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 Math Help > Physics > Centrifugal Force, Speed of Sound

David asks, is there any truth to the statement in Dan Brown's "Angels and Demons", that flying in a Mach 15 jet at 60,000', one would feel 30% lighter?

### Speed of Sound

Interestingly, the speed of sound in air depends on the temperature, but not the pressure, of the air.  The formula for the speed of sound, a, depends on the medium and the temperature of the medium, according to the formula,

a = sqrt(γRT), where

γ = ratio of specific heats (1.4 for air at STP)
R = gas constant (286 m²/s�/K for air)
T = absolute temperature (273.15+°C)

At 60,000', the temperature is about -56.5�C, so T=216.65�K, so a=sqrt(1.4*286*273.15) = 294.53 m/s = 1060.3 km/h = 658.84 mph

Sticking with meters and seconds, Mach15 = 15*294.53 = 4418 m/s, which is the air speed.

Assuming the air rotates more or less with the Earth (that is, neglecting the effect of winds), then the actual velocity of the jet is increased or reduced by the rotation of the earth, depending on whether the jet is flying toward the East or West, respectively.   Hold that thought.  We'll get back to it shortly.

### Centrifugal Force

The "fictitious" centrifugal force, F, acting on an object in a rotating frame of reference, is given by

F = mΩ²r,

where Ω is the angular velocity, in radians (which are unitless) per second.  The apparent centrifugal acceleration is

a = F/m = Ω�r

If v is the velocity of an object moving in a circle, and r is the radius of the circle, then Ω=v/r, so

a = Ω�r = v�/r

The radius of the earth is about 6,378,100 m, and then 60,000 feet is about 18,288 m, so the radius of the orbiting jet is r=6,396,388 m,  and mach 15, v = 4418 m/s.

The air at 60,000 feet is rotating around the earth once every 24 hours (or 86400 seconds), and so its velocity is 6396388*2*π/86400 = 465 m/s near the equator.

So if the jet is flying toward the east, its velocity is as much as 4418+465=4483 m/s.  And if it's flying toward the west, its velocity is 4418-465=3953 m/s.  So, depending on which way it's flying, and how close to the equator it is.

aE = v�/r = 4883�/6396388 = 3.73 m/s� flying toward the East near the equator, or
aW = v�/r = 3953�/6396388 = 2.44 m/s� flying toward the West near the equator.

### Back to David's question

By convention, the acceleration of gravity at sea level is taken to be 9.80665, and it varies inversely as the square of the distance from the center of the earth, so the gravitational acceleration at 60,000 feet is

g' = 9.80665*(6378100/6396388)² = 9.75

The feeling of gravity aboard David's plane, if it's flying toward the East, is g'-aE = 9.75-3.73 = 6.02 m/s�, which is more than 38% less than g.  But if it's flying toward the West, the feeling of gravity is g'-aW = 9.75-2.44 = 7.31 m/s�, which is only about 25% less than g.

If the jet is flying in some other direction, or farther from the equator, then the feeling of gravity will be somewhere between 0.62 and 0.75 g's

### Internet references

NASA: Speed of Sound

Digital Dutch: 1976 Standard Atmosphere Calculator

Wikipedia: Centrifugal force and Mach (speed)

The webmaster and author of this Math Help site is Graeme McRae.