Goals: The student will learn|
Terms: weightlessness (or zero-g), artificial gravity, Coriolis force.
- Understand the "weightlessness" or "zero g" environment aboard spacecraft
- Realize that weightlessness will occur in any frame of reference which moves freely, subject only to gravity (e.g. "freely falling").
- Learn about the Coriolis force, another inertial force not covered previously. Like other inertial forces, it too appears only in a rotating (or otherwise accelerating) frame. While the centrifugal force acts on any object in the rotating frame, the Coriolis force only acts on objects that move in that frame, in a way that affects their distance from the rotation axis.
Stories and extensions: The "Vomit Comet" airplane used by NASA to simulate weightlessness; "artificial gravity" in a rotating space station; the swirl of draining sinks and of storms, north and south of the equator. (And in this lesson plan--an amusement park ride which includes free fall.)
Starting the lesson:
This lesson continues the exploration of rotating frames of reference, in particular two of their features: weightlessness and the Coriolis force.
Every one here must have already heard of "weightlessness" in space, and has probably seen on TV astronauts floating weightlessly in the "zero g" environment of the space shuttle. Clearly, Earth's gravity does extend to the shuttle's altitude--without gravity, how could it stay in its orbit?
So what is happening to gravity on the shuttle? To get a better understanding, let us look at a few examples of gravity in action:
Example 1: When you jump down from a truck or a wall, gravity pulls your body down and gives it an acceleration g--until your feet are stopped by the ground.
Example 2: When you stand on the floor, gravity also pulls your body down. But you do not move and do not accelerate, because the floor won't let you. It produces an opposing force which prevents any motion, and your body is now in an equilibrium.
In example 1, gravity is the only force present, and you accelerate.
In example 2, two opposing forces cancel each other, and nothing moves.
Now, Example 3: You watch on TV an astronaut inside the orbiting shuttle, standing in front of the camera. The way gravity acts on that astronaut's body--which of the preceding two examples does it resemble? The person jumping from a truck, or the one standing on the floor?
[The class may discuss the question briefly.]
It sure looks like example 2, the person standing on the floor.
But actually it is much closer to example 1, the jump from the truck.
Once again: In example 2, two opposing forces cancel each other, and nothing moves.
In example 1, gravity is the only force present, and your body is accelerating.
The only reason motion is not evident is that the shuttle and camera are also moving and accelerating, just as fast. Gravity is already fully employed, it is producing all the acceleration it can, and there remains no additional force that would push the astronaut against the floor. So to the person in space, this feels like weightlessness.
- The astronaut in front of the camera is moving, at the orbital velocity of 8 km/sec.
- The astronaut is accelerating, because motion in a circular orbit (or in any non-straight orbit!) is accelerated. And
- Gravity is the only force present.
If the orbit is an exact circle, gravity supplies the centripetal force needed to maintain it, so we have....
[continue with the first equation in section #24, up to "The Coriolis Force" which is introduced below.]
The second half of this unit is concerned with the Coriolis force.
The centrifugal force, you will recall, is an inertial force--one which only enters the calculation if the motion is described in a rotating frame of reference. In that frame, if the centrifugal force on some body is balanced by some other force, that body is at rest.
If however the body is is moving in the rotating frame, and we want to use its equations of motion in the rotating coordinates, we not only need to add the centrifugal force, but may also need another inertial force--the Coriolis force. Unlike the centrifugal force, it only comes into play when an object moves in the rotating frame, in such a way that it changes its distance from the rotation axis.
[The discussion that follows can use a sketch on the board.]
At any point in a rotating system, you move around the axis with a certain velocity. The greater your distance from the axis, the greater the velocity. And when you move in the rotating frame to a greater or a smaller distance from the axis, your velocity should also increase or decrease. Wouldn't you think that inertia would resist such changes? It does--and that is the essence of the Coriolis force.
[The formula for the Coriolis force is too involved, so all that will be given here is a qualitative description. In principle, we do not have to use it in any calculation--we have always the option to solve the motion in a non-rotating frame!]
