Question

Determine the ratio β = v/c for each of the following.

Answer 1

Answer:

a) $\beta = 1.111\times 10^{-7}$, b) $\beta = 9\times 10^{-7}$, c) $\beta = 3.087\times 10^{-6}$, d) $\beta = 2.5\times 10^{-5}$, e) $\beta = 0.5$, f) $\beta = 0.877$

Explanation:

From relativist physics we know that $c$ is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) A car traveling 120 kilometers per hour:

At first we convert the car speed into meters per second:

$v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 33.333\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ($v = 33.333\,\frac{m}{s}$, $c = 3\times 10^{8}\,\frac{m}{s}$)

$\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 1.111\times 10^{-7}$

b) A commercial jet airliner traveling 270 meters per second:

The ratio $\beta$ is now calculated: ($v = 270\,\frac{m}{s}$, $c = 3\times 10^{8}\,\frac{m}{s}$)

$\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 9\times 10^{-7}$

c) A supersonic airplane traveling Mach 2.7:

At first we get the speed of the supersonic airplane from Mach's formula:

$v = Ma\cdot v_{s}$

Where:

$Ma$ - Mach number, dimensionless.

$v_{s}$ - Speed of sound in air, measured in meters per second.

If we know that $Ma = 2.7$ and $v_{s} = 343\,\frac{m}{s}$, then the speed of the supersonic airplane is:

$v = 2.7\cdot \left(343\,\frac{m}{s} \right)$

$v = 926.1\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ($v = 926.1\,\frac{m}{s}$, $c = 3\times 10^{8}\,\frac{m}{s}$)

$\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 3.087\times 10^{-6}$

d) The space shuttle, travelling 27000 kilometers per hour:

At first we convert the space shuttle speed into meters per second:

$v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 7500\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ($v = 7500\,\frac{m}{s}$, $c = 3\times 10^{8}\,\frac{m}{s}$)

$\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 2.5\times 10^{-5}$

e) An electron traveling 30 centimeters in 2 nanoseconds:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 0.3\,m$ and $t = 2\times 10^{-9}\,s$, then speed of the electron is:

$v = \frac{0.3\,m}{2\times 10^{-9}\,s}$

$v = 1.50\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ($v = 1.5\times 10^{8}\,\frac{m}{s}$, $c = 3\times 10^{8}\,\frac{m}{s}$)

$\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.5$

f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 10^{-14}\,m$ and $t = 0.38\times 10^{-22}\,s$, then speed of the electron is:

$v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}$

$v = 2.632\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ($v = 2.632\times 10^{8}\,\frac{m}{s}$, $c = 3\times 10^{8}\,\frac{m}{s}$)

$\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.877$