# Gravity and centripetal acceleration on a space station

Okay, so I’m standing on a space station that is a rotating toroid so I experience One gravity under my feet. I’m holding a coffee cup and let it drop. Isn’t its drop going to follow a V*cos function? Unless the space station is really small it’s going to look like it accelerates at one G, but what is the force it experiences on impact?

It seems that that force won’t be the same as if it were truly accelerating. Now add a little push, in the direction of spin, and anti-spin. Won’t the behavior vary?

I’m writing a science fiction book and would like to get it right.

Robert blanton

To answer this question, it is important to emphasize the difference between gravity per se and centripetal acceleration.

For some gory detail on how gravity and centripetal acceleration interplay on the Earth, see Gravity and centripetal acceleration on Earth. TL:DR; gravity is king, centripetal acceleration can mostly be ignored.

On the space station, the situation is exactly reversed. Centripetal acceleration is king, and gravity can mostly be ignored.

Let’s define a “day” as the amount of time it takes for our space station to undergo a single complete rotation. We’ll also define t as the length of the “day”, in seconds.

Let r be the radius of the space station, in meters.

We can manipulate the formula a = v2/r to relate the length of the day and the radius of the space station. Start with a = v2/r and substitute both a = 9.8 m/s2 (which is what humans on Earth are used to) and v = 2πr/t. Then do some manipulation to get:

9.8t2 = 4π2r

The radius is proportional to the square of the orbital time. That is, smaller space stations need to spin faster. The bigger the space station, the longer the day.

For story purposes, if we want a particular size, we can calculate the required length of the day. Solving for t:

t = √(4π2r/9.8) = 2.0√r

For example, a space station of radius 150 meters would need to have a day of 24 seconds; a space station of radius 1,500 meters would need to have a day of 77 seconds.

Or if we want a particular day length, we can calculate the required size. Solving for r:

r = 9.8t2/(4π2) = 0.25t2

So if we want a day to last 60 seconds, we need a space station of radius 900 m; if we want a day to last 120 seconds, we need a space station of radius 3.6 km.

The acceleration varies as you wander throughout the space station. The closer you are to the rotational axis, the weaker the effect. So things near the ceiling will be accelerated slightly less than things on the floor.

The variation will be minimal if the hole in the toroid is large; that is to say, if the ceiling and the floor are both quite far from the rotational axis.

So if the space station is large, the coffee cup will behave almost exactly as it does on the Earth. If you add a little push, the coffee cup will follow a parabolic path to the floor. If the floor is hard, the cup will break.

If the space station is small, and you had a fairly precise scale, you would notice that the cup weighs less on the top shelf than it does on the floor.

Just for fun, suppose that in addition to the inhabited toroidal part of the space station, there was a service area in the center, on the rotational axis. People would be weightless here.

Perhaps the service area is connected to the toroidal part by spokes, so that it can be reached without going outside. You would experience a gradual transition between full Earth gravity and complete weightlessness as you went along the spoke. Perhaps you have a ladder on the spoke which has rungs quite close together near the edge, where there is full gravity; and as you ascend toward the hub, the rungs get farther apart, as you move from climbing, to jumping, to swimming, to floating.

# Gravity and centripetal acceleration on Earth

Earth accelerates you in two different directions at once. Earth’s gravity pulls you one way, and its rotation pushes you another. We consider the direction, magnitude, and variation of these forces at various points on the Earth’s surface.

## Gravity

Direction: gravity pulls you toward the center of the Earth.

Magnitude: if you are on or above the Earth’s surface, the pull’s strength follows the “Acceleration due to Earth’s gravity” formula from “Formulas and facts” below.

Variation: the pull’s strength is inversely proportional to the square of your distance from the center of the Earth, so long as you’re on or above the surface. The closer you get to the center, the stronger gravity is; the further away you get from the center, the weaker gravity is.

• The Earth is shaped like a sphere. Well… not quite like a sphere. It’s flattened a bit, like an M&M. The “M”s are printed at the North and South Poles, and the Equator goes around the edge.
This means that when you’re at the Poles, you’re closer to the center of the Earth than you are at the Equator. So gravity is stronger at the Poles, and weaker at the Equator.
• Also, but to a much lesser degree, the Earth’s surface is uneven. There are bumps (mountains) and crannies (valleys.)
At a given latitude, when you’re at the bottom of a valley, you’re closer to the center of the Earth than you are at the top of a mountain. So gravity is weaker on mountains. (It is tempting to say that gravity is stronger in valleys, but valleys are a little complicated, because the higher land around you pulls you in the opposite direction than the rest of the Earth. It is true to say that gravity is stronger on low-laying plains, though.)
• The strongest gravity is at the North Pole, which is pretty much at sea level. (The South Pole is on a plateau.)
• The weakest gravity is at the peak of Mt. Chimborazo in Ecuador.
• What about Mt. Everest, you say? Well, Mt. Everest is the highest mountain above sea level, but the sea follows the M&M shape, so Mt. Chimborazo is further from the center of the Earth.

