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The Schwarzschild metric, describing the exterior gravitational field of a planet of mass $M$ and radius $R$, is given by$$ds^2 = -(1 - 2M/r)\,dt^2 + (1 - 2M/r)^{-1}\,dr^2 + r^2(d\theta^2 + \sin^2\theta \,d\phi^2).$$A tower has its base on the surface of this planet ($r = R$) and its top at radial coordinate $r = R_1$. A ball is held at rest by an observer at the top of the tower. It is then dropped and caught by an observer at the bottom of the tower.

What is the acceleration of the ball before it is dropped, i.e., if the ball has unit mass, what force would have to be exerted on the ball to hold it in place?

Here, we are not assuming that $R \gg 2M$ or that $R_1 - R \ll R$.

Qmechanic
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1 Answers1

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The acceleration should be

$$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$

with $r$ as the height above the center of mass and the Schwarzschildradius

$$r_s = \frac{2\cdot G\cdot M}{c^2}$$

The force to hold the ball at rest is

$$F=m\cdot a$$

As one can see it now takes an infinite force and energy to keep a body at a fixed height when $r=r_s$.

Yukterez
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