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
  • #36
When I look up brightness for Jupiter it comes up as -2.7 m. Is that something I could use. Could that be interpreted as the "apparent brightness near) and then I solve for the "apparent brightness far"? Or maybe that is all wrong, I don't know..
 
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  • #37
It is the brightness I am trying to solve for right?
 
  • #38
astronomystudent said:
It is the brightness I am trying to solve for right?

Yes. I'm a little concerned about your approach to all this. What are you using for a text? Have you read it?
 
  • #39
Yes, I have read the text. We get problems that our teacher makes up b/c he thinks the ones in the text are really too easy. These problems are much more dififcult than the examples given in the book. We are also given a supplement packet that our teacher authored full of facts, and defintions. It has a whole section on planets: that is where I got the brightness for Jupiter to be -2.7m. Do I use that?
 
  • #40
astronomystudent said:
Yes, I have read the text. We get problems that our teacher makes up b/c he thinks the ones in the text are really too easy. These problems are much more dififcult than the examples given in the book. We are also given a supplement packet that our teacher authored full of facts, and defintions. It has a whole section on planets: that is where I got the brightness for Jupiter to be -2.7m. Do I use that?

No, you need to use the inverse square law. Do your notes or textbook cover this anywhere? What is the name of the text?
 
  • #41
Textbook: Astronomy Today
Chaisson - McMillan
 
  • #42
I used the inverse square law to find the answer for the other part of this problem.

I can find the ratio if I apply the 1.0 A.U. distance from the Earth ---- Sun
If I do the following: (7.52 A.U./1.0 A.U.)^2 = 56.5504
Could this then be how much brighter it would appear from Earth if it was at a farther distance? Or am I totally off? I really am trying, I feel like you think I'm not.
 
  • #43
astronomystudent said:
Textbook: Astronomy Today
Chaisson - McMillan

Read (or reread) section 10.4 and see if you can answer the question:

How does the sun's apparent brightness change if you double the Earth's distance from the sun?
 
  • #44
F.Y.I. we never really use our textbook we use a program called Voyager and do most of our lessons on there (it is a computer program). It comes with a book entitled: voyages through space and time - project for Voyager II by Jon K. Wooley. This teaches us a lot of stuff.
 
  • #45
My section 10.4 has nothing to do with that. That entire chapter is about Mars and section 10.4 is entitled: The surface of Mars. Nothing in that section refers to what you are talking about, it talks about canyons and other things. Maybe you have a different edition. I'm not really sure..
 
  • #46
astronomystudent said:
I really am trying, I feel like you think I'm not.

No, I do think you're trying...the fourth problem is very difficult at this level. I think you should consider stepping back, though, cause you'll need a more complete understanding of what an inverse square law is before you can do this problem. Try starting with the question in my previous post.
 
  • #47
I just left you a note about that.
 
  • #48
astronomystudent said:
I just left you a note about that.

Please try to treat this more like a message board, not a chat room. It's awkward to do this when you're posting new things every other minute. If you have several things to say, put them all in one post. Here is a link to an explanation of the 1/r^2 law that I wrote a while back:

https://www.physicsforums.com/showthread.php?t=70634"

Read it and see if you can answer the question in my last post.
 
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  • #49
This isn't directly related to your problem, but if you know flux, #4 will be easier. Since your teacher didn't talk about it, I'll give you a quick review.

Flux is the amount of something (in this case, energy) that is transmitted or received by an area of surface. I like to think of it as departing and arriving flux, just like the airport terminals.

Imagine you have a light bulb that is spherical, and 10 cm in diameter. It is 100 W (Watts, also called Joules / S). How many Watts does each square centimeter of this light bulb emit?

What's the surface area of a sphere? 4 pi * r^2. You'll find the formula for surface area of a sphere poping up in lots of flux-related problems.

So the 10 cm diameter light bulb has a surface area of:
(Diameter is divided by 2 since the forumula wants radius.)

surface area = 4 * pi * r^2
surface area = 4 * pi * (10cm/2)^2 = 314.16 cm^2

Since the light bulb emits 100 W, each square cm emits 100W / 314.16 cm^2 = 0.318 W/cm^2. This is the departure flux from the lightbulb.

Now imagine the light bulb is in a spherical room 10 meters in diameter. How much energy does each square inch of this sphere receive from the light bulb?

What is the surface area of this sphere?

surface area = 4 * pi * r^2

surface area = 4 * pi * ( (10m/2) * (100cm/m) )^2 = 3141592.7 cm^2

So each square centimeter of this room-sized sphere would receive 100W / 3141592.7 cm^2 = 3.18*10^-5 W/cm^2. This is the arriving flux.

When you want to compute something such as how much flux of sunlight does the Earth receive, you just treat the Earth as a point on the surface of a sphere whose radius is 1 AU.

Notice a similarity in the answers? The departing flux, from d=10 sphere is 0.318. The arriving flux from d=1000m is 3.18*10^-5 W/cm^3.

d2 is 100 times more than d1.
f2 is (1/100^2) that of flux 1. (inverse square law)

This is why you only need inverse square law to solve for #4. You'll need flux later, but knowing it now will help you wrap your mind around this problem a little better.
 
  • #50
Ok, I have looked at these sites/notes on the inverse square law. However, these seem just like a string of examples. I don't have anything that is in WATTS, nor do I have luminosity. I just have the distance. I am still confused as to what it is here that you are suggesting I do to solve this problem. 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... maybe I am just dumb/oblivious. But I am REALLY confused...
 
  • #51
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?
 
  • #52
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?
 
  • #53
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.
 
  • #54
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!
 
  • #55
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:

[tex]L = \frac{C}{d^2}[/tex]

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?
 
  • #56
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.
 
  • #57
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!
 
  • #58
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?
 
  • #59
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.
 
  • #60
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?
 
  • #61
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?
 
  • #62
The length of the circumference of a circle is 2 [itex]\pi[/itex] r - and a circle has 360°.

In general the length of a circular segment, s, is given by s = r [itex]\theta[/itex], where [itex]\theta[/itex] 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 [itex]\pi[/itex] radians is equivalent to 360°.
 
  • #63
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?
 
  • #64
Soh Cah Toa?
 
  • #65
astronomystudent said:
Soh Cah Toa?

Which is it?
 
  • #66
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.
 
  • #67
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.
 
  • #68
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?
 
  • #69
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?
 
  • #70
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
 

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