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Physics · Ch 6 — Gravitation

Weightlessness

6.4.4

Weightlessness

Your everyday sense of "weight" and the actual gravitational force pulling on you are subtly different things, and understanding that difference explains the jerk you feel starting or stopping in a lift, and why astronauts float freely in orbit.

Defining weight. The gravitational force on a mass mm is mgmg, always pointing straight down. But your weight WW is formally defined as the magnitude of the upward force that must be applied to hold you at rest (or at constant velocity) -- in ordinary situations this equals the normal reaction force NN from whatever surface you are standing on, so W=N=mgW=N=mg. Weight and gravitational force happen to have the same numerical value at rest, but they are conceptually different quantities.

Apparent weight in a lift. Consider a person of mass mm standing on a scale inside a lift, with gravity FG=−mgj^F_G=-mg\hat{j} acting down and the scale's normal reaction Nj^N\hat{j} acting up.

  • At rest, or moving at constant velocity: net acceleration is zero, so N−mg=0⇒N=mgN-mg=0 \Rightarrow N=mg (Eq. 6.67) -- apparent weight equals true weight.
  • Accelerating upward with acceleration aa: applying Newton's second law along the vertical gives N=m(g+a)N=m(g+a) (Eq. 6.68) -- apparent weight is greater than true weight (the familiar "heavy" feeling as a lift starts moving up).
  • Accelerating downward with acceleration aa: similarly, N=m(g−a)N=m(g-a) (Eq. 6.69) -- apparent weight is less than true weight (the "light" feeling as a lift starts moving down).
  • Free fall (a=ga=g, e.g. the lift cable is cut): N=m(g−g)=0N=m(g-g)=0. The scale reads exactly zero -- this is the state of weightlessness. …
Figure 6.23Apparent weight of a person in a lift, four cases

What this figure shows. Four small panels (a)-(d) each show a person standing on a weighing scale inside a lift. In (a) the lift is at rest or moving at constant velocity, and the scale reads the true weight N = mg. In (b) the lift accelerates upward, and the scale reading N = m(g+a) is drawn larger than mg. In (c) the lift accelerates downward, and the reading N = m(g-a) is drawn smaller than mg. In (d) the lift is in free fall (cable cut, a = g), and the scale reads exactly zero, depicting the state of …