# Astronomy Problems: Please Look/help!

#### SpaceTiger

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astronomystudent said:
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

Gold Member
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

Staff Emeritus
Gold Member
astronomystudent said:
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

Gold Member
astronomystudent said:
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

Staff Emeritus
Gold Member
astronomystudent said:
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

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Gold Member
astronomystudent said:
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

Staff Emeritus
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

Staff Emeritus
Gold Member
astronomystudent said:
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?

Soh Cah Toa?

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#### 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

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Gold Member
astronomystudent said:
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

Gold Member
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?

Last edited:

#### astronomystudent

So I am gonna 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

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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?

#### 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.