How fast will 1g get you there?

How fast will 1g get you there?

1.75 hours to the moon!Β 

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

“>𝑑=(1/2)π‘Žπ‘‘2d=(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

“>𝑑=(1/2)π‘Žπ‘‘2d=(1/2)at2.

t=2d/a

“>𝑑=2𝑑/π‘Žβ€Ύβ€Ύβ€Ύβ€Ύβˆšt=2d/a

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

“>π‘Ž=9.8π‘šπ‘’π‘‘π‘’π‘Ÿπ‘ /𝑠𝑒𝑐2a=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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