find an expression for the magnitude of the initial velocity that a projectile must possess in order to leave the earth when air friction is negelcted.

The gravitational constant is given as G= 6.67428*10^-11 m^3 kg^-1 s^-2

Please give lots of details in answer

To find the expression for the magnitude of the initial velocity of a projectile to leave the Earth when air friction is neglected, we can use the concept of escape velocity.

Escape velocity is the minimum velocity an object needs to escape the gravitational pull of a planet or celestial body. Therefore, to determine the magnitude of the initial velocity, we need to find the escape velocity.

The escape velocity is calculated using the formula:

v = sqrt((2GM)/r)

Where:
- v is the escape velocity
- G is the gravitational constant (G = 6.67428 * 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth
- r is the distance from the center of the Earth to the object's initial position

Since we are neglecting air friction, the mass of the projectile doesn't play a role in this calculation. Therefore, we can ignore it when calculating the escape velocity.

The mass of the Earth (M) is approximately 5.972 × 10^24 kg, and the average radius of the Earth (r) is approximately 6,371,000 meters.

Plugging in the values, we have:

v = sqrt((2 * 6.67428 * 10^-11 * 5.972 × 10^24) / 6,371,000)

Let's simplify this expression step-by-step:

v = sqrt((2 * 6.67428 * 10^-11 * 5.972 × 10^24) / 6,371,000)
v = sqrt((7.98451 * 10^14 * 2) / 6,371,000)
v = sqrt(3.17535 * 10^5)

Finally, calculating the square root, we find:

v ≈ 5645 m/s

Therefore, the expression for the magnitude of the initial velocity (v) that a projectile must possess in order to leave the Earth when air friction is neglected is approximately 5645 m/s.