Astronomy Problems: Please Look/help

[astronomystudent]

Yeah, he said it is too difficult so now I don't have to do it. I am still interested in trying to solve it for extra credit or something.

But I have to do the other problem for sure.

Arcseconds help with distance, right and one arcsecond is equal=1/3600 of a degree. How does that apply?

 

[Astronuc]

The length of the circumference of a circle is 2 \pi r - and a circle has 360°.

In general the length of a circular segment, s, is given by s = r \theta, where \theta is the angle (expressed in radians) subtended by the arc of length s. For very large r or very short s, s can be treated as a straight line segment.

2 \pi radians is equivalent to 360°.

 

[SpaceTiger]

Arcseconds help with distance, right and one arcsecond is equal=1/3600 of a degree. How does that apply?

See if you can draw a triangle, where one side is the diameter of the object and one side is the distance. Can you see how the diameter might be related to the angle and the distance?

 

[astronomystudent]

Soh Cah Toa?

 

[SpaceTiger]

Soh Cah Toa?

Which is it?

 

[astronomystudent]

I don't know though. The only thing I am certain of is where the angle goes. I assume the diamter would be opposite of the angle. But I don't know where to place the distance of 42.00 AU which I assume I am going to need to convert to km at some point.

 

[SpaceTiger]

I don't know though. The only thing I am certain of is where the angle goes. I assume the diamter would be opposite of the angle.

That's right. It turns out that it doesn't matter whether you call the distance the adjacent or the hypotenuse when your angle is very small (as Astronuc was saying). Both the tangent and the sine function give the same value in this limit.

Given that, can you figure out the relationship between the angle, the distance, and the diameter?

...which I assume I am going to need to convert to km at some point.

As long as your diamter and distance are in the same units, it doesn't matter what you convert to.

 

[astronomystudent]

Okay, so if I use TANGENT. Then my opposite: is the diameter, the adjacent is my distance which i have converted from AU ---- KM now it is 6283110582 km.

I converted arcseconds to degrees by dividing 0.0150/3600 = 4.16666667X10^-6 degreees.

Using tangent I got the diameter to be: 20.5883110582 km
Is this right. Now can I use these numbers to solve for mass and density?

 

[astronomystudent]

Then I found the mass by using the equation for orbital period
mass = 2.54894772 x10^-21 kg

i calculated volume = 4569.416923
and then solved for density = 5.5782778x10^-25

this doesn't seem right? where did i go wrong?

 

[astronomystudent]

I have looked at this problem again. Time is running out this is due tomorrow. But from what I worked. I think I need to convert to meters so that it cancels out with the other labels specifically, the gravitational constant that is in meters. Also do i need to change T= which is given to me in hourst to seconds and then reconvert b/c seconds is the only thing that doesn't cancel in the problem? Maybe this is all wrong. But i really need help. And this is all i got

 

[tony873004]

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?

 

[astronomystudent]

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..

 

[astronomystudent]

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.

 

[SpaceTiger]

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?

 

[astronomystudent]

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.

 

[astronomystudent]

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.

 

[SpaceTiger]

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.

 

[tony873004]

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.

 

[SpaceTiger]

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.

 

[tony873004]

Oops. I didn't notice he recomputed from his original 20 km answer. 456 km is correct.

 

[astronomystudent]

Isn't it 456 meters, not km? I am finishing the problem. It is due in an hour..

 

[SpaceTiger]

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.

 

[astronomystudent]

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.

 

[astronomystudent]

Invert the equation, I solved for M so I could isolate the x, are you saying that is wrong?

 

[SpaceTiger]

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

P^2=\frac{4\pi^2 a^3}{GM}

Solving for M gives:

M=\frac{4\pi^2 a^3}{GP^2}

 

[astronomystudent]

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?

 

[astronomystudent]

And no idea as to what it is composed of..

 

[SpaceTiger]

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.

 

[astronomystudent]

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?

 

[tony873004]

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.