Astronomy Problems: Please Look/help

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In summary: The units of the Gravitational constant are (AU)^3/(years)^2. You can use that to convert from period in years to period in hours. You'll need the mass of Mars in addition to the semimajor axis. Don't forget to convert units.4.) The energy is inversely proportional to the square of the distance, and we can find the ratio of the two energies by squaring the ratio of the distances. For the apparent size, the apparent size is proportional to the ratio of the distances. You can use these ratios to find the difference in size and apparent magnitude.3.) P(years)=R(A.U.)^3/2You wrote a
  • #71
You're diameter is wrong. It's not 20.558 km. How did you get that?
opposite = tan(4.1666x10^-6 deg) * 6283110582 km does not equal 20.558 km.

You solved for volume and density properly, but since your diameter is wrong, you got the wrong answers here.

For mass, yes, you need to convert hours to seconds and km to m to use 6.67x10^-11 for G. But you got the wrong answer.

How did you re-write the formula to solve for M?
 
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  • #72
So I am going to convert those things and see if I can't get the right answer.

I rewrote the original formula so I could isolate the x which in this case was m.

The original formula was: T= 2(pi) square root of (a^3/u) where u = gm

therefore I isolated m so the new formula is as follows:
g(T/2(pi))^2/a^3=M

Maybe that is wrong, but I think I did it right.. please let me know if it is wrong..
 
  • #73
I think I finally got it. But let me know. Okay so I converted 0.0150 arc seconds to degrees = 4.16666666x10^-6. Then I converted 42.00 AU to meters = 6283110582000. I also converted 12500.0 km to meters = 12,500,000. Then I also converted the 5.50 hours to seconds = 19,800. Thus all my labels should match what is on the other side.

For the mass I got = 5.29890904 x 10^-11 kg
Diameter = 456920.6955 meters
Volume = 4.994834581 X 10^16m^3
Density = 1.06087778x10^-27 kg/m^3

How does that look? Am I right maybe, I converted everything and double checked my math. Let me know please, this is due tomorrow.
 
  • #74
astronomystudent said:
For the mass I got = 5.29890904 x 10^-11 kg

That's about a ten billionth of a kilogram. Does that make sense to you?
 
  • #75
I did it again, maybe it was a math error and I got the mass to
Mass = 5.16169132 x 10^-9 kg.

Is that right? That seems better.
 
  • #76
Sorry, that was wrong too. I wasn't cubing the bottom. So the new mass is:
Mass = 3.30343624 x 10^-23 kg.

That must be right. I have double-checked and looked over my work/calculations.
 
  • #77
astronomystudent said:
Sorry, that was wrong too. I wasn't cubing the bottom. So the new mass is:
Mass = 3.30343624 x 10^-23 kg.

That must be right. I have double-checked and looked over my work/calculations.

First, I would suggest reflecting on how much a kilogram actually is. If necessary, do a google search on the masses of various objects and see if you can find something as massive as your Kuiper Belt object. If you're not sure what a Kuiper Belt object is, you might also want to look that up. Second, you may want to consider writing out your calculation. I can only guess as to where you are going wrong.
 
  • #78
You still have your diameter wrong. **Edit... Never mind, you got it right**. You'll need to fix that before you can get your volume and density.

Your formula for mass is wrong. Let me give you an example of how to extract M from that formula. I'll use the escape velocity formula to illustrate. (You don't need this formula for your question).

Ve = 1.4142 * sqrt (GM/r)

M is inside a root symbol. So you have to get it out. Square everything:

Ve^2 = 1.4142^2 * sqrt (GM/r)^2

The sqrt and the ^2 cancel out leaving you with:

Ve^2 = 1.4142^2 * GM/r

Cross multiply:

Ve^2 x r = 1.4142^2 x GM

Divide to isolate M:

M = (Ve^2 x r) / (1.4142^2 x G)

Use the same procedure to isolate M from your orbital period formula.
 
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  • #79
tony873004 said:
You still have your diameter wrong. You'll need to fix that before you can get your volume and density.

Are you sure about that? I'm getting the same result.
Your formula for mass is wrong.

Haw. Yeah, I just noticed that his/her formula for mass is inverted. His/her method for obtaining it from Kepler's Third Law was probably alright. I suspect he/she just made an algebra mistake.
 
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  • #80
Oops. I didn't notice he recomputed from his original 20 km answer. 456 km is correct.
 
  • #81
Isn't it 456 meters, not km? I am finishing the problem. It is due in an hour..
 
  • #82
astronomystudent said:
Isn't it 456 meters, not km? I am finishing the problem. It is due in an hour..

No, what you wrote in your other post is right -- 456 km and ~456000 meters. For the mass, just invert the calculation you were doing before.
 
  • #83
Okay I did it for the last time I think. And the numbers are worse, but I doubled checked it.

diameter = 456927.1982 meters
mass = 3.3913079 x 10^-25 kg
volume = 4.995047837 X 10^16 m^3
density = 6.78932795 x 10^-42 kg/m^3

I mean seriously.. i have no idea what is wrong.
 
  • #84
Invert the equation, I solved for M so I could isolate the x, are you saying that is wrong?
 
  • #85
astronomystudent said:
Invert the equation, I solved for M so I could isolate the x, are you saying that is wrong?

Yes, your algebra was probably wrong. Kepler's Third Law is

[tex]P^2=\frac{4\pi^2 a^3}{GM}[/tex]

Solving for M gives:

[tex]M=\frac{4\pi^2 a^3}{GP^2}[/tex]
 
  • #86
ok so the mass is 2.36 x10^16 kg and the density is 4.72269734.. so can i say the density = 4.72 kg/m^3?
 
  • #87
And no idea as to what it is composed of..
 
  • #88
astronomystudent said:
ok so the mass is 2.36 x10^16 kg

You're getting there, but this number is still much too small. Check to make sure your distances are in meters, your time is in seconds, and G=6.67e-11.
 
  • #89
You're right. I squared instead of cubed a
the mass = 2.95x10^24 kg
and the density = 59032871.67 kg/m^3
how can i rewrite the density, and how do i figure out what that is composed of?
 
  • #90
What it's made of will be your guess. Your options are pretty much: ice, rock, metal. It would be odd to find something of that size be pure metal, and being that it's in the Kuiper Belt, should give you a clue. But this is hypothetical, so don't rule anything out until you compare your values to the density of ice, rock and metal (iron). And it can be a combination of these things. Just take your best guess. That's all scientists do with this small amount of data.
 
  • #91
I am going to go with rock. Thank you very much for all of ya'lls help. It is greatly appreciated.
 
  • #92
astronomystudent said:
You're right. I squared instead of cubed a
the mass = 2.95x10^24 kg
and the density = 59032871.67 kg/m^3
how can i rewrite the density, and how do i figure out what that is composed of?

You could express it in grams / cm^3. Just divide by 1000.

You got the right answers.

This is where your teacher is weird! This object is more than half as massive as Earth, while only ~1/30 Earth's diameter. That's why it is so dense. It is a hypothetical question, so anything's possible. I'd like to know what your teacher says is the correct answer for its composition.

You should try to do #4 now, even though you don't need it. You almost had it and that's a good problem to understand.
 
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  • #93
Yeah, I went to class, and he told us the answer was METAL. And I asked why it was metal and he said it was much too dense to be rock or ice. Is taht true?
 
  • #94
astronomystudent said:
Yeah, I went to class, and he told us the answer was METAL. And I asked why it was metal and he said it was much too dense to be rock or ice. Is taht true?
Yes, but it's also too dense to be metal. Ask him what kind of metal.
 
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