## Centripetal acceleration

Direction: centripetal acceleration pushes you away from Earth’s axis of rotation.

If you’re at the Equator, this is straight up; if you are at the North Pole or South Pole, the only effect is to make you very slightly dizzy; and if you are somewhere in between, you can get a basic feel for the direction by looking at the nearest satellite dish.

You can get a more precise direction by knowing your latitude and compass heading:

1. Point an arrow straight up (that is, away from the center of the Earth.)
2. If your latitude is zero, you’re on the Equator. You’re done.
3. Bend it down, in the direction of the Equator, by an angle equal to your latitude.
• If you are in the Northern hemisphere, bend it South.
• If you are in the Southern hemisphere, bend it North.
4. The final direction of the arrow is the way centripetal acceleration is pushing you.

Magnitude: the pull’s strength follows the “Acceleration due to Earth’s rotation” formula.

Variation: the pull’s strength is proportional to the square of your distance from Earth’s axis of rotation. The closer you get to the axis, the weaker centripetal acceleration is. The further you get from the axis, the stronger centripetal acceleration is.

In particular, centripetal acceleration is strongest at the Equator. It is weakest at the Poles. (In fact, it is zero at the Poles.)

## Putting the two together

Direction: gravity is much stronger than centripetal acceleration, so the sum of the two vectors always points almost completely in the direction of the gravity vector.

Magnitude: the pull’s strength follows the “Combined acceleration due to Earth’s gravity and rotation” formula. The angle between the two vectors is always between 90° and 180°, so the forces counteract each other; the total acceleration is always a little bit less than gravity alone (or, in the case of the Poles, exactly equal to gravity alone.)

Variation: because of the M&M shape, centripetal acceleration (which counteracts gravity) is strongest in the same places where gravity is already weak – the Equator, and mountains. So centripetal acceleration just exaggerates the weakness of already-weak gravity.

Here’s a table. I throw in La Rinconada, in Peru, because it’s the highest permanently inhabited place, and it’s also quite near the equator. The table has the strongest combined acceleration at the top and the lowest at the bottom.

In conclustion, a varies across the Earth from 9.769 m/s2 to 9.863 m/s2, which is a variation of about 0.95%. About 0.84% of this comes from variation in gravity, and the other 0.11% comes from variation in centripetal acceleration.

## Facts and formulas

• Newton’s second law of motion
F is the force applied to an object
m is the mass of the object
a is the acceleration experienced by the object
F = ma
• Gravitational constant
G = 6.674e-11 m3/(kg s2)
• Earth mass
M = 5.9722e24 kg
• Earth radius (at sea level)
• At the North Pole or South Pole = 6.357e6 m
• Mean = 6.371e6 m
• At the Equator = 6.378e6 m
• Acceleration due to Earth’s gravity
G is the gravitational constant
M is Earth’s mass
r is the distance of the object from the center of the Earth
ag = GM/r2
• Centripetal acceleration
v is the speed of an object orbiting a fixed point
r is the radius of the orbit
t is the time to complete a single orbit; note that v = 2πr/t
a is the acceleration experienced by the object
a = v2 /r = 4π2r/t2
• Earth rotation
• Nominal time for one solar rotation: 24 * 60 * 60 = 8.6400e4 s
• Actual time for one solar rotation: a teensy bit more due to leap seconds
• Actual time for one sidereal rotation: T = 8.6164e4 s
• Acceleration due to Earth’s rotation
r is the distance between the object and the Earth’s axis of rotation (not the center of the Earth)
T is the time for one sidereal rotation
ar is the acceleration experienced by the object
ar = 4π2r/T2
a is a vector
b is another vector
θ is the angle between a and b when placed tail-to-tail
|a + b|2 = |a|2 + |b|2 + 2|a||b|cos θ
• Combined acceleration due to Earth’s gravity and rotation
ag is the acceleration due to Earth’s gravity
ar is the acceleration due to Earth’s rotation
L is the latitude of the object, in degrees; the angle of the two vectors in radians is π – (πL/180)
a is the combined acceleration
|a|2 = |ag|2 + |ar|2 + 2|ag||ar|cos (π – (πL/180))

# Why is 1 Pascal equal to 94 dB Sound Pressure Level? (1 Pa = 94 dB SPL)

Last time we talked about why a full-scale digital sine wave has a power measurement of -3.01 dB FS (Spoiler: because it’s not a square wave.)

