Deriving the equation for centripetal acceleration

I'm going to be completely honest about my last post in which I looked at how fast you would have to spin a spaceship to make artificial gravity.  I didn't remember how to calculate centripetal force, and I didn't have my phone or computer or physics books to look up the equation.  I was able to do it on one bar napkin though.

Assuming an object orbits a central point and has its position defined by the functions:

x(t) = r * cos(ωt)

y(t) = r * sin(ωt)

Take the first derivative to get velocity and you have:

vx(t) = x'(t) = -ω * r * sin(ωt)

vy(t) = y'(t) = ω * r * cos(ωt)

Take another derivative for acceleration:

ax(t) = vx'(t) = x''(t) = -ω² * r * cos(ωt)

ay(t) = vy'(t) = y''(t) = -ω² * r * sin(ωt)

So now we can just use the Pythagorean Theorem (x² + y² = r²) to find the magnitude of centripetal acceleration.  Rearranging:

a = √(ax² + ay²) 

It looks like it's going to get hairy with all the squaring and square rooting, but a quick substitution will simplify it. Substitute and multiply out:

a = √( (-ω² * r² * cos(ωt) )² + (-ω² * r² * sin(ωt) )² ) 

a = √(   ω⁴ * r² * cos²(ωt)   +   ω⁴ * r² * sin²(ωt)   )

Divide out ω⁴ * r²:

a = √(  ω⁴ * r² * ( cos²(ωt) + sin²(ωt)  )

a = ω² * r * √( cos²(ωt) + sin²(ωt) )

Remember the property of sin²ϑ + cos²ϑ = 1?  If we just apply that, we are left with the equation of a = ω² * r.  Derivation complete!!

Making fake gravity on the way to Mars

The New York Times had an article this week "Squashed Eyeballs Are a Danger for Astronauts" discussing how eyeballs deform in zero gravity and the complications it poses for long-term space missions, such as a two year trip to Mars.  It discusses rotating spaceships to produce artificial gravity.

My question is this: how fast would a spaceship have to rotate to simulate normal Earth gravity?  I grew up reading and watching a lot of sci-fi where this was done, such as in Babylon 5:

2001: A Space Odyssey:

I always accepted that rotation would work to provide artificial gravity, but never actually looked at the numbers involved with it.  Next time I watch some B5 I'll certainly be taking a stopwatch to the rotation of the space station.

We'll start with the equation for centripetal acceleration.

a = ω² * r 

Where ω is the angular velocity (speed of rotation) and r is the radius of the spaceship.

Centripetal force is that force which pulls on a rotating object to keep it in its curved trajectory.  If you were to swing a yo-yo around in the air over your head, the force you would pull on the string with would be centripetal force.  An equal force, centrifugal force, is felt pushing outward, and that would be perceived as gravity by the crew of the ship.

Since we already know the value of acceleration we want to simulate (9.8 m/sec² or 32 ft/sec²), we're going to rearrange the equation to be in terms of ω:

        a 

ω = √ (---)

        r    

I want my answer in rpm (revolutions per minute), so I'm going to add some conversion factors.  This equation will solve for ω in terms of radians per unit time, so I need to convert radians into revolutions (2π radians per revolution).

        a     60 sec      rev

ω = √ (---)*(--------)*(--------)

        r      min       2π rad

Now to plug some numbers in:

r (ft)	ω (rpm)
100	5.4
300	3.12
250	3.42
500	2.42
2640	1.05
5280	0.74

 or in metric:

r (m)	ω (rpm)
50	4.23
100	2.99
250	1.89
500	1.34
1000	0.95
5000	0.42

So all of these are pretty fast.  If I had a spaceship that was 1 mile across (r = 2640 ft), I would need to spin it around at just under one rpm to simulate normal gravity.

A good read I'm halfway through is Mary Roach's Packing for Mars, which takes a light-hearted look at some of the obstacles to tackle with space travel, specifically those relating to human limitations.  I had previously read her first book Stiff: The Curious Lives of Human Cadavers, which was also a great read.