Question
Formula for escape velocity of an object
Answers
GPT-4o mini
The escape velocity (\( v_e \)) of an object is the minimum velocity an object must achieve to break free from the gravitational pull of a celestial body, without any further propulsion. The formula to calculate escape velocity is given by:
\[
v_e = \sqrt{\frac{2GM}{r}}
\]
Where:
- \( v_e \) = escape velocity
- \( G \) = gravitational constant \( (6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) \)
- \( M \) = mass of the celestial body (for example, Earth)
- \( r \) = radius of the celestial body from its center to the point of escape (for example, the surface radius of Earth)
For Earth, with a mass of approximately \( 5.972 \times 10^{24} \, \text{kg} \) and a mean radius of about \( 6.371 \times 10^6 \, \text{m} \), the escape velocity is approximately \( 11,186 \, \text{m/s} \) or about \( 11.2 \, \text{km/s} \).
\[
v_e = \sqrt{\frac{2GM}{r}}
\]
Where:
- \( v_e \) = escape velocity
- \( G \) = gravitational constant \( (6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) \)
- \( M \) = mass of the celestial body (for example, Earth)
- \( r \) = radius of the celestial body from its center to the point of escape (for example, the surface radius of Earth)
For Earth, with a mass of approximately \( 5.972 \times 10^{24} \, \text{kg} \) and a mean radius of about \( 6.371 \times 10^6 \, \text{m} \), the escape velocity is approximately \( 11,186 \, \text{m/s} \) or about \( 11.2 \, \text{km/s} \).