This time we’ll discuss why an atmospheric sound which generates a root-mean-square pressure of 1 Pascal has a power measurement 94 dB SPL.

As before, dB is defined as 10 log10(PA2 / PB2) where PB is a reference level.

Before, we had a digital measurement with an obvious ceiling: sample values of -1 and 1. So the reference point 0 dB FS was defined in terms of the signal with the greatest possible energy.

In the analog domain, there isn’t an obvious ceiling. We instead consider the floor – the quietest possible signal that is still audible by human ears.

This is a rather wishy-washy definition, but the convention is to take PB = 20 μPa = 0.00002 Pa exactly.

So our 0 dB SPL reference point is when PA = PB: 0 dB SPL = 10 log10(0.000022 / 0.000022) = 10 log10(1) = 10 (0) = 0.

What if the pressure level is 1 Pascal? This is a quite loud sound, somewhere between heavy traffic and a jackhammer.

1 Pa in dB SPL =

10 log10(12 / PB2) =

20 log10(1 / PB) =

-20 log10(PB) =

-20 log10(2(10-5)) =

-20 (log10 2 + log10 10-5) =

-20 ((log10 2) – 5) =

100 – 20 log10 2 ≈ 93.9794 dB SPL

So 1 Pa is actually a tiny bit less than 94 dB SPL; it’s closer to 93.98 = (100 – 6.02) dB SPL.

# Weighing the Sun and the Moon

In an earlier post I mentioned how the Cavendish experiment allowed us to weigh the Earth – to determine the mass of the Earth mE.  Newton knew the acceleration due to gravity on the surface of the Earth and was able to use that to find the product G mE; Cavendish determined G directly, and was thus able to solve for mE.  He would also have been able to find the mass of the sun as follows:

mE aE = G mE mS / rE2

G, rE, and aE = vE2 / rE are known, so we can solve for mS.

But calculating the mass of the moon is trickier.

Once we were able to put a satellite around the moon we could measure its orbital radius and speed, deduce the acceleration, and use that plus the known G to calculate the mass of the moon.  But prior to that we were limited to techniques like:

The moon does not exactly orbit the Earth, but instead orbits the center of mass of the moon/Earth system.  By careful observation we can determine where this center of mass is.  We can then measure the distance between the center of mass and the Earth’s center.  This plus the known mass of the Earth and the distance of the Earth from the Moon allows us to determine the mass of the Moon.

If we’re lucky enough to see a foreign object come close to the moon, we can determine how much it is accelerated by the Moon.  This will allow us to determine the mass of the Moon using the technique above.  (We won’t be able to determine the mass of the foreign object, but we don’t need it.)

When the USSR launched Sputnik, American scientists really wanted to know what its mass was.  But because none of the techniques above were useful, they were unable to determine it.

# Newton: combining celestial and terrestrial mechanics

For as long as there have been humans, we have looked at the skies – and around us on Earth as well.  Things move in the skies; things move on Earth.  The way things move in the skies is called “celestial mechanics”; the way things move on Earth is called “terrestrial mechanics.”

The ancient Greeks thought there were five elements, corresponding to the five regular (Platonic) solids: the four terrestrial elements (air, fire, earth, and water) and the fifth celestial element which the Romans called quintessence.

All kinds of rules were discovered/proposed for how things move on Earth.  Galileo demonstrated that falling objects accelerate at the same rate regardless of their mass.  Similar rules were discovered/proposed for how things move in the heavens.  Kepler demonstrated that the planets move in ellipses around the sun, that they swept out equal areas in equal times, etc.  But everybody knew that things in the heavens were different than they were on Earth.

Then Newton, building on Hooke, came up with the crazy idea that perhaps things fall on Earth for the same reason that heavenly objects go in ellipses… that all objects, celestial or terrestrial, act on each other from a distance in a uniform fashion.  (He himself wrote that this was such a crazy idea that no-one should take it seriously.)

That one body may act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action and force may be conveyed from one another, is to me so great an absurdity that, I believe, no man who has in philosophic matters a competent faculty of thinking could ever fall into it. — Isaac Newton

But he had done some calculations and it seemed to work out… specifically, he worked out how much acceleration you would expect the Earth to exert on an object as far away as the Moon, and how much acceleration the Moon would need to stay in its orbit (and not wander off or spiral down into the Earth.)

