Question

Suppose your handheld calculator will show six places beyond the decimal point. How fast (in meters per second) would an object have to be traveling so that the decimal places in the value of \(\gamma\) can actually be seen on your calculator display? That is, how fast should an object travel so that \(\gamma=1.000001 ?\) [Hint: Use the binomial approximation.]

Short answer

Answer: An object must travel around \(1.34 × 10^6\) meters per second.

Step-by-step solution

Write down the given equation

We're given the equation using the binomial approximation: $$\gamma \approx 1 - \frac{1}{2}\frac{v^2}{c^2}$$ And we know that \(\gamma = 1.000001\). Let's substitute this value in our equation.

Substitute the given value of \(\gamma\)

Substituting \(\gamma = 1.000001\) in the equation, we have: $$1.000001 \approx 1 - \frac{1}{2}\frac{v^2}{c^2}$$ Now we need to solve the equation for the speed, v.

Solve the equation for the speed, v

Solving the equation for v, we get: $$1.000001 - 1 = - \frac{1}{2} \frac{v^2}{c^2}$$ $$0.000001 = \frac{1}{2} \frac{v^2}{c^2}$$ Now, we'll multiply both sides by 2 and by \(c^2\): $$2(0.000001)(c^2) = v^2$$ Then take the square root of both sides to find v: $$v = \sqrt{2(0.000001)(c^2)} $$

Substitute c and find v

Substituting the speed of light c (\(3.0 × 10^8\) m/s), we get: $$v = \sqrt{2(0.000001)\left(3.0 × 10^8\right)^2}$$ $$v \approx \sqrt{1.8 × 10^{12}}$$ $$v \approx 1.34 × 10^6 \text{ m/s}$$ So, an object must travel around \(1.34 × 10^6\) meters per second for the decimal places in the value of \(\gamma\) to be visible on a calculator display with six decimal places using the binomial approximation.