What is Escape Velocity - The Formula to Calculate Escape Velocity

 Escape velocity is the minimum speed needed for an object to "break free" from the gravitational attraction of a massive body, without further propulsion. In the context of sending a satellite to the Moon, escape velocity is crucial for understanding how much energy is needed to overcome Earth's gravitational pull and travel to the Moon.

To send a satellite to the Moon, scientists and engineers calculate the escape velocity required to break free from Earth's gravitational field. This calculation takes into account the mass of the Earth, the mass of the satellite, and the distance between them. 

The formula to calculate escape velocity

${�}_{�}=\sqrt{\frac{2��}{�}}$

• $�$ is the gravitational constant ($6.67430\times 10^{-11}{�}^3�{�}^{-1}{�}^{-2}$),
• $�$ is the mass of the Earth (approximately $5.972\times 10^{24}��$),
• $�$ is the distance from the center of the Earth to the satellite.

The escape velocity of Earth is approximately 11.2 kilometers per second (about 33 times the speed of sound in Earth's atmosphere). This means that an object must reach a speed of 11.2 km/s (or about 33,600 km/h) to break free from Earth's gravitational pull and enter space.

By using this formula, scientists can determine the minimum speed the satellite must reach to overcome Earth's gravity and travel towards the Moon. However, it's important to note that this is a simplified explanation. In reality, additional factors such as atmospheric drag, the rotation of the Earth, and the gravitational influence of other celestial bodies also need to be considered in the trajectory calculations for a successful lunar mission. Once the satellite reaches the calculated escape velocity, it can travel on its trajectory towards the Moon without the need for continuous propulsion, although adjustments might still be necessary during the journey due to various gravitational influences and other factors.

What is the Escape Velocity of Moon

To calculate the escape velocity of the Moon, you can use the same formula I mentioned earlier, but with the mass and radius of the Moon instead of Earth.

The mass of the Moon ($�$) is approximately $7.342\times 10^{22}��$, and the radius of the Moon ($�$) is approximately $1.737\times 10^6�$. The gravitational constant ($�$) remains the same.

Using the formula for escape velocity:

${�}_{�}=\sqrt{\frac{2��}{�}}$

Substituting the values:

${�}_{�}=\sqrt{\frac{2\times 6.67430\times 10^{-11}{�}^3�{�}^{-1}{�}^{-2}\times 7.342\times 10^{22}��}{1.737\times 10^6�}}$

Calculating this out:

${�}_{�}=\sqrt{\frac{9.97333\times 10^{12}{�}^3�{�}^{-1}{�}^{-2}}{1.737\times 10^6�}}$

${�}_{�}=\sqrt{5.7415\times 10^6{�}^2{�}^{-2}}$

${�}_{�}\approx 2396.8�\mathrm{/}�$

Therefore, the escape velocity of the Moon is approximately $2396.8�\mathrm{/}�$. This means that for an object to leave the Moon's surface and never return, it needs to achieve a speed of about $2396.8�\mathrm{/}�$.