Урок 63. Перегрузка. Вес тела на полюсе и на экваторе

Scientific concepts and nature phenomena presented

Weight, “apparent weight,” and free fall

• Weightlessness (weight = 0) occurs when a body is in free fall: the support (or suspension) provides no normal force because the support acceleration matches (g).
• Free fall: motion of a body under the influence of gravity alone.
• The key distinction emphasized is between:
  • the direction of acceleration of the support
  • whether the body separates from the support (support accelerates downward) or presses harder on it (support accelerates upward).

Overload (пере-грузка, increased apparent weight)

• When the support accelerates upward, the body’s resultant force is upward.
• Then the reaction force from the support exceeds gravity, so the apparent weight (the normal force) becomes larger than (mg).
• Overload state: a condition where apparent weight is greater than the gravitational force (mg).

• Overload factor (K): [ K=\frac{|P|}{|mg|} ]

  • In overload, (K>1).
  • Biological/medical implications described:
  • High overload can strain bones (via elastic forces), causing injury.
  • Very large overload can affect circulation (blood pressure changes), potentially leading to fainting/loss of consciousness.
  • Mitigation suggestion: being in a horizontal position.

Overload in aviation/cosmonautics (centripetal acceleration and G-forces)

• In a curved flight path (an arc), at the lowest point the body experiences acceleration directed upward toward the center.

• Centripetal acceleration magnitude: [ a=\frac{v^2}{R} ]

  • This produces high G-forces on the pilot.
  • A “red/blank screen” effect in simulators is attributed to reduced blood flow to the brain, described as a “pink veil” effect.
  • A critical maximum overload is mentioned:
  • About 10 (no irreversible consequences, for trained people per the narration).

Body weight at the Pole vs at the Equator (Earth rotation)

Two cases on Earth’s surface:

• At the pole:

  • (Idealized) the body is stationary, so there is no centripetal acceleration from rotation.
  • (a=0) → weight (P=mg)

• At the equator:

  • due to Earth’s rotation, the body has centripetal acceleration toward Earth’s center.
  • the support reaction becomes smaller than (mg).
  • therefore apparent weight at the equator is less than at the pole.

Apparent weight at the equator (as described): [ P_{\text{equator}}=m\,(g-a) ] with centripetal acceleration expressed using angular speed: [ a=\omega^2R,\quad \omega=\frac{2\pi}{T} ] Hence: [ P’ = m\left(g-\frac{4\pi^2R}{T^2}\right) ]

Apparent gravity at the equator: [ g’ = g-\frac{4\pi^2R}{T^2} ] (interpreted as the acceleration relative to Earth’s rotating surface)

Why the difference from a simple sphere-model exists

• The ideal model predicts only the rotation effect.
• Additional differences arise because Earth is not a perfect sphere:
  • described as non-spherical (tangerine-like) and not round
  • polar radius is stated to be about 21 km smaller than the equatorial radius
• Earth shape terminology mentioned:
  • geoid (Earth’s actual gravitational/shape surface), approximated as an ellipsoid of rotation.

Measuring weight / “weighing” methods

Two weighing approaches contrasted:

• Spring scales (dynamometers)

  • Measure normal force; under equilibrium: weight equals spring elastic force.
  • Drawback: the reading depends on local apparent gravity, so results change with rotation/latitude.
  • Conclusion noted: spring scales would read slightly differently at pole vs equator, and would not work correctly on other planets without recalibration.

• Beam scales

  • Compare masses by balancing weights on two pans.
  • In principle, they are more suitable for determining mass, since both sides experience the same overall conditions, even if gravity/acceleration differs locally.

Calculation problem: changing Earth’s rotation rate to halve equatorial weight

• Problem (summarized from the Gelfgat problem book):
  • By what factor must Earth’s rotation speed increase so that bodies at the equator weigh half as much as at the pole?
  • Assumption: Earth’s shape does not change.
• Relations used:
  • At the equator: (P’ = m(g-a))
  • At the pole: (P=mg)
  • Condition: (P’=\frac{1}{2}P)
• With (a=\omega^2R) and (\omega=\frac{2\pi}{T}), a new rotation period (T’) is derived.
• Narration’s conclusion (qualitative/numerical):
  • Earth’s rotation period would need to be about 12 times shorter than 24 hours (roughly 12 hours, with the narration emphasizing a much faster rotation corresponding to the “half-weight at the equator” scenario).

Researchers / sources mentioned (featured)

• Galileo (reference to Galileo’s principle of relativity)
• Foucault (Foucault’s pendulum)
• Gelfgat (problem book author/source: “Gelfgat’s problem book”, problem 1528)
• Andrey (speaker mentioned; not explicitly identified as a researcher)
• Ilya (speaker mentioned; not explicitly identified as a researcher)