The Missing Zero in Zero Gravity

By Jim Plaxco

In my younger days, I used to think that there was no gravity in space. The evidence was right there for me to see on the television: astronauts floating around, food hanging in mid-air. Then there were the commentators always talking about zero gravity, weightlessness, or microgravity. I was sure that once you got into space, gravity disappeared. Boy was I wrong.

The terms zero gravity, weightlessness, and microgravity are all misleading. They are misleading because they are taken literally. Zero gravity must mean no gravity. Weightlessness must mean no weight. Micro means one part in a million (10^-6); therefore the force of gravity must be only one millionth of its Earth normal level. Wrong, right, and wrong.

The terms zero gravity and weightlessness are used because the conditions that exist inside a spacecraft have all the appearances of the absence of gravity. Microgravity is an even worse description. It attempts in one word to reconcile the appearance of zero gravity with the fact that there is no such thing as zero gravity.

To prove that I am not leading you astray, I am going to call upon Sir Isaac Newton and that most feared of human creations: math. Please don't duck and cover - if you can multiply and divide, you can understand Newton's Theory of Gravitation. In 1665, Sir Isaac Newton developed an equation that accurately describes how the force of gravity behaves. That equation is:

Equation 1.

What the equation says is that the force of gravity between two objects depends on their respective masses and the distance separating them. For this example lets say that M represents the mass of the Earth and m represents my mass. The mass of the Earth is around 5.98 * 10²⁴ kilograms (kg). My mass is around 77 kg. The mass of the Earth is contained in a sphere with an average radius (the distance from the center to the surface) of 6378 kilometers (km). For the sake of simplification, we can think of the total mass of the Earth as being concentrated at a point 6378 km away from us. For the equation, the distance is expected to be in meters rather than kilometers, so we need to multiply the radius by 1000. The value of the constant G is really small: 6.67 * 10^-11. So the force of gravity that exists between myself and the Earth is equal to the amount F. This works out to almost 700 newtons. Just as distance is measured by meters and mass is measured by kilograms, here force is measured by a term called newtons.

Now, let's say that I have just won a contest to take a trip into space aboard the Shuttle and that my mission will be in an orbit 250 km above the surface of the Earth. That's 250,000 meters. Lets call this altitude by the name h. The new force of gravity that exists between myself and the Earth can be stated as:

Equation 2.

The only difference between this equation and the previous one is that my distance from the surface of the Earth has been increased by the amount h. At this new distance, the force of gravity between myself and the Earth will be f. But how does this compare with the force of gravity that you and I experience right now?

To find out how the amount of this new force compares with the Earthbound force, let's compare Equations 1 and 2 as shown by the ratio in Step 1.

Step 1.

The first thing is to simplify this equation by throwing out everything we don't need. In Step 2, I have separated out those items that remain the same in both the original equation and the Shuttle orbit equation. These items are the Gravitational Constant G and the masses of the Earth M and myself m.

Step 2.

The result of this simplification is seen in Step 3. We are left with only those items that change: the distances d vs d+h and the forces f vs F.

Step 3.

Let's simplify the equation some more by getting rid of some of those fractions. Step 4 shows how it is done. The term in the parenthesis has a value of 1 and doesn't alter the value of the equation. What it does is to make the equation take on a simpler appearance.

Step 4.

so that

Step 4.

The next step is to isolate the value of the force f. To do this requires two steps. In Step 5, the two force variables are first put on opposite sides of the equation.

Step 5.

so that

Step 6.

The final step is to isolate the force f. This is done in Step 7.

Step 7.

so that

Step 8.

Actually what we have just done is to construct an equation so that for any distance anyone is above the Earth's surface, we can figure out what percentage of the Earth's gravity is being exerted on them. We can do this by assuming that the force F that exists between a person and the Earth while on the surface is equal to 1. In fact, in this form it is not even necessary to convert kilometers to meters. So if the force of gravity between myself and the Earth is 1 while I am standing on its surface, then when I am 250 km above the surface, the force of gravity will be:

Solution.

That's right - an astronaut in a 250 km orbit is subject to 92 percent of the gravitational force that you and I experience every day.

So, how is it that the astronauts can float? Well, they are actually in a state of continuous free fall, which to us looks like an absence of gravity, i.e. zero gravity. Recall that the Shuttle is in an orbit around the Earth. In an orbit that is 250 km above the Earth, the Shuttle is traveling at 27,900 kilometers per hour. Earth's gravity is pulling the Shuttle back to Earth, but the surface of the Earth curves away from the Shuttle. The Shuttle's forward speed causes it to fall at a rate that matches this curvature. The Shuttle is always falling, but from its perspective, so is the surface of the Earth. To the astronauts aboard the Shuttle, it seems as though gravity has disappeared even though it is still very much present. It is the Shuttle's motion acting as a counterbalance to Earth's gravity that makes it possible for astronauts to float and a wide variety of experiments, impossible to perform on Earth (with the exception of the very brief periods of "zero gravity" that can be achieved in drop towers and in airplanes in a parabolic flight pattern), to be carried out.

This article originally appeared in the October-December 1994 issue of PSF News.