In my classes, I like to bring up the question:

*Why do astronauts float around in space?*

The most common response to this question is that they float around because there is no gravity in space. Some people take this a small step further and say that there is no gravity in space because there is no air in space. This is why they claim there is no gravity on the moon (even though there is – more on this later).

I like to start off with the concept of gravity. Gravity is an attractive force between any two objects with mass. Your pencil and your dog both have mass so there is a force pulling your dog and your pencil (that is if you have a pencil) together. This force turns out to be extremely small. So small that you would never notice it. However, if one of the masses is very large, it is noticeable. An expression for the gravitational force was first determined by Newton. He came up with the following (turning off vector notation for simplicity).

![Page 25 1](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-1.jpg)

Where G is the universal gravitational constant, m1 and m2 are the two masses in the interaction and r is the distance between their centers. This force (as are all forces) is really a vector that points from one mass to the other.

![Page 25 2](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-2.jpg)

Here is an astronaut in the International Space Station. Let me use the above expression and calculate the gravitational force on him both on the surface and in orbit. Suppose he is 70 kg. On the Earth, the gravitational force would be:

![Page 25 3](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-3.jpg)

Now, in orbit (typically 360 km above the surface of the Earth, the same calculation (with a different r):

![Page 25 4](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-4.jpg)

Well, that is odd. The picture clearly shows the guy “floating around” but the gravitational force is not that much less. How can this be? It turns out it is possible to be “weightless” without even going into space. Here is a picture:

![Page 25 5](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-5.jpg)

These guys also look like they are floating around, but they are not in space. They are in an airplane commonly called “The Vomit Comet” – I didn’t make that up, that is really what they call it ([its even in wikipedia – so it must be true](http://en.wikipedia.org/wiki/Vomit_comet)).

So how does this work? Let me start by pretending a person is in an elevator. Here is a force diagram for the person if the elevator is stationary and not accelerating:

![Page 25 6](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-6.jpg)

Here, there are two forces acting on the person: the gravitational force from the Earth and the contact force of the floor pushing on the person. Since the person is not changing motion, these two forces must add up to zero (from Newton’s second law) $$/vec{F}_\text{net} = m\vec{a}$$ in the vertical directions would be:

$${F_\text{net}}_y = F_\text{Floor} – F_\text{Earth} = 0\\

F_\text{floor} = F_\text{Earth}$$

Everything looks ok, everything would feel normal in this case.

Now, suppose the elevator is accelerating upwards. This would happen if you got in an elevator and pressed the “up” button. In order to move in the upward direction, you would need to accelerate upwards at first. Here is the force diagram:

![Page 25 9](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-9.jpg)

Comparing this diagram to the previous, you should notice that the gravitational force is the same as before. This is because your mass did not change nor did the mass of the Earth. However, the floor must push harder. This is because:

![Page 25 10](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-10.jpg)

In order to accelerate upwards, the total force must be up. The only way this can happen is for the floor to be exerting a force greater than gravity . I assume you have been in an elevator before. This effect can be more easily felt if you have a very tall building that has a fast elevator. If the elevator is only going up one floor, it really doesn’t need to go too fast and thus does not have a large acceleration. How do you feel when you push the up button on an elevator? You should feel a little heavier – yes? But did your weight change? No. (extra credit: under what other circumstances would you feel heavier in an elevator?)

**KEY POINT: You feeling heavier is not due to your weight changing (because your weight did not change).**

Now suppose you accelerate down in an elevator. This would happen if you press the “down” button. Here is the force diagram for that case:

![Page 25 11](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-11.jpg)

Now the force of the floor is LESS than gravity for the following:

![Page 25 12](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-12.jpg)

And here you FEEL lighter (but you are not). So, when the force the floor pushes on you is large, you feel heavy. When the force the floor pushes on you is small, you feel light. What would you feel like if the floor didn’t push on you at all? How would you make this happen? If the force of the floor is zero, the above equation becomes:

![Page 25 13](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-13.jpg)

I could plug in the entire formula for the gravitational force (from above) but the radius of the Earth and the mass of the Earth are essentially unchanging. This gives an expression for the gravitational force on the surface of the Earth of $$F_\text{Earth} = mg$$ where g = 9.8 N/kg = 9.8 m/s^{2}. (Note: I already stated that the gravitational force was down, so here I am just declaring the magnitude of the gravitational force). So the force the floor exerts on you will be zero if the elevator accelerates downward at 9.8 m/s2. What would this diagram look like?

