Key 1

Up a level : The Special Theory of Relativity - an introduction
  1. We have

\displaystyle a=\frac{{{{v}^{2}}}}{r}=\frac{{{{{465}}^{2}}}}{{6378000}}=0.0034\text{ m/}{{\text{s}}^{2}}

    1. This is about 0.0034/9,8=0.34% of the free fall acceleration.
    2. I would say no, and that centripetal acceleration is already built into the free fall acceleration value at the place you are. It’s one of the two reasons why the free-fall acceleration is lower at the equator than further away from it. The other reason is the longer distance from the centre of the Earth at the equator than at the poles.
  2. 1 AU =1.5·1011 m.

\displaystyle v=\frac{s}{t}=\frac{{2\pi \cdot 1.5\cdot {{{10}}^{{11}}}}}{{365.24\cdot 24\cdot 60\cdot 60}}=29900\text{ m/s}

so,

\displaystyle =\frac{{{{v}^{2}}}}{r}=\frac{{{{{29900}}^{2}}}}{{1.5\cdot {{{10}}^{{11}}}}}=5.93\cdot {{10}^{{-3}}}\text{ m/}{{\text{s}}^{2}}

So, virtually nothing. After all, a full rotation takes a year.

Up a level : The Special Theory of Relativity - an introductionLast modified: Sep 23, 2025 @ 09:19