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The following quote is from Epstein's book "Thinking physics". It is part of the answer to the classical horse and buggy problem:

If the horse pulls the buggy a foot forward, how far backwards does the earth move? Suppose the buggy weighs 1000 pounds. The mass of the earth is $10^{22}$ times greater. So the planet moves a $10^{22}$th part of a foot backwards, which is hard to notice!

How does he calculate this? If I start with momentum conservation and use $F=ma$, I get $$ \frac{a_\text{horse+Earth}(t)}{a_\text{buggy}(t)} = - \frac{m_\text{buggy}}{m_\text{horse+Earth}} $$

How do I get a distance from this?

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Using F=ma is the correct; due to newton's second law the force exerted on the buggy and horse by the earth is the same as the force exerted by the buggy and horse on the earth. The time over which this force acts is also the same for all objects since if the horse stops pushing the earth will stop pushing back. A this means that the impulse which is equal to the force multiplied by the time is the same for both objects. aka $$M_{earth}\left(\frac{S_{earth}}{t}\right)=M_{horse+buggy}\left(\frac{S_{horse+buggy}}{t}\right)$$

Which is rearranged to show that $$\frac{M_{earth}}{M_{horse+buggy}}=\frac{S_{horse+buggy}}{S_{earth}}$$ therefore $$10^{22}=\frac{1 foot}{S_{earth}}$$ or $$S_{earth}=\frac{1}{10^{22}}^{th}$$ of a foot

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  • $\begingroup$ oh my bad I didn't realise it was only the buggy that was 1000 pounds, that does sound pretty heavy though $\endgroup$ – cal Dec 26 '17 at 0:05
  • $\begingroup$ Ah, ok, so you are neglecting the acceleration phase and deceleration phase and treat it as a uniform motion during which the buggy moves one foot. Yes, this may be what the author had in mind. $\endgroup$ – Marc Dec 27 '17 at 14:04

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