Let us throw a particle of mass m from the north pole with a velocity v_0 forming an angle α with the surface of the earth. Considering only the effect of gravity and neglecting air friction...
a) What is the initial angular momentum l_0 when the particle is launched?
b) Compute the value of v_0 so that the particle will land precisely on the south pole. Express your result in terms of α, G (Gravitational constant), M the mass of the earth, and R its radius.
c) What is the eccentricity of the trajectory in terms of α only?
Hint: Use the equation of the trajectory r(θ) = (l_0^2/mk)/(1 - εcos(θ - θ_0)), and evaluate the initial angular momentum l_0. What is r(θ = ±π/2)? To compute the eccentricity, use the formula ε = √1 + (2El_0^2/mk^2) from the lecture notes.
a) To find the initial angular momentum (l_0) when the particle is launched, we need to consider the definition of angular momentum:
l = r x p
Here, r is the position vector (from the axis of rotation to the particle) and p is the linear momentum of the particle. In this case, the particle is launched from the north pole, so the position vector is perpendicular to the velocity vector (since it forms an angle α with the surface). Let's assume the particle is launched along the x-axis. In this case, the position vector r points along the positive y-axis.
The linear momentum p is defined as:
p = mv_0
where m is the mass of the particle and v_0 is the initial velocity of the particle.
Since r and p are perpendicular to each other, their cross product becomes:
r x p = r * p * sin(90 degrees)
= r * p
Thus, the magnitude of the angular momentum is given by:
l_0 = |r| * |p|
= |r| * m * v_0
b) To compute the value of v_0 so that the particle lands precisely on the south pole, we need to analyze the trajectory of the particle. Since we are neglecting air friction and considering only the effect of gravity, the only force acting on the particle is the gravitational force.
The trajectory of the particle can be described using the equation:
r(θ) = (l_0^2 / mk) / (1 - ε * cos(θ - θ_0))
Here, r(θ) represents the distance of the particle from the axis of rotation (Earth), θ is the angle the radius vector of the particle makes with a reference direction, l_0 is the initial angular momentum, m is the mass of the particle, k is the spring constant, ε is the eccentricity, and θ_0 is the angle at which the particle is launched.
The south pole is located at θ = π/2. So, to land precisely on the south pole, we need r(π/2) to be equal to the radius of the Earth (R).
Substituting θ = π/2 in the trajectory equation, we have:
R = (l_0^2 / mk) / (1 - ε * cos(π/2 - θ_0))
Since cos(π/2 - θ_0) = sin(θ_0) and ε is small (as the trajectory is close to a circle), we can assume that ε * sin(θ_0) ≈ 0.
Therefore, the equation becomes:
R = (l_0^2 / mk)
Rearranging the equation, we can solve for v_0:
v_0 = √[(R * g * m) / sin(α)]
where g is the acceleration due to gravity.
c) To find the eccentricity of the trajectory in terms of α only, we use the formula:
ε = √[1 + (2E * l_0^2 / mk^2)]
In this case, the term E represents the total mechanical energy of the particle. Since we are neglecting air friction and only considering the effect of gravity, the mechanical energy of the particle is given by:
E = 0.5 * m * v_0^2 - G * M * m / R
Substituting the value of v_0 from part b, the equation becomes:
E = 0.5 * m * [(R * g * m) / sin(α)]^2 - G * M * m / R
Now, we can substitute E and solve for the eccentricity ε in terms of α.