A good example is given by a rotating space station, the kind that was first proposed in the early 1950s. It was then suggested that the centrifugal force might provide an "artificial gravity" in space, inside a wheel-shaped rotating space station...
[Continue in section #24, from the heading "The Coriolis Force"]
Are astronauts weightless in space because they have gone beyond the reach of Earth's gravity?
No. Even as far as the Moon, Earth's gravity is still felt, since that is what keeps the Moon in its orbit.
If astronauts are not free from the Earth's gravity, why do they feel
weightless in orbit?
The way we feel gravity is by the way we interact with our surroundings. When an astronaut is inside an orbiting spacecraft, both of them move with the same acceleration g (or g(rE/r)2 at distance r). No other force acts, and therefore no force exists that pushes the astronaut towards the floor--or makes standing up require an effort.
In a high dive into a pool of water, in the time between leaving the diving
board and entering the water, is the diver weightless?
Essentially, yes (air resistance interferes--but more so with skydivers). That is one reason why divers can perform intricate twists and somersaults. .
One of the rides in the "Great America" amusement park in Santa Clara, California, is the "Drop Zone" (other parks have similar rides), centered on a 220-foot tower with vertical rails running down its side. Riding on each rail is a frame holding several seats. At the foot of the tower, riders are strapped into those seats and and then the frame is pulled up to the top of the tower.
At the appropriate moment, the frame is released and plunges down, in what is essentially a free fall. At 1/3 of the tower's height, brakes become active and break the fall, so that by the time the ground is reached, all the falling speed is again lost. How many gravities do the riders feel (1) in the free-fall phase, (2) in the braking phase (assuming a constant rate (–a) of deceleration)?
(1) Zero gravities. Free fall is a weightless state.
(2) You start by defining the quantities you will work with.
Suppose the tower's height is 3h: the riders fall a distance 2h with acceleration g, then stop within distance h with acceleration –a (here a negative number). You would guess the riders would have to decelerate twice as fast
a = 2g
producing on them an additional inertial force 2mg. That, however, is in addition to their own weight mg, making the total force 3mg, or 3 gravities.
[ Optional] That guess is correct and here is a formal calculation:
Suppose V is the velocity achieved at the end of the free fall. From conservation of energy
2hmg = mV2/2 hence V2 = 4hg
The formulas for accelerated motion are like the ones for free fall, but with the acceleration (–a) in place of g. Suppose the braking motion takes a time t. Then
vfinal = vstarting – at or 0 = V – at
From which: at = V
For the stopping distance, using an initial velocity V = at
h = vstartingt – (1/2)at2 = Vt – (1/2)Vt
h = 1/2 Vt
Replace t = V/a to get h = V2/2a or V2 = 2ah
and since also
V2 = 4hg
it follows that
a = 2g
A space payload must achieve 8 km/sec to go into orbit. Can NASA get a "head start" on that motion by launching in the direction of the Earth's rotation? Is this equivalent to starting out with a velocity equal to the velocity of rotation at the Earth's surface, about 400 meter/sec near the equator? If not--why not?
Launching in the direction of the Earth's rotation (namely, eastward) does provide an initial velocity equal to the surface velocity of rotation. If the Earth did not rotate, NASA would have to furnish all the 8 km/s, but because it does, rockets launched eastward get a head start. That is one reason why Cape Canaveral was chosen as the prime launching site, as was the European site at Kourou, French Guiana.
Satellites launched into a polar orbit (e.g. from Vandenberg, California) lack this advantage. Israel's satellites have to be launched westward over the Mediterranean sea and require a boost larger than 8 km/sec.
A large low pressure area in the northern hemisphere has air flowing into it and swirling (as was seen in the lesson) counterclockwise. Suppose instead we have a high pressure area, with air flowing out. Will it swirl? And if so, in what direction? Give reasons.
Arguments similar to the ones used for low pressure areas show that here swirling also does occur and is (north of the equator) clockwise. In meteorology, low pressure areas are sometimes called cyclones and their circulation is "cyclonic," while high pressure areas are anticyclones and their circulation is "anti-cyclonic" in the opposite sense.