Calculate the escape velocity on surface of earth of radius of moon is 1740km and Massis7.4 *10to the power 22
F = - G m M/r^2
work done from R to infinity
F dr = G m M/R
so
G M m/R^2 = Ke at surface = (1/2)m v^2
v^2 = 2 G M/R
G = 6.67 * 10^-11
M = 7.4 * 10^22 kg
R = 1.74^10^6 meters
plug and chug
To calculate the escape velocity on the surface of Earth, we need to use the following equation:
Escape velocity (Ve) = sqrt(2 * gravitational constant * mass of the celestial body / radius of the celestial body)
First, let's clarify the values we have:
Mass of the Earth (M) = 5.972 × 10^24 kg
Radius of the Earth (R) = 6,371 km (or 6,371,000 meters)
Gravitational constant (G) = 6.67430 × 10^-11 m^3 kg^-1 s^-2
Now, we need to convert the given values for the Moon to SI units:
Radius of the Moon (Rm) = 1,740 km (or 1,740,000 meters)
Mass of the Moon (Mm) = 7.4 × 10^22 kg
Using the formula, we can now calculate the escape velocity on the surface of the Earth:
Ve = sqrt(2 * G * M / R)
Ve = sqrt(2 * (6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / (6,371,000 meters))
Calculating this expression will give us the escape velocity on the surface of the Earth, which is approximately 11.186 km/s.
Please note that the given values for the radius and mass of the Moon are not relevant in this calculation. The escape velocity calculation here is only considering the Earth and its parameters.