I… compared the force requisite to keep the moon in her orb with the force of gravity at the surface of the earth and found them to answer pretty nearly. — Isaac Newton

Note Newton’s use of the word “orb” to describe the motion of the moon, a direct reference to celestial mechanics.

Let’s see how compelling those calculations were, shall we?

His first idea was that there was this thing called “force” which was equal to the mass of an object multiplied by the acceleration it was undergoing.  F = m a.  Fine.

His second idea (which he stole, at least partially, from Hooke) was that any two objects have a force that attracts them, which is proportional to both of their masses, and inversely proportional to the square of the distance between them (that’s the part he stole from Hooke.)  F = G m1 m2 / r2.  Fine.

His third idea was that an object that goes around a circle of radius r at a constant speed v is undergoing a constant centripetal acceleration of a = v2 / rThis follows pretty quickly from the definitions with a little geometry, trigonometry, and a dash of calculus.  Fine.

He also knew some facts.  I’m going to state them in metric because, hey, this is the twenty-first century.  He knew:

• The distance from the center of the Earth to the surface (where the apple trees are) is 3,960 miles: 6370 km.
• Falling objects on the surface of the Earth accelerate at 32.2 ft/s2: 9.81 m/s2.
• The distance from the center of the Earth to the center of the Moon is 239,000 miles: 384,000 km.
• The Moon takes 27.3 days to make a single sidereal orbit around the Earth: 236,000 s. Coarse estimate time…

If the Moon is roughly 50 times further away from the center of the Earth than an apple is, and “gravitational” acceleration is inversely proportional to the square of the distance between two objects, then the Moon should be accelerated roughly 1/2500th as much as the apple is: 0.004 m/s2 instead of 10 m/s2,

More precisely…

Consider the apple case (mE is the mass of the Earth, mA the mass of the apple:)

FA = mA aA = G mE mA / rA2
a
A rA2 = G mE
9.81 m/s2 (6.37e6 m)2 = G mE
G
mE = 3.98e14 m3/s2

Now consider the moon case:

FM = mM aM = G mE mM / rM2
aM = G mE / rM2
aM = 3.98e14 m3/s2 / (3.84e5 m)2
aM = 0.00269 m/s2

So Newton’s proposed formula predicts that the Earth’s gravity causes the Moon to accelerate at 0.00269 m/s2.  This is, indeed, roughly 0.004 m/s2 so it jives with our coarse estimate.

Let’s see how that compares to the necessary centripetal acceleration to achieve a closed orbit: a = v2 / r.

The distance the moon travels in its sidereal orbit is 2π * 3.84e5 m = 2.42e9 m, and it does so in 27.3 days; so its velocity is 2π * 3.84e5 m / (27.3 days * 24 (hours / day) * 60 (minutes / hour) * 60 (s / minute)) = 1020 m/s.  This is roughly Mach 3 in our atmosphere.

The centripetal acceleration is thus measured to be (1020 m/s)2 / 3.84e8 m = 0.00273 m/s2. Newton’s prediction is about 1.5% off, which isn’t bad, considering.

Note that although Newton knew the value of the product G mE, he didn’t know either G or mE seperately.  It wasn’t until Cavendish took two known masses and measured the gravitational pull between them directly that G was evaluated; we could then calculate the mass of the earth as G mE / G.

Once G was known we could also evaluate the mass of the Sun based on Earth’s orbit.  However, we still could not determine the mass of the Moon… in fact, back in the 1950s, American scientists were unable to determine the mass of Sputnik (which would have been very helpful) even though we could see its orbit.

Which raises the question… how did we know the mass of the Moon?

# Deriving the centripetal acceleration formula

A body that moves in a circular motion (of radius r) at constant speed (v) is always being accelerated.  The acceleration is at right angles to the direction of motion (towards the center of the circle) and of magnitude v2 / r.

The direction of acceleration is deduced by symmetry arguments.  If the acceleration pointed out of the plane of the circle, then the body would leave the plane of the circle; it doesn’t, so it isn’t.  If the acceleration pointed in any direction other than perpendicular (left or right) then the body would speed up or slow down.  It doesn’t.

Now for the magnitude.  Consider the distance traveled by the body over a small time increment Δt: We can calculate the arc length s as both the distance traveled (distance = rate * time = v Δt) and using the definition of a radian (arc = radius * angle in radians = r Δθ) The angular velocity of the object is thus v / r (in radians per unit of time.)

The right half of the diagram is formed by putting the tails of the two v vectors together.  Note that Δθ is the same in both diagrams. Note the passing from sin to cos is via l’Hôpital’s rule.

QED.