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 required for a projectile to leave the Earth's surface without considering air friction, we need to use the concept of escape velocity.
Escape velocity is defined as the minimum velocity required for an object to escape the gravitational pull of a planet or celestial body. In this case, we are looking for the escape velocity from Earth.
The equation to calculate escape velocity is given by:
v = sqrt((2 * G * M) / r),
where:
- "v" represents the escape velocity,
- "G" is the gravitational constant, which was provided as 6.67428 * 10^-11 m^3 kg^-1 s^-2,
- "M" is the mass of the Earth, which is approximately 5.972 x 10^24 kg,
- "r" is the radius of the Earth, which is approximately 6.371 x 10^6 m.
Plugging in these known values into the formula, we get:
v = sqrt((2 * (6.67428 * 10^-11) * (5.972 x 10^24)) / (6.371 x 10^6)).
Now let's simplify the calculation step by step:
Step 1: Calculate the product of G, M, and 2:
(6.67428 * 10^-11) * (5.972 x 10^24) * 2 = 7.97377545616 x 10^14.
Step 2: Calculate the result of the numerator:
7.97377545616 x 10^14 / (6.371 x 10^6) = 1.25 x 10^8.
Step 3: Take the square root of the result:
sqrt(1.25 x 10^8) = 1.11803398875 x 10^4.
Therefore, the magnitude of the initial velocity needed for the projectile to leave the Earth without considering air friction is approximately 1.11803398875 x 10^4 m/s.