of paper would you need to include the Sun at this scale?
  • What if you wanted to space the planets out in correct horizontal scale also – how big of a paper would you need?
  • Do Randall’s sample escape speed calculations work?
  • What if you wanted to redo the whole image and include the rotational effects of the planets AND the orbital effects. What would it look like?

Update

Well, maybe this isn’t an update, but I thought I would share the python code I used to plot the potential well. Perhaps someone will find my sloppy code useful.

gravity_well_plot.py

If you don’t have pylab module installed, the easiest thing to do is to get the Enthought Python Distro

Comments

  1. #1 Greg Laden
    January 4, 2010

    Poor Pluto.

  2. #2 Rhett Allain
    January 4, 2010

    @Greg,

    Pluto had it coming. Pluto should have known it was a planet on borrowed time.

  3. #3 Vorn
    January 4, 2010

    I did some of the calculations in the LJ RSS feed for XKCD (click my name): the Sun’s well is 53ish times deeper than Jupiter’s. As for horizontal distance, well – the Sun’s well is only 9.7 gigameters deep, but Neptune is 4.5 terameters from the Sun, nearly 500 times as long as the deepest well. Adding in the dwarf planets requires that we go out to 14.6 terameters (for Eris), or about 1500 times the deepest well.

  4. #4 Rhett Allain
    January 4, 2010

    @Vorn,

    You get bonus points for being the first one to turn in your homework.

  5. #5 Rhett Allain
    January 4, 2010

    @Vorn,

    Actually, it looks like you turned in your homework before it was assigned. You must have a time traveling machine. Do you mind if I borrow it sometime?

  6. #6 CCPhysicist
    January 4, 2010

    If you included the tangent line (from the inset) on the full graph of U/m, it should cross U=0 just shy of 1 earth radius (2 on your horizontal axis) given xkcd’s value (which I have not bothered to check).

    What bothered me about the earth part of xkcd’s drawing was that it obscures the difference between being at a certain height (such as the shuttle’s orbital altitude) and being in orbit at that height. The K needed to be in orbit is what puts you “halfway to anywhere”.

  7. #7 √Ądamas hinton
    January 5, 2010

    is the module in the python script free and if it is not are there any alterunitons

  8. #8 David
    January 5, 2010

    Buying Vorn’s numbers, the slope of the solar system (bottom of the Sun up to Neptune) is 1/500 = 0.002.
    That’s pretty flat – so why can’t I just walk there?

  9. #9 Rhett Allain
    January 5, 2010

    @Adamas,

    pylab is a free module. It is part of the matplotlib – or you can get all that stuff installed with enthought.

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