Show that the kinetic energy of an object moving with velocity \(v\) can be approximated by the formula \(K \approx \frac{1}{2} m_0v^2\) when \(v \ll c\text{.}\)
Hint.
We can use the Taylor series for \(f(x) = \frac{1}{\sqrt{1 - x}}\) at \(x=0\) to approximate \(K\text{.}\) Use \(x = \frac{v^2}{c^2}\text{.}\)
Solution.
We can rewrite the kinetic energy as follows:
\begin{equation*}
K = mc^2-m_0c^2 = m_0c^2 \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right)
\end{equation*}
Let \(x = \frac{v^2}{c^2}\text{.}\) Then we have:
\begin{equation*}
K = m_0c^2 \left( \frac{1}{\sqrt{1 - x}} - 1 \right)
\end{equation*}
Now we can use the Taylor series for \(f(x) = \frac{1}{\sqrt{1 - x}}\) at \(x=0\) to approximate \(K\text{.}\) The Taylor series for \(f(x)\) at \(x=0\) is given by:
\begin{equation*}
f(x) = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 + \cdots
\end{equation*}
Therefore, we have:
\begin{equation*}
K = m_0c^2 \left( \frac{1}{\sqrt{1 - x}} - 1 \right) = m_0c^2 \left( \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 + \cdots \right)
\end{equation*}
Substituting \(x = \frac{v^2}{c^2}\text{,}\) we get:
\begin{align*}
K \amp = m_0c^2 \left( \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \frac{5}{16}\frac{v^6}{c^6} + \cdots \right) \\
\amp = \frac{1}{2} m_0v^2 + \frac{3}{8} m_0\frac{v^4}{c^2} + \frac{5}{16} m_0\frac{v^6}{c^4} + \cdots
\end{align*}
When \(v \ll c\text{,}\) the higher order terms in the series become negligible, and we can approximate the kinetic energy as:
\begin{equation*}
K \approx \frac{1}{2} m_0v^2.
\end{equation*}
