Excerpt from textbook:

According to Archimedes' law the weight of a body of mass $m$ and density $\rho$ inside air is: $$G=mg\left(1 - \frac{\rho_v}{\rho}\right)$$ Where $\rho_v$ is the density of air.
A "weight" of mass $m_t$ and density $\rho_t$ in air weights $$G_t = m_tg\left(1-\frac{\rho_v}{\rho_t}\right)$$ If using a balance scale one determined $G=G_t$ one would get: $$m=m_t\frac{(1 - \rho_v/\rho_t)}{(1-\rho_v/\rho)}$$ i.e. The mass of the weight and the mass of the body are not equal. In real measurments $\rho_v/\rho_t\ll1$ and $\rho_v/\rho\ll1$ so the previous equation can be approximately written as: $$m=m_t\left(1+\frac{\rho_v}{\rho} - \frac{\rho_v}{\rho_t}\right)$$

How was the last approximation made? I tried deriving the 4th equation starting from the 3rd equation:

$$m- m_t - m\frac{\rho_v}{\rho} + m_t\frac{\rho_v}{\rho_t} = 0$$ Using approximation magic (perhaps the fact $\rho_v/\rho_t\ll1$ and $\rho_v/\rho\ll1$) I may turn $m$ into $m_t$ in the 3rd addend and factor $m_t$ : $$m- m_t - m_t\frac{\rho_v}{\rho} + m_t\frac{\rho_v}{\rho_t} = 0$$ $$m- m_t\left(1 + \frac{\rho_v}{\rho} - \frac{\rho_v}{\rho_t}\right) = 0$$ That gives me the formula I wanted.
But it doesn't make any sense and probably another line of logic was used.

  • 3
    $\begingroup$ Guess you've not heard of the binomial approximation? Direct application of it leads to your solution. $\endgroup$ – Kyle Kanos Aug 6 '17 at 16:45

For x<<1 the following common approximation is true:


So for y<<1 we get:


The -xy component in the last equation is negligible, as it contains a product of two small values.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.