# Troubleshooting Interplanetary Cruise Calculations

• Viii
In summary: Apparently, it was moved there. I have a post with the same title but a little bit more understanding. Hope this helps.No, what would HELP would be if you would define your variables and separate out your equations from the rest of the lines they are on so that things read more clearly (both in terms of meaning and in terms of typography)
Viii said:
Hello. I assumed the z axis. I chose it randomly, since it's not stated anywhere which one shall I take. By the way, love your answer. It's very detailed.

The first attachment shows a set of coordinate axes (x, y and z) drawn isometrically beside an isometric drawing of a portion of the skeleton of a cube. Unfortunately, the axes are drawn slanted differently from the cube. However we can see that the "z" direction is normal to the "bottom" face of the cube.

So we know the direction that the axis of rotation points -- vertical. But that does not answer the question I was trying to ask.

Are we considering rotation around a vertical axis through the centers of the top and bottom faces?

Are we considering rotation around a vertical axis along the center line of the right hand face?

Are we considering rotation around a vertical axis along one vertical edge?

It matters because the moment of inertia of a cube is different depending on what axis you are using.

I had provided a clickable link with my question. If you had clicked on that link, you would have been taken to a page that gives two different formulas for the moment of inertia of a cube. Let me extract the information from that link and provide it directly here.

https://www.vedantu.com/iit-jee/moment-of-inertia-of-a-cube said:
I = 1/6 ma2 = ma2/6 when the axis of rotation passes through the center
I = 2 mb2 / 3 when the axis of rotation passes through its edge

Way back in the original post in this thread, you wrote that the moment of inertia of a cube is:
"Iz=(2*m*(b^2))/3". That looks very much like the second of the formulas above.

That means that it is the wrong formula. Because it assumes the wrong axis of rotation.

Viii
Viii said:
Since now we have 180 degrees, we need the 2 factor. Right?
Yes, if we are solving for the time required to accelerate all the way through 180 degrees then the angle ##\theta## is ##\pi## and we need the factor of two in the formula.

Viii
jbriggs444 said:

The first attachment shows a set of coordinate axes (x, y and z) drawn isometrically beside an isometric drawing of a portion of the skeleton of a cube. Unfortunately, the axes are drawn slanted differently from the cube. However we can see that the "z" direction is normal to the "bottom" face of the cube.

So we know the direction that the axis of rotation points -- vertical. But that does not answer the question I was trying to ask.

Are we considering rotation around a vertical axis through the centers of the top and bottom faces?

Are we considering rotation around a vertical axis along the center line of the right hand face?

Are we considering rotation around a vertical axis along one vertical edge?

It matters because the moment of inertia of a cube is different depending on what axis you are using.

I had provided a clickable link with my question. If you had clicked on that link, you would have been taken to a page that gives two different formulas for the moment of inertia of a cube. Let me extract the information from that link and provide it directly here.
Way back in the original post in this thread, you wrote that the moment of inertia of a cube is:
"Iz=(2*m*(b^2))/3". That looks very much like the second of the formulas above.

That means that it is the wrong formula. Because it assumes the wrong axis of rotation.
You're right. I had the wrong formula all this time. Thank you!

Hello Viii, I would like to know what are your correct answers for this problem

Hello Viii, I would like to know what are your correct answers for this problem
Do you have the same problem in your homework?

berkeman said:
Do you have the same problem in your homework?
Hello, yes, it's the same problem, but my answers are different from the ones that viii obtained, so I would like to know which ones are the right ones

I understand that the moment of inertia is I=1/6 ma2 = ma2/6, where m= 3000kg, and a=2m, that gives a value of 2000m² for the moment of inertia, the, when applying the formula for t=sqrt(2*θ*Iz/n*F*L), where θ=180 degrees=pi, Iz=2000 m², n=4 (because you will use four attitude thrusters per burn, F=2 N, L=1m, that gives a value of 39.633 s, then, this "t" that you just obtained is the t_b: time per burn, but you want to know the maneuver time, so, as there are 4 thrusters per burn, then you just have to multiplie the time per burn for the number of thrusters per burn, this gives a total of 158.3 s, but it its incorrect.

Now for the second question, i think there´s an error, the An (area perpendicular to the sun) its not 2, its 4, its the area of a square, then replacing this new An, the temperature gives a value of 260.42 K, or -13.15 °C.

As for the 3rd question, i don´t get why is wrong, but i think i might have an idea, can it the the answer it´s the substraction of the initial Pe (Power emited) that you have with the temperature that you obtained in the second question and the one that you calculated for T=0?

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