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Apparent Weight on Earth

Level 1 - Physics topic page in Gravitation.

Principle

Apparent weight on Earth is the scale reading, so it is the contact force needed after Earth's rotation has supplied the required centripetal acceleration.

Notation

\(m\)
person or object's mass
\(\mathrm{kg}\)
\(N\)
normal reaction or scale reading
\(\mathrm{N}\)
\(g\)
gravitational field magnitude near Earth's surface
\(\mathrm{N\,kg^{-1}}\)
\(\omega\)
Earth's angular speed
\(\mathrm{rad\,s^{-1}}\)
\(R\)
Earth's radius
\(\mathrm{m}\)
\(\lambda\)
latitude
rad or deg
\(r_\perp\)
distance from the rotation axis
\(\mathrm{m}\)

Method

Derivation 1: Find the rotation radius

A point on Earth's surface moves in a circle about the rotation axis, not about the local vertical. At latitude \(\\lambda\), the circular-motion radius is the perpendicular distance to the axis.

Axis distance
\[r_\perp=R\cos\lambda\]
Centripetal acceleration magnitude
\[a_c=\omega^2r_\perp=\omega^2R\cos\lambda\]

Derivation 2: Extract the local vertical effect

Only the component opposite the local upward direction changes the scale reading. For a spherical-Earth model, that upward component is reduced by another factor of \(\\cos\\lambda\).

Upward component of the rotation effect
\[a_{c,\mathrm{up}}=\omega^2R\cos^2\lambda\]
Balance along the local vertical
\[N= m\left(g-a_{c,\mathrm{up}}\right)\]
Latitude model
\[N=m\left(g-\omega^2R\cos^2\lambda\right)\]
This ignores Earth's small equatorial bulge and the fine detail of local gravity variation.

Derivation 3: Check the extreme cases

The formula is easiest to interpret at the equator and the poles.

Equator
\[N_{\mathrm{eq}}=m\left(g-\omega^2R\right)\]
Poles
\[N_{\mathrm{pole}}=mg\]
Difference
\[N_{\mathrm{pole}}-N_{\mathrm{eq}}=m\omega^2R\]

Earth's rotation therefore makes the scale reading slightly smaller at low latitude, with the largest reduction at the equator.

Rules

These are the compact results from the method above.

Latitude radius
\[r_\perp=R\cos\lambda\]
Latitude model
\[N=m\left(g-\omega^2R\cos^2\lambda\right)\]
Equator
\[N_{\mathrm{eq}}=m\left(g-\omega^2R\right)\]
Poles
\[N_{\mathrm{pole}}=mg\]

Examples

Question
Estimate the equatorial scale reading for a
\[70\,\mathrm{kg}\]
person using
\[g=9.81\,\mathrm{N\,kg^{-1}}\]
\[\omega=7.29\times10^{-5}\,\mathrm{rad\,s^{-1}}\]
and
\[R=6.37\times10^6\,\mathrm{m}\]
Answer
\[N_{\mathrm{eq}}=m\left(g-\omega^2R\right)\]
\[\omega^2R=(7.29\times10^{-5})^2(6.37\times10^6)=3.39\times10^{-2}\,\mathrm{m\,s^{-2}}\]
\[N_{\mathrm{eq}}=70(9.81-0.0339)=6.84\times10^2\,\mathrm{N}\]
The reduction is about
\[2.4\,\mathrm{N}\]
compared with \(mg\).

Checks

  • Apparent weight is the normal reaction \(N\), not the gravitational force \(mg\).
  • The rotational reduction is largest at the equator and vanishes at the poles.
  • The correction is small compared with \(g\), so apparent weight on Earth is close to but not exactly \(mg\).
  • A scale reading can change even when the object's mass does not.

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Spherical Masses

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Black Holes