Hardest and Trickiest Question I've Encountered.

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Homework Help Overview

The problem involves an alien spaceship in a low circular orbit around a spherical asteroid, with a focus on the dynamics of a rock launched through a tunnel drilled along the asteroid's diameter. The period of the spaceship's orbit is given as 15 minutes, and the challenge is to determine the time it takes for the rock to return to its launch point after being launched with the same speed as the spaceship.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to relate the period of the orbit to the speed of the rock and expresses confusion about the number of variables involved. Some participants suggest the need for Newton's law of gravitation and the assumption of uniform density to analyze the problem correctly. Others explore the relationship between distance, speed, and time using basic equations.

Discussion Status

Participants are actively engaging with the problem, with some providing calculations based on their interpretations. There is recognition that gravity plays a crucial role in the dynamics of the situation, and while some computations have been shared, there is no explicit consensus on the correct approach or final answer.

Contextual Notes

There is an indication that the problem may involve complexities related to gravitational effects, which have not been fully addressed in the initial attempts. The original poster expresses uncertainty about the necessary variables and equations, highlighting potential gaps in the information provided.

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Hardest and Trickiest Question I've Encountered. Please Help!

Homework Statement


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?




Homework Equations


None that I know of



The Attempt at a Solution



Here is what I think, since T= 2piR/v, I thought I could get v. However, I don't even know R and I don't even know distance if I am to use v as in d/t. There seems to be too many variables involved and too little equations to use. I believe that words, very low circular orbit, could mean something. I guess I only know that the period is 15 mintues and that the rock is launched at an initial speed equal to the orbiting space ship. That means that the velocity of the rock is equal to 2piR/15 minutes. however, i still don't know r.
 
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Your equations are seriously underpowered. You need to know Newton's law of gravitation and assume that the body is uniformly dense. Then you need to compute the orbital period at the surface and compare it with the time it takes on a trajectory through the center. If the only equation you have is T=2piR/v, this is way too hard for you.
 
no idea if this is right but yeah...

T=(2*PI*r)/v
V=(2*Pi*r)/T

T=15min = 900sec

V=(2*pi*r)/900 _____(1)

Distance= speed*time
t=D/S _______________(2)

distance is the diamater which is 2r

equation 1 into 2

t=2r/((2*Pi*r)/900)

after abit of algebra

t=900/pi
t= 286.48 sec (2dp)
t= 4min 46 sec

dunno if its right but yeah there it is

cheers
 
You just computed 15min/pi. That's the time required to pass through the hole one-way ignoring gravity. That much is right. But if you ignore gravity nothing will orbit and the object shot down the hole will never come back. That's no fun!
 

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