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 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
I will be happy to critique your work.
You need to relate initial KE to total gravitational PE from the surface of Earth to infinity.
Can you please give me a website that would help answer this question
To calculate the magnitude of the initial velocity required for a projectile to leave the Earth's gravitational field when air friction is neglected, we need to consider the Law of Gravitation and the concept of escape velocity.
The Law of Gravitation, formulated by Sir Isaac Newton, states that the force of gravitational attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically, this can be expressed as:
F = (G * m1 * m2) / r^2,
where F is the force of gravitational attraction, G is the gravitational constant (6.67428 * 10^-11 m^3 kg^-1 s^-2), m1 and m2 are the masses of the two objects in question, and r is the distance between their centers.
For a projectile to escape Earth's gravitational field, its initial kinetic energy must be equal to the gravitational potential energy it possesses at an infinite distance from the Earth.
The kinetic energy of an object is given by the equation:
KE = (1/2) * m * v^2,
where KE is the kinetic energy, m is the mass of the object, and v is its velocity.
Similarly, the gravitational potential energy of an object near the surface of the Earth is given by:
PE = -(G * m * M) / r,
where PE is the gravitational potential energy, M is the mass of the Earth, and r is the distance from the center of the Earth to the object.
At an infinite distance from the Earth, the gravitational potential energy becomes zero. Therefore, we can set the gravitational potential energy equal to zero and solve for the escape velocity.
-(G * m * M) / r = (1/2) * m * v^2.
Canceling out the mass term, we get:
v^2 = (2 * G * M) / r.
Taking the square root of both sides, we find:
v = sqrt((2 * G * M) / r),
where v is the escape velocity, G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.
Substituting the radius of the Earth (r) for the distance from center (R), we get:
v = sqrt((2 * G * M) / R).
By considering R as the sum of the radius of the Earth (r) and the height (h) from the surface on which the projectile is launched (R = r + h), we can rewrite the equation as:
v = sqrt((2 * G * M) / (r + h)).
In this equation, h represents the height from the Earth's surface, and v is the magnitude of the initial velocity required for the projectile to escape the Earth's gravitational field when air friction is neglected.