Why might exposing astronauts to weightlessness be harmful?

1 answer:

5 0

The body doesn't have to work as hard when there's no gravity for it to work against, so it becomes accustomed to a much lower work load on every level. It leads to lower bone mass and weaker muscles, including the heart, leading to a drop in blood pressure that can eventually build up to create problems with cognitive function. After so long, minor accidents can lead to major, even life threatening problems. A simple bump that would do little more than leave a bruise on you and I can result in a broken femur bone or broken neck on an astronaut who has been exposed to a weightless environment for too long.

This is one of the several hurdles that must be overcome in order for a manned mission to Mars to succeed. Exposure to a weightless environment on the order of roughly two years for a manned Mars mission would be so degrading to the body that the rough, turbulent re-entry into Earth's atmosphere might prove to be too violent for an astronaut to survive.

The problem is bones.

On Earth, every time you do something with "impact" (like walking), there are microcracks in your bones. Calcium is used by the body to fix these cracks... and that is how the bones grow and become strong.

No weight = no impact = no cracks = no "repairs" being done by the body = the body gets rid of un-neede calcium and bones become brittle and weak.

There are some other operations in the body that require gravity as a "director", or resistance to movement as a driver of change (think of muscles in the legs, when there is no need to walk).

The organ themelves are (generally) OK since many things can work in any orientation.

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2). If you record a wave going 4m/s and the Period of the wave was<br> 60s, how long was the wave?

Taya2010 [7]

Answer:240 meters

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3 years ago

A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment wit

xenn [34]

To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.

Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by

$z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}$

Where,

$\epsilon_i =$ energy of that state

k = Boltzmann's constant

T = Temperature

With our values we have that

T= 250K

$k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}$

$\epsilon_1=0J$

$\epsilon_2=1.6*10^{-21}J$

$\epsilon_3=1.6*10^{-21}J$

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

$\beta = \frac{1}{kT}$

$\beta = \frac{1}{(1.381*10^{23} m^2)(250)}$

$\beta = 2.9*10^{20}J$

Therefore the average energy would be,

$\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}$

Replacing with our values we have

$\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}$