a space launch vehicle has a mass of 500,000 kg at liftoff. if it achieves a velocity of 7,500 m/s, what is its kinetic energy?
The kinetic energy of the launch vehicle is 1.41×10131.41\times 10^{13}1.41×1013 joules.
Quick Scoop 🚀
We use the kinetic energy formula for a moving object:
KE=12mv2KE=\tfrac{1}{2}mv^2KE=21mv2
where mmm is mass and vvv is velocity. This is the standard expression for translational kinetic energy in physics and is widely used in textbooks and learning resources.
Given:
- Mass m=500,000textkgm=500{,}000\\text{kg}m=500,000textkg
- Velocity v=7,500textm/sv=7{,}500\\text{m/s}v=7,500textm/s
Step-by-step:
- Square the velocity:
v2=7,5002=56,250,000v^2=7{,}500^2=56{,}250{,}000v2=7,5002=56,250,000
- Multiply by the mass:
mv2=500,000×56,250,000=2.8125×1013mv^2=500{,}000\times 56{,}250{,}000=2.8125\times 10^{13}mv2=500,000×56,250,000=2.8125×1013
- Take half of that:
KE=12×2.8125×1013=1.40625×1013textJKE=\tfrac{1}{2}\times 2.8125\times 10^{13}=1.40625\times 10^{13}\\text{J}KE=21×2.8125×1013=1.40625×1013textJ
Rounded to three significant figures:
Kinetic energy ≈ 1.41×1013textJ1.41\times 10^{13}\\text{J}1.41×1013textJ
That’s about 14 trillion joules of kinetic energy — an enormous amount, reflecting just how energetic orbital-class rockets are.
Information gathered from public forums or data available on the internet and portrayed here.