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Artificial Gravity Equations
by Ron Kurtus (updated 29 May 2023)
Artificial gravity can be established in a spacecraft by employing linear acceleration or using centrifugal force, where the whole spacecraft or parts of it can be continually rotated, creating the effect of gravity, depending on the rate of rotation.
Simple equations are used to calculate the requirements to simulate gravity.
Questions you may have include:
- What linear acceleration is needed to simulate gravity?
- What are the equations for using centrifugal force to simulate gravity
- How fast must a circular space station rotate to simulate gravity?
This lesson will answer those questions. Useful tool: Units Conversion
Linear acceleration relationship
When a spaceship accelerates, the crew can feel a force pressing on them that simulates the force of gravity. That force is the resistance to overcoming inertia and is shown by the equation:
F_{a} = ma
where
- F_{a} is the force pushing on the astronaut in newtons (N)
- m is the mass of the astronaut in kilograms (kg)
- a is the linear acceleration of the spacecraft in meters per second squared (m/s^{2})
For an object or person close to Earth, the force of gravity is:
F_{g} = mg
where
- F_{g} is the force caused by gravity in newtons (N)
- m is the mass of the astronaut in kilograms (kg)
- g is the acceleration due to gravity in meters per second squared (g = 9.8 m/s^{2})
At some acceleration, the force pushing on the astronaut equals the force from gravity:
F_{a} = F_{g}
ma = mg
Thus
a = g
In otehre words, when the linear acceleration of the spaceship is 9.8 m/s^{2}, the astronaut will feel a force similar to that of gravity.
Centrifugal force relationship
Unfortunately, linear acceleration has its limits. It is better to provide the acceleration through circular motion at a constant rate. In other words, you can use centrifugal force to simulate gravity.
A better way to create this artificial gravity than constant acceleration is to use centrifugal force. This outward force is caused by an object being made to follow a curved path instead of a straight line, as dictated by the Law of Inertia.
(See Centrifugal Force Caused by Inertia for more information.)
Centrifugal force equation
When you swing an object around you that is tied to a string, the outward force is equal to:
F = mv^{2}/r
where
- F is the outward force of the object in newtons (N)
- m is the mass of the object in kilograms (kg)
- v is the linear or straight-line velocity of the object in meters/second (m/s)
- r is the radius of the motion or the length of the string in m
Note: 1 N = 1 kg-m/s^{2}
Angular velocity equation
A better way to write the force equation is to use angular velocity, which will then lead to revolutions per minute.
ω = v/r
where
- ω (lower-case Greek letter omega) is the angular velocity in radians per second
- v is the linear velocity in m/s
- r is the radius of the curve in meters
Note: A radian is the distance along a curve divided by the radius
Also:
v = ωr
Substituting for v in F = mv^{2}/r, you get the equation of the centrifugal force as a function of the mass, angular velocity, and radius:
F = mω^{2}r
Centrifugal force and artificial gravity
Since the centrifugal force is F = mω^{2}r and the force due to gravity is F = mg, you can combine the two equations to get the relationship between the radius, rate of rotation, and g:
mg = mω^{2}r
g = ω^{2}r
Solving for ω:
ω = √(g/r)
Also, solving for r:
r = g/ω^{2}^{}
Convert radians per second to rpm
The units for ω are inconvenient for defining the rate of rotation of the space station. Instead of radians per second, it would be better to state the units as revolutions per minute (rpm). Conversion factors are:
1 radian = 1/2π of a full circle (π is "pi", which is equal to about 3.14)
ω radians per second is ω/2π is revolutions per second
ω/2π revolutions per second is 60ω/2π revolutions per minute
60ω/2π = 9.55ω rpm
Let Ω (capital Greek letter omega) be the rate of rotation in rpm.
Ω = 9.55ω rpm
Thus:
Ω = 9.55√(g/r)
and
r = 91.2g/Ω^{2}
Applications
You can use these equations to determine the size of the space station and the rate of rotation needed to simulate artifical gravity.
Rate of rotation example
Suppose the space station had a radius of r = 128 ft. How fast would it have to turn to create an acceleration due to gravity of g = 32 ft/s^{2}?
Ω = 9.55√(g/r)
Ω = 9.55√(32/128) rpm
Ω = 9.55√(1/4) rpm
Ω = 9.55/2 rpm
Ω = 4.775 rpm
Radius example
If you wanted the space station to rotate at only 2 rpm, how many meters must the radius be to simulate gravity?
r = 91.2g/Ω^{2}
r = (91.2)(9.8)/(2^{2}) meters
r = 233.44 m
Summary
Artificial gravity can be established in a spacecraft by employing linear acceleration or using centrifugal force, where the whole spacecraft or parts of it can be continually rotated. The rate or rotation necessary to duplicate the Earth's gravity depends on the radius of the circle.
Simple equations are used to calculate the requirements to simulate gravity.
Be methodical in your methods
Resources and references
Websites
Artificial gravity - Wikipedia
Simulating Gravity in Space - From Batesville, Indiana HS Physics class
Artificial Gravity and the Architecture of Orbital Habitats - Theodore W. Hall - Space Future; detailed technical paper
Artificial Gravity - Technical resources from Theodore W. Hall -
The Physics of Artificial Gravity - Popular Science magazine
Simulated Gravity with Centripetal Force - Oswego City School District Exam Prep Center, New York
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Top-rated books on Simple Gravity Science
Top-rated books on Advanced Gravity Physics
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Artificial Gravity Equations