# Question from a book: “If the horse pulls the buggy a foot forward, how far backwards does the earth move?”

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?

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