Astronomy Problems: Please Look/help

[SpaceTiger]

Still the only part of the problem I have done is (7.52/5.20)^2= 2.09. And I don't even know what that means...

That was the purpose of the link I sent you, as well tony's last post. You really don't see any connection between those things and the above calculation?

 

[astronomystudent]

Okay, well I assume I am going to disregard the work that I have done for this problem. I guess you are saying it doesn't really apply to this problem.

The flux from a point source falls off as 1/r2.
So the brightness of Jupiter or any other planet has is related to the solid angle subtended by the planets area as suggested by flux? As the the distance to Jupiter increases, the solid angle subtended by its area, decreases?

For the first part of the problem I need to find the ratios of the spheres that correspond to Jupiter.

In the first part, simple find the ratio of the spheres corresponding to Jupiter orbit, and the sunlight should decrease by the square of the ratio of the closer/further, or (5.2/7.52)2 = 0.478, which is the inverse of your 2.09.

As for the earth, one has to consider the position of the Earth at 1 au from the sun. So the light from the sun is less, and the reflected light received is less?

Is this right, do I even need calculations for the second part of this question?

 

[tony873004]

Try drawing a picture.

Draw a horizontal line across a piece of paper.

Draw a small circle on this line on the left edge of the paper to represent the Sun.

Draw another small circle on this line 1 inch from the left edge and label it Earth.

Draw another circle on this line 5.2 inches from the left edge. Label it Jupiter's actual position.

Draw another circle 7.52 inches from the left edge. Label it Jupiter's hypothetical position.

What are the ratios of real Jupiter and hypothetical Jupiter's distances from the Sun?

What are the ratios of real Jupiter and hypothetical Jupiter's distances from Earth?

I think it's safe to assume that the teacher meant when the planets are at their closest to each other (conjunction / opposition) or you'd need to express your answer as an integral which seems way beyond the scope of this question.

 

[astronomystudent]

WHAT??

I drew the picture, but I don't understand how this is solving the problem. I have been working this problem for the past three days. I am in INTRO ASTRONOMY. Am I just making it more difficult than it is.. here is the problem again:

4.) If Jupiter were really 7.52 AU from the Sun instead of 5.20 AU, what would the affect be on the energy it gets from the Sun? What would be the affect on its appearence (size and brightness) from Earth? Careful, these are two different questions!

 

[SpaceTiger]

In the first part, simple find the ratio of the spheres corresponding to Jupiter orbit, and the sunlight should decrease by the square of the ratio of the closer/further, or (5.2/7.52)2 = 0.478, which is the inverse of your 2.09.

Excellent!

As for the earth, one has to consider the position of the Earth at 1 au from the sun. So the light from the sun is less, and the reflected light received is less?

That's right. Now, the last step is to combine those two facts. If the total luminosity reflected by the planet is is proportional to the light it receives from the sun:

L = \frac{C}{d^2}

where C is some constant that we don't need to worry about for the problem and d is the planet's distance from the sun. How, then, does the total flux we receive from the planet depend on its distance from the sun?

 

[tony873004]

WHAT??
I drew the picture, but I don't understand how this is solving the problem. I have been working this problem for the past three days. I am in INTRO ASTRONOMY. Am I just making it more difficult than it is.. here is the problem again:
4.) If Jupiter were really 7.52 AU from the Sun instead of 5.20 AU, what would the affect be on the energy it gets from the Sun? What would be the affect on its appearence (size and brightness) from Earth? Careful, these are two different questions!

Yes, you're making it more difficult than it needs to be, but that's ok. After you get it you'll wonder how it ever confused you.

You already got the 2.02. You made a math mistake by putting an exponent after it. 2.02 does not equal 2.02 x 10^9. It's just 2.02. Real Jupiter receives 2.02 times as much light from the Sun as hypothetical Jupiter. Hypothetical Jupiter receives 1/2.02 or 0.495 times as much energy as real Jupiter. Just to visualize, round off these numbers to 2 and 0.5. Real Jupiter receives twice as much energy as hypothetical Jupiter. Hypothetical Jupiter receives half as much energy as real Jupiter.

And you did this with the numbers 5.2 and 7.52 which are the real Jupiter and hypothetical Jupiter distances from the Sun.

Now what are real Jupiter's and hypothetical Jupiter's distance from Earth? Apply the same logic.

 

[astronomystudent]

Okay so I found the ratio, it was the inverse of my old calculation so it is 0.478? Okay so you are saying I use this new equation to solve for the rest of the probelm. Don't I need to know what "C" is, if it is a constant? And what do I plug in for L or D. I am a little confused again. But it looks like I am finally getting somewhere.. YES!

 

[SpaceTiger]

Okay so I found the ratio, it was the inverse of my old calculation so it is 0.478? Okay so you are saying I use this new equation to solve for the rest of the probelm. Don't I need to know what "C" is, if it is a constant?

Normally, yes, but remember that you're not solving for the brightness itself, you're solving for the ratio of brightnesses. How does brightness depend on luminosity?

 

[astronomystudent]

I emailed my teacher, and he said he made a mistake in this problem. You were right, it would involve integrals. In that case, I have to solve this porblem:
If the size of a new Kuiper object is 0.0150 arc sec in angular size as seen from a distance of 42.00 AU, what is the true diameter? If it has a satellite with a period of 5.50 hours at a semimajor axis of 12500.0 km, what is the mass? What is the resulting density? What do you it is composed of?
I know that I use V=4/3(pi)r^3 to get the density and previous equations to solve for mass. I do n't know how I am supposed to apply the 0.0150 arc seconds or distance to solve for the diameter.

 

[SpaceTiger]

I emailed my teacher, and he said he made a mistake in this problem.

You mean #4? I don't think it's necessary to do an integral...but anyway, does this mean you don't have to do it?

I do n't know how I am supposed to apply the 0.0150 arc seconds or distance to solve for the diameter.

What are arcseconds a measure of?

 

[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.

 

[astronomystudent]

I am going to go with rock. Thank you very much for all of ya'lls help. It is greatly appreciated.

 

[tony873004]

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.

 

[astronomystudent]

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?

 

[tony873004]

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.