Question
What is escape velocity?
Answers
GPT-4o mini
Escape velocity is the minimum speed an object must reach to break free from a celestial body's gravitational pull without any additional propulsion. This concept is most commonly applied in astrodynamics when launching spacecraft from planets, moons, or other astronomical bodies.
The escape velocity depends on the mass \( M \) of the celestial body and its radius \( R \). The formula for escape velocity \( v_e \) is derived from the principles of gravitational potential energy and kinetic energy:
\[
v_e = \sqrt{\frac{2GM}{R}}
\]
Where:
- \( v_e \) = escape velocity,
- \( G \) = gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3/\text{kg/s}^2\)),
- \( M \) = mass of the celestial body,
- \( R \) = radius of the celestial body from its center to the point of escape.
For Earth, the escape velocity is approximately 11.2 kilometers per second (about 33 times the speed of sound). This value changes for other celestial bodies depending on their mass and size; for instance, the escape velocity from the Moon is significantly lower due to its smaller mass and radius.
The escape velocity depends on the mass \( M \) of the celestial body and its radius \( R \). The formula for escape velocity \( v_e \) is derived from the principles of gravitational potential energy and kinetic energy:
\[
v_e = \sqrt{\frac{2GM}{R}}
\]
Where:
- \( v_e \) = escape velocity,
- \( G \) = gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3/\text{kg/s}^2\)),
- \( M \) = mass of the celestial body,
- \( R \) = radius of the celestial body from its center to the point of escape.
For Earth, the escape velocity is approximately 11.2 kilometers per second (about 33 times the speed of sound). This value changes for other celestial bodies depending on their mass and size; for instance, the escape velocity from the Moon is significantly lower due to its smaller mass and radius.