![Page 25 15](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-15.jpg)

If this happened, you would feel weightless. Maybe you have been in this situation. Hopefully it was not in an elevator but maybe something like this

![Page 25 16](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-25-16.jpg)

This ride has the riders “falling” down the track. Instead of crashing into the ground, the track curves away. This is called the “Tower of Terror”.

**Key Point:** How you feel depends on the force a floor (or something supporting you) and not the gravitational force. Some call this your apparent weight. For the elevator accelerating down at 9.8 m/s2, your apparent weight would be zero. For the Vomit Comet, your apparent weight is zero. For the astronaut in orbit, his apparent weight is also zero. Why is the apparent weight what you feel? Why do people experience the normal force, but not the weight? I don’t think I have the exact answers here, but rather a plausible argument.

First, let me model a person as the following:

![Page 26 1](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-1.jpg)![Page 26 2](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-2.jpg)

Wait. That doesn’t look anything like a person. I said it was a model, so its ok. This is a model of a person’s spine. The model consists of 5 masses (the red balls) connected by springs (the white lines). This person is standing on the surface of the Earth. I have also represented the gravitational force on each part of the person’s spine as yellow vectors. Note that they are all the same length. The gravitational force on each one is the same. The compressions of the springs are not the same. The lower springs are compressed more than the higher springs because they have to exert more force on the lower masses. Take the top mass (assume all masses are 1 kg). The gravitational force on this top mass is Fg = (1 kg)(9.8 N/kg) = 9.8 Newtons.

So, for this mass to be in equilibrium, the spring must also push up with 9.8 Newtons. If that spring connected the top mass and the next mass is the same spring, then this same spring should push DOWN on the next lower mass with 9.8 Newtons (Newton’s 3rd law says forces come in pairs). So, the second lower spring would have to push up 18.6 Newtons. 9.8 Newtons to counter act the weight of the second mass, and another 9.8 Newtons to counter at the spring above it pushing down. Since that spring must provide more force, it will be compressed more (the more you stretch or compress a spring, the more force it exerts).

**Key point:** The gravitational force on each piece of this ‘person’ is the same, but lower parts are compressed more. In my model of a person, gravity is “sensed” by how body is compressed.

Suppose I now put my “person model” in an elevator that accelerates upward at 3 m/s2. Here is what that would look like:

![Page 26 3](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-3.jpg)

The red arrow represents the acceleration. Notice that the gravitational force is the same as it was before, but the springs are compressed more. In this case, you would FEEL heavier. You are NOT heavier. Notice also that the blue arrow representing the force the floor pushes on the person is greater.

What if the ‘person’ is in an elevator that is accelerating down? (at 3 m/s^{2})

![Page 26 4](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-4.jpg)

Now there is less compression, smaller force of the floor pushing on the person but the SAME gravitational force. Here you would feel lighter (but you are not).

Next case, accelerating down at 9.8 m/s^{2}:

![Page 26 5](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-5.jpg)

Once again, gravity is the same. But here, there is NO force of the floor pushing on the person and the masses are equally spaced. This is because there is no need for the spring to exert a force on each mass since the following is true

![Page 26 6](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-6.jpg)

Since the acceleration is the same as the gravitational field (g), no other force is needed to make it move this way. Here you would feel weightless, but again – NOT.

Ok, here is my final case. What if there was zero gravity (as would happen if you were very far from any massive objects like planets or stars). Is it possible to make it “feel” like you have weight? Here is such a case:

![Page 26 7](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/page-26-7.jpg)

In this case, the masses seem to be compressed the same as they were with gravity, but no acceleration. But here there is no gravitational force. How would this happen? If the floor is accelerating upwards at 9.8 m/s2 then the floor would have to push on the person exactly the same as when the person was at rest in gravity.

Apparent weight is related to the force that is pushing on a person from an external object (this is what you feel). You don’t really feel gravity because it pulls equally on all parts of you.

[Here is the vpython program I used to create this model (in case you want it)](http://www.dotphys.net/page1/gravity/gravity/gravity2_assets/zerog2.py)

If you don’t know about vpython – go here [http://www.vpython.org](http://www.vpython.org)