The Earth has an angular speed of 7.272 10-5 rad/s in its rotation. Find the new angular speed if an asteroid (m = 1.07 1022 kg) hits the Earth while traveling at a speed of 1.30 103 m/s (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases.
The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation.
The new angular speed of the Earth would be 7.272 10-5 rad/s. The angular momentum of the Earth is conserved, so the angular speed of the Earth does not change.
To find the new angular speed of the Earth after the asteroid hits nearly tangentially in the direction of Earth's rotation, we can use the law of conservation of angular momentum. The equation for angular momentum is given by:
L = Iw,
where L is the angular momentum, I is the moment of inertia, and w is the angular speed.
Since the asteroid is a point mass compared to the radius of the Earth, we can simplify the moment of inertia equation to:
I = m * r^2,
where m is the mass of the Earth and r is the radius of the Earth.
We can assume the initial angular momentum of the Earth is:
L_initial = I_initial * w_initial,
where w_initial is the initial angular speed of the Earth.
After the asteroid hits, the new angular momentum of the Earth is:
L_final = I_final * w_final,
where w_final is the new angular speed of the Earth.
Since angular momentum is conserved, we can set L_initial equal to L_final:
L_initial = L_final.
Since the asteroid hits the Earth nearly tangentially in the direction of Earth's rotation, there is no external torque acting on the system. Therefore, the initial and final moments of inertia are the same:
I_initial = I_final = I.
We can rearrange the equation to solve for w_final:
L_initial = I_initial * w_initial,
L_final = I_final * w_final,
I_initial * w_initial = I_final * w_final,
w_final = (I_initial * w_initial) / I_final.
Let's plug in the given values:
Given:
w_initial = 7.272 x 10^-5 rad/s (angular speed of the Earth initially)
m = 1.07 x 10^22 kg (mass of the asteroid)
v = 1.30 x 10^3 m/s (speed of the asteroid)
r = radius of the Earth (we can assume it is 6.37 x 10^6 m)
To find the radius of the asteroid, we'll use the equation for kinetic energy:
KE = (1/2) m v^2,
KE = GPE,
(1/2) m v^2 = G * (m * M) / r_a,
(1/2) v^2 = G * M / r_a,
r_a = G * M / (v^2 / 2),
where G is the gravitational constant and M is the mass of the Earth.
Let's calculate the radius of the asteroid:
r_a = 6.67 x 10^-11 m^3 kg^-1 s^-2 * (5.97 x 10^24 kg) / ((1.30 x 10^3 m/s)^2 / 2) = 3.71 x 10^6 m.
Substituting the values into the equation for w_final:
w_final = (I_initial * w_initial) / I_final,
w_final = (I_initial * w_initial) / (m_a * r_a^2),
w_final = (I_initial * w_initial) / ((m_a * r_a^2) + I_initial),
where m_a is the mass of the asteroid and r_a is the radius of the asteroid.
Let's calculate the new angular speed w_final:
w_final = (5.97 x 10^24 kg * (6.37 x 10^6 m)^2 * 7.272 x 10^-5 rad/s) / ((1.07 x 10^22 kg * (3.71 x 10^6 m)^2) + (5.97 x 10^24 kg * (6.37 x 10^6 m)^2) * 7.272 x 10^-5 rad/s),
w_final ≈ 7.272 x 10^-5 rad/s.
Therefore, the new angular speed of the Earth after the asteroid hits nearly tangentially in the direction of Earth's rotation is approximately 7.272 x 10^-5 rad/s.
To find the new angular speed when an asteroid hits the Earth nearly tangentially in the direction of Earth's rotation, we need to apply the principle of conservation of angular momentum.
Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (ω). Mathematically, it is represented as:
L = I * ω
Since the Earth's moment of inertia remains constant during the collision, the initial angular momentum (L_initial) before the collision is equal to the final angular momentum (L_final) after the collision.
Therefore, we can write:
L_initial = L_final
Now, let's calculate the initial and final angular momenta.
The initial angular momentum (L_initial) of the Earth can be calculated by multiplying the moment of inertia of the Earth (I_earth) with its initial angular speed (ω_initial).
L_initial = I_earth * ω_initial
The moment of inertia of a rotating body can be calculated using the formula:
I = (2/5) * m * r^2
Where:
m = mass of the body
r = radius of rotation
Given that the mass of the Earth (m_earth) is approximately 5.97 * 10^24 kg and the radius of the Earth (r_earth) is approximately 6.37 * 10^6 m, we can substitute these values into the formula to find the initial moment of inertia (I_initial) of the Earth.
I_initial = (2/5) * m_earth * r_earth^2
Next, substitute the given values into the formula to find the initial angular momentum of the Earth (L_initial).
L_initial = I_initial * ω_initial
Now, let's calculate the final angular momentum (L_final) after the collision.
The final angular momentum (L_final) can be calculated by adding the angular momentum of the Earth (L_earth) and the angular momentum of the asteroid (L_asteroid).
L_final = L_earth + L_asteroid
Since the asteroid is a point mass compared to the radius of the Earth, its moment of inertia (I_asteroid) can be approximated as:
I_asteroid = m_asteroid * r_earth^2
Substitute the given values into the formula to find the moment of inertia (I_asteroid) of the asteroid.
I_asteroid = m_asteroid * r_earth^2
Next, calculate the angular momentum of the asteroid (L_asteroid) using the formula:
L_asteroid = I_asteroid * ω_asteroid
Finally, substitute the calculated values of L_initial, L_earth, and L_asteroid into the equation L_initial = L_final, solve for the final angular velocity (ω_final), and calculate the new angular speed.
ω_final = L_final / I_earth
Since the values for mass (m_earth), radius (r_earth), initial angular velocity (ω_initial), mass of the asteroid (m_asteroid), and the velocity of the asteroid (v_asteroid) are not provided, you would need to input these numerical values into the appropriate formulas to obtain the final result.