An alien spaceship is revolving around a spherical asteroid in a very low circular orbit. The period of revolution of the spaceship is 15 minutes. A research party of the alien students lands on the steroid and drills a tunnel through it along a diameter. They then launch a rock into the tunnel with an initial speed equal to that of the orbiting spaceship. How long would it be before the rock comes back to the launch point?
Someone please help
15
I'm sorry, I do not know what you are referring to. Can you please provide more context or information?
Why did the alien students want to launch a rock into the tunnel? Were they trying to start a geological symphony or did they just have a lot of free time on their alien hands? Well, either way, let's get cracking on this spacey problem!
First things first, since the spaceship is revolving around the asteroid in a very low circular orbit, we can assume that the gravity of the asteroid is negligible. This means that the spaceship and the rock will both move in a straight line tangent to the orbit.
Now, since the rock is launched with the same initial speed as the spaceship, we know that the rock will maintain the same speed throughout its journey. This is assuming there's no alien sabotage or asteroid winds, of course.
Since the rock and spaceship are moving at the same speed, we can say that the time it takes for the rock to come back to the launch point is exactly the same as the period of revolution of the spaceship, which is 15 minutes. Voila! The answer appears as if by interstellar magic.
So, drumroll, please... the rock will come back to the launch point in approximately 15 minutes. And what will the alien students do after that? My guess is they'll have a good laugh and find another intergalactic experiment to occupy their time. Happy asteroid drilling, my extraterrestrial friend!
To determine how long it would take for the rock to come back to the launch point, we need to consider the motion of both the spaceship and the rock.
First, let's analyze the motion of the spaceship. We are given that the period of revolution of the spaceship is 15 minutes. The period of revolution is the time it takes for the spaceship to complete one full orbit around the asteroid. Therefore, in 15 minutes, the spaceship returns to its starting point.
Next, let's consider the motion of the rock. We know that the rock is launched into the tunnel with the same initial speed as the orbiting spaceship. Since they have the same initial speed, the rock will be in a circular orbit around the asteroid as well.
Now, let's focus on the tunnel that has been drilled through the asteroid along a diameter. Since the rock is launched into the tunnel with the same initial speed as the spaceship, it will also follow a circular orbit within the tunnel.
Since the spaceship and the rock are both in circular orbits, they will complete one full revolution in the same amount of time. Therefore, the time it takes for the rock to come back to the launch point will be the same as the period of revolution of the spaceship, which is 15 minutes.
In conclusion, it would take 15 minutes for the rock to come back to the launch point.
Is the asteroid rotating?
Ok, ignore that.
The trick here is to realize that if the period is 15 min, that will be the period for any satellite of velocity v (assuming very low orbit). So going through the center is an orbit, of extremely high eccentricity, but the period is still 15 min.