Q&A

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posted 1y ago by celtschk‭  ·  edited 1y ago by celtschk‭

• If you are accelerating while running on Earth, actually you are also accelerating Earth in the opposite direction. However for a given force, the acceleration is inversely proportional to mass, therefore when some $60\\, m kg$ person accelerates by, say, $1\\, m m/s^2$, then Earth, which has a math of about $6\cdot 10^{24}\\, m kg$, will only accelerate with $10^{-23}\\, m m/s^2$, far to low to actually notice (if accelerating for a second, the change of speed of the Earth will be so that in about three years it moved by the diameter of a proton — provided you kept running at the final speed in the same direction for the whole three years).
• If you are accelerating while running on Earth, actually you are also accelerating Earth in the opposite direction. However for a given force, the acceleration is inversely proportional to mass, therefore when some $60\\, m kg$ person accelerates by, say, $1\\, m m/s^2$, then Earth, which has a mass of about $6\cdot 10^{24}\\, m kg$, will only accelerate with $10^{-23}\\, m m/s^2$, far to low to actually notice (if accelerating for a second, the change of speed of the Earth will be so that in about three years it moved by the diameter of a proton — provided you kept running at the final speed in the same direction for the whole three years).
If you are accelerating while running on Earth, actually you are also accelerating Earth in the opposite direction. However for a given force, the acceleration is inversely proportional to mass, therefore when some $60\\,\rm kg$ person accelerates by, say, $1\\,\rm m/s^2$, then Earth, which has a math of about $6\cdot 10^{24}\\,\rm kg$, will only accelerate with $10^{-23}\\,\rm m/s^2$, far to low to actually notice (if accelerating for a second, the change of speed of the Earth will be so that in about three years it moved by the diameter of a proton — provided you kept running at the final speed in the same direction for the whole three years).