Zero Gravity: ISS Orbit & Neil Degrasse Tyson

  • Thread starter Thread starter kghosh
  • Start date Start date
  • Tags Tags
    Gravity Zero
AI Thread Summary
The discussion clarifies that the term "zero gravity" is misleading when describing the environment of the International Space Station (ISS), which is actually in free fall, creating a sensation of weightlessness for astronauts. Despite the term, gravitational forces are still present, enabling the ISS to maintain its orbit around Earth. The confusion arises because "zero gravity" refers to the absence of normal force felt by astronauts, not the absence of gravitational force itself. Neil Degrasse Tyson's comments on bone loss in space relate to this weightlessness, highlighting the physiological effects of living in a low-gravity environment. Ultimately, understanding the distinction between gravity and the sensation of weightlessness is crucial for comprehending the conditions in space.
kghosh
Messages
4
Reaction score
0
I read in a physics textbook that it is not zero gravity in ISS but the fact that the ISS is in "free fall" makes astronauts float around. They do not feel any normal force because the vehicle they are in are also in free fall along with them. Then I watched Neil Degrasse Tyson talk about bone loss in ISS because astronauts are in zero gravity. If it is zero gravity then why is the ISS in orbit? There should be no centripetal force to keep it in orbit. I am confused!
 
Physics news on Phys.org
The term 'zero gravity' is often used when referring to things in free fall.

It is inaccurate.
 
The gravitational force on the Space Station obviously is not zero. It is gravity after all that makes the Space Station orbit the Earth.

The term "zero gravity" does not refer to the force due to gravity. It refers instead to all forces except gravity. When you go to an amusement park and take a ride that bills itself as having a zero g roll, or a zero g drop, gravity is not changing during the course of the ride. Your uneasy stomach tells you that something is changing. That something is your "apparent weight" or "scale weight".

You can't feel the force due to gravity. You can't conduct a local experiment that measures gravity. An example: An accelerometer placed at rest on the surface of the Earth registers an upward acceleration of 9.8 meters/second2. Why? The answer is that the accelerometer does not (and cannot) measure gravity. It measures everything but gravity. In this case, it measures the normal force from the ground that pushes the accelerometer upward. The accelerometer does not sense the downward force from gravitation, so even though the accelerometer is at rest with respect to the Earth it registers an upward acceleration with respect to the Earth.

When NASA and Dr. Tyson speak of the problems such as bone loss due to the zero g environment of space, they are referring to weightlessness (scale weight) rather than gravity.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Hello! I am generating electrons from a 3D gaussian source. The electrons all have the same energy, but the direction is isotropic. The electron source is in between 2 plates that act as a capacitor, and one of them acts as a time of flight (tof) detector. I know the voltage on the plates very well, and I want to extract the center of the gaussian distribution (in one direction only), by measuring the tof of many electrons. So the uncertainty on the position is given by the tof uncertainty...
Back
Top