What would be some benefits of living on a planet with less surface gravity

2 answers:

4 0

This is a tricky one.

-- IF I evolved on that planet, then my body is perfectly matched to the
conditions and the environment over there, and I am approximately just
as happy and care-free la de da over there as I really am on Earth,
and I can't imagine living anywhere else.

-- IF I evolved on Earth, and I move this body to a planet with less
surface gravity, then there are all kinds of complications.

. . . Even if the atmosphere over there has the same gases in the
same ratios as Earth's atmosphere, the lower gravity means that the
atmosphere has lower weight. In turn, that means that the surface
pressure is lower, and I have to work harder to take in enough oxygen to
keep my Earth body functioning. Breathing alone could be a full-time job.

. . . The acceleration of gravity is less than 9.8 m/s² over there.
This means that when I drop something, it falls slower than I'm used to,
and I can usually grab it before it hits the floor.
When I lift something with the normal force of my Earth arm, it jumps up
faster than I expect. Until I get used to things, I'll probably overshoot,
lift things too high and too fast, maybe spill the coffee etc.

. . . With my Earth muscles, almost everything takes less force.
It's a lot easier to walk around. I bob up and down more than usual,
and my steps are longer. When I run, it's really funny. Every time
I take a step, I sail several feet into the air, and come down farther
than I ever did, but it all happens in slow motion.

All of this is fun while it lasts, but it doesn't last long. My Earth body
gets adjusted to the new planet. Before long, I actually lose muscle mass
AND bone, so that if I were to go back to Earth, I would be so weak that
I'd have to be carried around for a while.

We know this from what happens to the astronauts who spend 6 months
or a year in zero G on the International Space Station. Even when they
do resistance exercises for an hour a day, they come back with less
muscle strength and less bone mass.

zepelin [54]3 years ago

3 0

You could jump high!

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A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

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For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. 

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2)

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m

Answer:

1.925m

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