Law of Gravitation- 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 neglected.
The gravitational constant for g is, G = 6.67428*10^-11 m^3 Kg^-1 s^-2
Please answer the question fully and please give any helpful websites, thankyou.
Do you know calculus?
Yes
To find the expression for the magnitude of the initial velocity that a projectile must possess in order 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 field of a planet.
The equation we use is:
v = sqrt((2 * G * M) / r)
Where:
v is the escape velocity,
G is the gravitational constant (6.67428 * 10^-11 m^3 kg^-1 s^-2),
M is the mass of the Earth (5.972 × 10^24 kg),
and r is the distance between the center of the Earth and the object.
Since we want to find the initial velocity, which is the velocity needed at the surface of the Earth, r is equal to the radius of the Earth (approximately 6,371 km or 6,371,000 m).
Plugging in the values, we get:
v = sqrt((2 * 6.67428 * 10^-11 * 5.972 × 10^24) / 6,371,000)
Calculating this expression gives us the magnitude of the initial velocity required to leave the Earth when air friction is neglected.
Here is a website that provides further information on the escape velocity and related concepts:
- NASA's page on escape velocity: https://www.grc.nasa.gov/www/k-12/rocket/escvel.html
- Khan Academy lesson on escape velocity: https://www.khanacademy.org/science/ap-physics-1/ap-circular-motion-and-gravitation/ap-escape-velocity-and-orbital-velocity/a/escape-velocity