5.6

Apparent Weightlessness and Artificial Gravity

The idea of life on board an orbiting satellite conjures up visions of astronauts floating around in a state of “weightlessness,” as in Figure 5-17. Actually, this state should be called “apparent weightlessness,” because it is similar to the condition of zero apparent weight that occurs in an elevator during free-fall. Conceptual Example 12 explores this similarity.

Figure 5-17 In a state of apparent weightlessness, astronaut and pilot Susan L. Still floats in the orbiting Spacelab Module. (Shuttle Mission Imagery/NASA Media Services)

| Conceptual Example 12 | | | Apparent Weightlessness and Free-Fall

Figure 5-18 shows a person on a scale in a freely falling elevator and in a satellite in a circular orbit. In each case, what apparent weight is recorded by the scale?

Figure 5-18 (a) During free-fall, the elevator accelerates downward with the acceleration due to gravity, and the apparent weight of the person is zero. (b) The orbiting space station is also in free-fall toward the center of the earth.

Reasoning and Solution As Section 4.8 discusses, apparent weight is the force that an object exerts on the platform of a scale. In the freely falling elevator in Figure 5-18a, the apparent weight is zero. The reason is that both the scale and the person on it fall together and, therefore, cannot push against one another. In the orbiting satellite in Figure 5-18b, both the person and the scale are in uniform circular motion. Objects in uniform circular motion continually accelerate or “fall” toward the center of the circle, in order to remain on the circular path. Consequently, the scale in Figure 5-18b and the person on it both “fall” with the same acceleration toward the center of the orbit and cannot push against one another. Thus, the apparent weight in the satellite is zero, just as it is in the freely falling elevator. The only difference between the satellite and the elevator is that the satellite moves on a circle, so that its “falling” does not bring it closer to the earth. In contrast to the apparent weight, the true weight is the gravitational force () that the earth exerts on an object and is not zero in a freely falling elevator or aboard an orbiting satellite.

The physics of apparent weightlessness.

The physiological effects of prolonged apparent weightlessness are only partially known. To minimize such effects, it is likely that artificial gravity will be provided in large space stations of the future. To help explain artificial gravity, Figure 5-19 shows a space station rotating about an axis. Because of the rotational motion, any object located at a point P on the interior surface of the station experiences a centripetal force directed toward the axis. The surface of the station provides this force by pushing on the feet of an astronaut, for instance. The centripetal force can be adjusted to match the astronaut’s earth-weight by properly selecting the rotational speed of the space station, as Examples 13 and 14 and 13 and 14 illustrate.

Figure 5-19 The surface of the rotating space station pushes on an object with which it is in contact and thereby provides the centripetal force that keeps the object moving on a circular path.

The physics of artificial gravity.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 13 | | | Artificial Gravity

At what speed must the surface of the space station (r = 1700 m) move in Figure 5-19, so that the astronaut at point P experiences a push on his feet that equals his weight on earth? Reasoning The floor of the rotating space station exerts a normal force on the feet of the astronaut. This is the centripetal force () that keeps the astronaut moving on a circular path. Since the magnitude of the normal force equals the astronaut’s weight on earth, we can determine the speed v of the space station’s floor.
Knowns and Unknowns The data for this problem are given in the following table:

Description | Symbol | Value
Radius of space station | r | 1700 m
Unknown Variable |  |
Speed of space station’s floor | v | ?

Modeling the Problem

| Step 1 Speed and Centripetal Force
| The centripetal force acting on the astronaut is given by Equation 5.3 as

Solving for the speed v gives Equation 1 at the right. Step 2 considers the centripetal force Fc that appears in this result.

| (1)

| Step 2 Magnitude of the Centripetal Force
| The normal force applied to the astronaut’s feet by the floor is the centripetal force and has a magnitude equal to the astronaut’s earth-weight. This earth-weight is given by Equation 4.5 as the astronaut’s mass m times the magnitude g of the acceleration due to the earth’s gravity. Thus, we have for the centripetal force that

which can be substituted into Equation 1, as indicated at the right.

Solution Algebraically combining the results of each step, we find that
The astronaut’s mass m is eliminated algebraically from this result, so the speed of the space station floor is

Related Homework: Problem 44

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 14 | | | A Rotating Space Laboratory

A space laboratory is rotating to create artificial gravity, as Figure 5-20 indicates. Its period of rotation is chosen so the outer ring (rO = 2150 m) simulates the acceleration due to gravity on earth (9.80 m/s2). What should be the radius rI of the inner ring, so it simulates the acceleration due to gravity on the surface of Mars (3.72 m/s2)?

Figure 5-20 The outer ring (radius = rO) of this rotating space laboratory simulates gravity on earth, while the inner ring (radius = rI) simulates gravity on Mars.

Reasoning The value given for either acceleration corresponds to the centripetal acceleration (Equation 5.2) in the corresponding ring. This expression can be solved for the radius r. However, we are given no direct information about the speed v. Instead, it is stated that the period of rotation T is chosen so that the outer ring simulates gravity on earth. Although no value is given for T, we know that the laboratory is a rigid structure, so that all points on it make one revolution in the same time. This is an important observation, because it means that both rings have the same value for T. Thus, we will be able to use the fact that (Equation 5.1) together with in order to find the radius of the inner ring.
Knowns and Unknowns The data for this problem are shown in the following table:

Description | Symbol | Value | Comment
Radius of outer ring | rO | 2150 m |
Centripetal acceleration of outer ring | ac, O | 9.80 m/s2 | Acceleration due to gravity on earth
Centripetal acceleration of inner ring | ac, I | 3.72 m/s2 | Acceleration due to gravity on Mars
Unknown Variable |  |  |
Radius of inner ring | rI | ? |

Modeling the Problem

| Step 1 Inner Ring
| According to Equation 5.2, the centripetal acceleration of the inner ring is , which can be solved for the radius rI:

The speed vI is given as (see Equation 5.1). With this substitution, the expression for the inner radius becomes

Solving for the radius rI gives Equation 1 at the right. The term T2 in Equation 1 is unknown, but it can be determined, as Step 2 discusses.

| (1)

| Step 2 Outer Ring
| The inner and outer rings both have the same period. Thus, we can obtain the period by considering the outer ring. The same approach used in Step 1 for the inner ring can be applied to the outer ring, with the result that the radius rO is given by an expression analogous to Equation 1:

Note that, except for T, all the variables in this result are known (see data table). Solving for T2 gives

As shown at the right, this expression can be substituted for T2 in Equation 1.

Problem solving insight All points on a rotating rigid body have the same value for the period T of the motion.

Solution Algebraically combining the results of each step, we find that
Related Homework: Problems 28, 35

Check Your Understanding 4

The acceleration due to gravity on the moon is one-sixth that on earth. (a) Is the true weight of a person on the moon less than, greater than, or equal to the true weight of the same person on the earth? (b) Is the apparent weight of a person in orbit about the moon less than, greater than, or equal to the apparent weight of the same person in orbit about the earth? (The answers are given at the end of the book.)
Background: The solution to this question involves an understanding of the difference between the concepts of true weight and apparent weight. In particular, pay attention to the apparent weight of a person in orbit.
For similar questions (including calculational counterparts), consult Self-Assessment Test 5.2. This test is described at the end of Section 5.7.

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