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How fast will 1g get you there?

1.75 hours to the moon!ย 

The concept of 1 g acceleration providing a comfortable environment for travel is enticing. But even if you have the energy it’s not so clear where you get the propellant throw mass.

Assuming acceleration is constant, d=(1/2)at2

“>๐‘‘=(1/2)๐‘Ž๐‘ก2d=(1/2)at2

. So plotted over time, distance traveled is a nice parabola.

If you want the time it’d take for a specific distance, it’s easy to manipulate

d=(1/2)at2

“>๐‘‘=(1/2)๐‘Ž๐‘ก2d=(1/2)at2.

t=2d/a

“>๐‘ก=2๐‘‘/๐‘Žโ€พโ€พโ€พโ€พโˆšt=2d/a

If you’re using meters and seconds as your units, a=9.8meters/sec2

“>๐‘Ž=9.8๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ๐‘ /๐‘ ๐‘’๐‘2a=9.8meters/sec2

To travel half the distance to the moon would take about 1.75 hours. The other half distance spent decelerating would take the same amount of time.

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