Submitted by starfyredragon t3_zmt3lg in askscience
OathToAwesome t1_j0fgncy wrote
Reply to comment by alukyane in Does rotation break relativity? by starfyredragon
really pedantic note: it's "zero-g" (as in g-forces, the force you feel from acceleration) and not "zero-gravity"
[deleted] t1_j0fvj6l wrote
[deleted]
[deleted] t1_j0geaya wrote
[removed]
rexsilex t1_j0g0sxw wrote
Weightlessness is called all those things. They're interchangeable terms. Though none imply that gravity isn't still working on you
Alis451 t1_j0ghk8i wrote
The Space station is at 0g, but they aren't far enough outside of the Earth's Gravity well to be at Zero Gravity, because if it were it would start sticking itself, Earth's Gravity supersedes your own in relation to other nearby objects. So YES, there is a VAST difference between 0g and Zero Gravity. The space station is in a constant freefall and needs to continuously adjust.
Max-Phallus t1_j0gk2us wrote
If anyone is interested, gravity on the IIS is still 88.33% of earth's gravity. The IIS orbits around 408KM from sea level.
Here is a PowerShell script I wrote that will calculate earth's gravity at any given altitude (if you have windows, you can open it and copy&paste it in):
function Get-GravityAtAltitude
{
[CmdletBinding()]
Param
(
[double]$AltitudeKm,
[switch]$ReturnStats
)
BEGIN
{
$EarthParams = @{
MeanRadius = 6371.009 # KM
SeaLevelGravity = 9.80665 # m/s^2
}
}
PROCESS
{
$AltLevelGravity = $EarthParams['SeaLevelGravity'] * [Math]::Pow($EarthParams['MeanRadius'] / ($EarthParams['MeanRadius'] + $AltitudeKm), 2)
if($ReturnStats.IsPresent)
{
$EarthParams.Add('Altitude', $AltitudeKm)
$EarthParams.Add('AltLevelGravity', $AltLevelGravity)
$EarthParams.Add('GsAtGravity', ($AltLevelGravity / $EarthParams['SeaLevelGravity']).ToString('.00%'))
return [PsCustomObject]$EarthParams
}
else
{
return $AltLevelGravity
}
}
END {}
}
Get-GravityAtAltitude -ReturnStats -AltitudeKm 408
Viewing a single comment thread. View all comments