Relativistic Energy (AQA A Level Physics): Revision Note

A spacecraft with a proper mass of 450 kg accelerates to a speed of 0.7c. Calculate the difference in its relativistic kinetic energy and its kinetic energy calculated using Newtonian physics.

Answer:

Step 1: List the known quantities:

• Rest mass, m₀ = 450 kg

• Speed, v = 0.7c

• Speed of light, c = 3.0 × 10⁸ ms^-1

Step 2: List the relevant equations:

• Newtonian kinetic energy, E_kn = ½m₀v²

• Relativistic kinetic energy, E_kr = γm₀c² − m₀c²

Step 3: Calculate the Newtonian kinetic energy:

• Substitute the known quantities:

$E_{kn}=\frac{1}{2}\times 450\times {\left(0.7\times \left(3.0\times {10}^8\right)\right)}^2=9.9\times {10}^{18}\mathrm{J}$

Step 4: Calculate the gamma term:

• Substitute the speed into the gamma term:

$\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

$\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.7c\right)}^2}{c^2}}}}=\frac{1}{\sqrt{1-0.7^2}}=1.40$

Step 5: Substitute the gamma term into the relativistic kinetic energy equation:

• Substitute the rest mass, speed of light and speed as well:

$E_{kr}=\gamma m_0{c}^2-m_0{c}^2=m_0{c}^2\left(\gamma -1\right)$

$E_{kr}=450\times {\left(3.0\times {10}^8\right)}^2\times \left(1.40-1\right)=1.6\times {10}^{19}\mathrm{J}$

Step 6: Calculate the difference between the two kinetic energies:

• Subtract the smaller Newtonian kinetic energy from the larger relativistic energy

$\mathrm{\Delta }{E}_k=\left(1.6\times {10}^{19}\right)-\left(9.9\times {10}^{18}\right)=6.1\times {10}^{18}\mathrm{J}$

• As you can see, the relativistic kinetic energy is almost twice as large as the Newtonian kinetic energy!