Q.Using $P=1.0\times10^{5}\ \text{Pa}$, $\rho=1.293\ \text{kg m}^{-3}$, and $\gamma=1.4$ for air, calculate

Given: $P=1.0\times10^{5}\ \text{Pa}$, $\rho=1.293\ \text{kg m}^{-3}$, $\gamma=1.4$.

1. Newton's formula:

   $v_N = \sqrt{\frac{P}{\rho}} = \sqrt{\frac{1.0\times10^{5}}{1.293}} = \sqrt{77338.7} \approx 278.1\ \text{m s}^{-1}$

2. Laplace's formula: since $v_L=\sqrt{\gamma P/\rho}=v_N\sqrt{\gamma}$,

   $v_L = 278.1\times\sqrt{1.4} = 278.1\times1.1832 \approx 329.0\ \text{m s}^{-1}$

3. Percentage increase:

   $\frac{v_L-v_N}{v_N}\times100\% = \left(\sqrt{1.4}-1\right)\times100\% \approx 18.3\%$

   ✓Final answer

   Newton's formula gives about $278.1\ \text{m s}^{-1}$; Laplace's corrected formula gives about $329.0\ \text{m s}^{-1}$ -- about $18.3\%$ higher, purely a factor of $\sqrt{\gamma}$, independent of the particular $P$ and $\rho$ used.

Compute $v_N=\sqrt{P/\rho}$ directly, obtain $v_L=v_N\sqrt{\gamma}$ (since Laplace's formula differs from Newton's only by the factor $\sqrt{\gamma}$ inside the square root), then find the percentage increase, which is simply $(\sqrt{\gamma}-1)\times100\%$.

• Recomputing $v_L$ from scratch with $\sqrt{\gamma P/\rho}$ and making an independent arithmetic slip instead of using the simpler $v_L=v_N\sqrt{\gamma}$ relation as a check.
• Computing the percentage change relative to $v_L$ instead of relative to $v_N$ (the question asks by how much Laplace's exceeds Newton's, so the base of the percentage is Newton's value).