Kvant

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(outside the ball) acceleration a=−a0(rr0)2a = -a_0 \left( \frac{r}{r_0} \right)^2.

The negative sign of the acceleration means that it is always directed towards the center of the ball.

When a ball moves without acceleration, its center is the point of equilibrium. For small deviations, a restoring force occurs. When accelerating the ship ww, we change a0a_0, there is always a cabin position at which the free fall acceleration is equal to ww. To find this position on the graph, draw a straight line below the rr axis at a distance ww. Two intersection points will appear, their coordinates give the required distances. The point inside the ball corresponds to the source equilibrium. With a small change in the acceleration of the ship, the cabin will oscillate around the equilibrium point, itself monitoring the consistency of the accelerations, and all this time the cabin will remain inert.

When braking with the same acceleration, the equilibrium point will be on the other side of the center of the ball.

We have proposed a radical way to combat overloads. What “ifs” stand in the way of realizing this fantastic project?

It does not contradict the laws of nature. The main difficulties lie in the power source: this is a common difficulty of any intercity ship project. But where did the massive body come from? You can either take a body of very large volume or use a substance of very high density. To make it easier to think about this, understand the exercises below.

Exercises

  1. If a jet plane dives with an acceleration of 1.5 g, what is the weight of the pilot and where is the force of the weight directed?
  2. Find the relative radius and mass of a ball of the same density as the Earth to the radius and mass of the Earth if the acceleration on the surface of the ball is 1000 g.
  3. What if the density of the ball is 10^6 times greater than that of the earth? (Under some physical conditions, for example, on stars called white dwarfs, the density of matter significantly exceeds our values.)