**2.5 Freely Falling Bodies**

This section’s pretty straightforward, but introduces a symbol you’ll use a lot: g.

g is usually pronounced “little gee” since there is the more important Big G, which we’ll get to when we talk about force.

Little g isn’t a fundamental constant of the universe – it’s the average acceleration something will experience if you drop it close to the surface of the Earth. Later, we’ll understand this through more basic principles. For now, think of it way: in your equations that contain an *a* for acceleration, if they deal with freely falling bodies, you can change *a *to g.

g is usually given as 9.8 meters per second per second. You’ll note the units are the same as acceleration, because little g is just a particular rate of acceleration that we humans care about.

For convenience, unless your teacher is anal about these things, you’re better off just saying g=10. It’ll save time on your calculations, and won’t make you any less intelligent. Remember, “9.8″ is bullshitty too. You’ll have a slightly different lower little g on Mount Everest than you will near the Dead Sea. We’ll get more into the specifics later, but here’s a simple way to think about it: You know if you drop something in space that it experiences almost no gravity. So, in space, little g=~0. It seems reasonable therefore to suppose that as you go higher up, little g will get lower, and this is indeed the case.

There are two other good takeaway facts here that are simple but important:

1) Little g itself is always positive. Let me be clear here, because I think the book might be a tad confusing: This DOES NOT mean you never put a minus sign in front of g. In fact, you often do. It means the value of g is 9.8 m/s^2, NOT -9.8 m/s^2. Students often screw this up because they know gravity usually “points down.” The way we handle that, by convention, is by writing “-g” for the acceleration. That keeps everything nice and clear.

2) The quadratic formula is your friend. If you don’t have it memorized, you should. It’s pretty simple, and you’ll need it any time you encounter a quadratic polynomial that doesn’t have obvious solutions. This is physics, not pure math, so your solutions are not going to have pretty answers unless the professor is being nice to you. So, you’re going to want the quadratic formula. If you don’t already know how to use it, learn.

Since this is not pure math, you should also be aware that you can just throw out answers to the quadratic formula that don’t make any sense. Because you’re dealing with quadratic roots, you’ll always get two answers. Sometimes you won’t understand why, and then later it’ll turn out the math is telling you something you didn’t realize. But, often, it’s just giving you answers that don’t make physical sense. Remember, the math is just a way to solve equations that are similar to reality – it doesn’t mean all answers are equal. If your math tells you the solution occurred at t=20 and t=-10, and the question doesn’t involve a time machine, you can be pretty certain t=20 is the correct solution.

I know you know what you mean when you say “You know if you drop something in space that it experiences almost no gravity.” but I worry that some readers trying to teach themselves physics from these posts won’t know what you mean.

What you mean is that “far away from any massive bodies, objects experience almost no gravity.” You’re not talking about the “zero g” of the space station astronauts or that sort of thing, where they are accelerating a lot, but they and their toys and their station are all accelerating the same amount.

Maybe there’s another interpretation of what you’re saying from different reference frames, one way or another.

You know if you drop something in space that it experiences almost no gravity. So, in space, little g=~0. It seems reasonable therefore to suppose that as you go higher up, little g will get lower, and this is indeed the case.This quoted bit has the potential to mislead the reader, by reinforcing a common misconception.

You are correct, that if you get very far from the earth, the gravitational acceleration due to the earth (“Little g”) approaches zero. However, when you’ve seen movies of people in the Space Shuttle or on the International Space Station, they’re not “very far” from the earth. They’re in low earth orbit, which is only a little bit further from the center of the earth than you are now. You’re about 12,750 km from the center of the earth, and they’re another 350 km further out.

When you see their Tang floating in front of them, it’s not because the earth is exerting a much smaller gravitational force on the Tang than it would if the Tang were in your glass at home. Instead, it’s because the Space Shuttle, the astronaut, and the Tang are all “falling” around the earth in unison. That’s what an orbit is. The gravitational force that’s making them “fall” is nearly as strong as it would be on the surface. But they also have a horizontal speed that keeps them from hitting the earth… hence they keep falling around it.

If you cut the cable of an elevator, you and your Tang would float inside the elevator in the same way. At least briefly. It’s not because g is smaller, but because you’re all in free-fall together.

I’ve been beaten to the punch with my initial comment. Though one of my thoughts was “it’s enough to hold the moon into place” as a demonstration of its near-space potency. But then running some quick numbers it looks like g ~ 0.002 m/s^2 by the time it gets to the moon. Though for the ISS the difference between it and the Earth’s surface is pretty negligible. So the numbers can be deceptive!

Besides, the way that the ISS can be so damn close to the Earth and yet be in perpetual free-fall, as though it’s constantly trying to crash land on the Earth’s surface but keeps on missing on account of moving so fast, it’s just a really cool thought and mechanism, a great little thought experiment for students.

While it’s not as misleading as ‘g=0′, I always found the ‘free falling’ depiction as a bit hard to grasp, intuitionally. In contrast, the forces picture is quite clear and the maths involved quite easy.

By correcting for the larger height of the space station/craft you can easily calculate the gravitational pull g. Running in a circular orbit around Earth, a centripetal force a_c will work on the object but in the opposite direction (away from the center of the orbit). Assuming that you’re looking at a stable orbit means both forces have to be the same,so g=a_c (otherwise the orbit would widen or shrink because you had an effective force towards or pointing away from Earth), so on the orbit the effective acceleration of objects with regards to each other becomes a_eff=0.

It’s a fun calculation for an undergraduate class of non-physics science majors – most will never have heard of it, in my experience, and it teaches a lot about the nature of forces.

I was taught g as acceleration due to gravity, but as a teacher, I’ve used g as the local strength of gravity, or the local gravitational field strength. Steers some learners away from misconceptions later down the line about the pull of gravity always being g*m