While visiting Planet Physics, you toss a rock straight up at 15 and catch it 2.4 later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet's radius every 240 min .

What is the mass of Planet Physics?
Express your answer using two significant figures.

What is the radius of Planet Physics?
Express your answer using two significant figures.

To solve this problem, we need to use two different equations from the laws of physics: the equation of motion for the rock and the equation for orbital motion.

For the rock's motion:
1. Let's assume the upward direction is positive. The initial velocity of the rock is 15 and the time it takes to reach its maximum height is 2.4 .
2. We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
3. At the maximum height, the final velocity is 0 m/s since the rock reaches its highest point and then starts falling back down.
4. Plugging in the values, we have 0 = 15 + (-9.8)t, where the acceleration due to gravity is approximately -9.8 m/s² (negative because it acts opposite to the upward direction).
5. Solving for t, we get t ≈ 1.53 s. This is the time it takes for the rock to reach its maximum height.

Now, for the orbital motion:
1. The cruise ship orbits at an altitude equal to the planet's radius, which means their distances from the planet's center are the same.
2. The time it takes for one complete orbit is given as 240 min. We need to convert this to seconds: 240 min × 60 s/min = 14,400 s.
3. We can use the equation for orbital period: T = 2π√(r³/GM), where T is the period, r is the radius, G is the gravitational constant, and M is the mass of the planet.
4. In this case, T = 14,400 s and r = R (planet's radius). Rearranging the equation, we have r = (T²GM)/(4π²).
5. To calculate the radius, we need to know the value of G. The gravitational constant is approximately 6.67430 × 10^-11 m³/(kg·s²).
6. Now, we can substitute the values and solve for r: r = (14,400² × (6.67430 × 10^-11) × M)/(4π²).
7. Dividing both sides by (6.67430 × 10^-11) × M, we get r/[(6.67430 × 10^-11) × M] = (14,400²)/(4π²).
8. Taking the reciprocal of both sides to solve for the mass, we have M/(r × (6.67430 × 10^-11)) = (4π²)/(14,400²).
9. Now we can rearrange this equation to solve for the mass M: M = (r × (6.67430 × 10^-11) × (4π²))/(14,400²).
10. Plug in the known value of r (the radius of Planet Physics) and calculate the mass M using a calculator. Round the final answer to two significant figures.

So, to summarize the steps:
1. Use the equation v = u + at to solve for the time it takes for the rock to reach its maximum height.
2. Use the equation r = (T²GM)/(4π²) to solve for the radius of Planet Physics.
3. Substitute the known values and solve for the mass, M.

Please note that without the values for T (orbital period) and r (radius), we cannot provide the specific mass and radius of Planet Physics.