I have a confession to make, I love kid shows, mostly the stuff from my time in Elementary School, like Doug, Hey Arnold, Bill Nye the Science Guy, and Magic School Bus and I still watch those shows when the opportunity presents itself; however, recently I have developed an unhealthy obsession with the Disney cartoon Phineas and Ferb. There are several reasons for this, but mostly it is because it is a smart, well-written show that doesn’t take itself too seriously and feels free to be really geeky.
Another thing I enjoy about the show is that they try to get science right (when it is brought up) sure they violate the laws of physics all the time but it is a cartoon about tweens building rockets and rollercoasters. However, when they show physics formulas they use the correct ones, not just complicated science babble (except when using a formula for a visual gag).
However, in one episode they demonstrate a common misconception between weight and mass, which I think is very common among people raised in the US (I single out my country because the Imperial System is stupid). In the scene the kids are floating through town in a helium-filled bouncy house when they pass over their destination. So the largest kid, Buford, jumps out of their floating vehicle and grabs onto a rope to allow the kids to sink to the ground. At this point another child points out that he doesn’t weigh any more at the lower height than when he was higher, and the two boy geniuses do not correct him. Well I thought I would, as there is a fundamental difference between mass and weight, in that mass is the amount of “stuff” in an object (usually measured in grams) and weight is the force due to gravity on an object thus it is a measurement of the interaction between the mass of an object and gravity (measured in pounds or Newtons). So what’s the deal his weight shouldn’t have changed just due to a difference in altitude right? Actually yes, yes it would have, you see the effect of gravity lessens the farther from the center of mass (center of the Earth) one is from it. So Buford’s weight would have changed when he lowered his elevation, as he would have experienced a greater force due to gravity.
How much you ask? Well currently I don’t know, but I do know that there is a physics formula for that:
- Where G is the gravitational constant (6.674*10^-11 m3/(kg*s2))
- Newton figured it out and he was one smart cookie, so we will go with it.
- The little m is the mass of Buford, which according to my fiancé (an Elementary Education major) would be about 50kg
- The big M is the mass of the Earth (5.974*10^24 kg)
- They used a really big scale to measure it
- And the r is the radius or distance from the center of the Earth (6.371*10^6 meters plus the elevation of the objects (or children) in question
- They used a really big tape measure
- Also the F stands for force, as in the force exerted on the objects by gravity, but we can also consider this weight
So first let’s talk about assumptions first off we are assuming that the Earth is a perfect sphere (not true) so that we won’t have to change the value for the radius as the jump house floats through the air, secondly we will assume that there is no change in mass for either object during this time period. So what happens?
Well first we need to figure out how high the jump house is above the ground and then how far below the jump house Buford is when he is hanging by the rope. So since this is a cartoon I’m going to make educated guesses rather than attempt to use some type of computer software to analyze the scene and get some hard numbers. I’m doing this for two reasons first I don’t have experience with that kind of software, and second it’s a cartoon and I’m just trying to make a point.
So let’s look at this video, the scene in question starts about at the 6:20 point and it looks as if the floating jump house is about 30 meters above the ground, and the rope Buford hangs off of is about 8 meters long, so we now have all the info we need.
Video url: http://www.youtube.com/watch?v=PIneRclgW8I
F1 is the original height in the jump house, and F2 will be Buford’s second lower height.
F1=(GmM)/r^2
F1=[(6.674*10^-11 m3/(kg*s2)) (50kg) (5.974*10^24 kg)]/(6.371*10^6m + 30m) ^2
F1=(1.99*10^16kg m3/ s2)/ (6.371*10^6 + 30m) ^2
F1=491.1364025 kg*m/s2 aka Newtons
Converts to 110.41 lbs
F2=(GmM)/r^2
F2= [(6.674*10^-11 m3/(kg*s2)) (50kg) (5.974*10^24 kg)]/(6.371*10^6m + (30-8)m) ^2
F2=(1.99*10^16kg m3/ s2)/ (6.371*10^6 + 22m) ^2
F2=491.1364025 kg*m/s2 aka Newtons
Also converts to 110.41 lbs
So for all practical purposes there is no change in weight, however if my calculator could show more decimal points you would be able to see a difference eventually. If you don’t believe me we can check Buford’s weight at vastly different elevations, to see if this changes at all.
Radius in Earth radii | Weight in Newtons |
R*1 | 491.141 |
R*1.25 | 314.330 |
R*1.5 | 218.285 |
R*1.75 | 160.373 |
R*2.0 | 122.785 |
R*2.25 | 97.016 |
R*2.5 | 78.583 |
Remember, all we changed was the distance from the center we moved Buford; we didn’t change the mass of any of the objects. The cool thing is that the first row shows that Buford is about 5 thousandths of a newton lighter 30 feet in the air than he is on the ground (or a 0.001 pound difference).
Now In Graph Form! (I labeled my axis when I made this, but they disappeared!) |
As you can see there is a downward trend as you move Buford farther from the surface of the Earth (note do not try to extrapolate to the left to R*0 as there the formula is undefined and also if you go below the Earth’s surface you have to start subtracting mass from your value of big M so the force of gravity becomes weaker—however if you had an Earth mass black hole you could find the force of gravity approaching zero *hint* that number would be really big! As in 49114.103 Newtons (11041.290 pounds) at R*.1)
So while they were right for all intensive purposes in the episode, two boys able to build an interstellar space ship should have known better.