Question

The escape speed from the surface of Planet Zoroaster is $12.0 \mathrm{km} / \mathrm{s} .$ The planet has no atmosphere. A meteor far away from the planet moves at speed \(5.0 \mathrm{km} / \mathrm{s}\) on a collision course with Zoroaster. How fast is the meteor going when it hits the surface of the planet?

Short answer

Answer: The final speed of the meteor when it hits the surface of Planet Zoroaster is 13.0 km/s.

Step-by-step solution

Identify the variables

We are given the escape speed from the surface of the planet (\(v_e = 12.0\,\mathrm{km/s}\)) and the initial speed of the meteor far away from the planet (\(v_i = 5.0\,\mathrm{km/s}\)). We need to find the final speed of the meteor \(v_f\) on the surface of the planet.

Express the energy conservation law

According to the conservation of mechanical energy, the sum of the kinetic energy and gravitational potential energy remains constant throughout the motion. Let \(K_i\) and \(U_i\) be the initial kinetic energy and gravitational potential energy, and \(K_f\) and \(U_f\) be the final kinetic energy and gravitational potential energy. Then, we can express the conservation of mechanical energy as: \(K_i + U_i = K_f + U_f\)

Write the equations for kinetic and gravitational potential energies

Now we can write the kinetic energies (\(K\)) and gravitational potential energies (\(U\)) as functions of the speeds (\(v\)) and distances (\(r\)) from the center of the planet. Use the mass of the meteor (\(m\)) and the mass of Planet Zoroaster (\(M\)). Note that escape velocity (\(v_e\)) is related to the planet's mass and radius (\(R\)) by \(v_e^2 = 2GM/R\): Initial kinetic energy: \(K_i = \frac{1}{2}m v_i^2\) Initial gravitational potential energy: \(U_i \approx 0\) (Since the meteor is far away from the planet) Final kinetic energy: \(K_f = \frac{1}{2}m v_f^2\) Final gravitational potential energy: \(U_f = -\frac{GMm}{R}\)

Substitute the energies into the conservation of mechanical energy equation

Now, we can substitute the expressions for kinetic and gravitational potential energies into the conservation equation: \(\frac{1}{2}m v_i^2 + 0 = \frac{1}{2}m v_f^2 - \frac{GMm}{R}\) Note that the mass of the meteor \(m\) can be canceled out from both sides of the equation: \(\frac{1}{2} v_i^2 = \frac{1}{2} v_f^2 - \frac{GM}{R}\)

Solve for final speed \(v_f\) using escape velocity relation

Recall the escape velocity relation \(v_e^2 = 2GM/R\), and use it to solve for final velocity \(v_f\): \(v_i^2 = v_f^2 - v_e^2\) Now solve for \(v_f\): \(v_f = \sqrt{v_i^2 + v_e^2}\)

Calculate the final speed of the meteor

Substitute the given initial speed \(v_i = 5.0\,\mathrm{km/s}\) and escape speed \(v_e = 12.0\,\mathrm{km/s}\) into the equation for the final speed \(v_f\): \(v_f = \sqrt{(5.0\,\mathrm{km/s})^2 + (12.0\,\mathrm{km/s})^2}\) \(v_f = \sqrt{169\,\mathrm{km^{2}/s^{2}}} = 13\,\mathrm{km/s}\) So, the final speed of the meteor when it hits the surface of Planet Zoroaster is \(13.0\,\mathrm{km/s}\).