Okay, I think I understand what is meant in problem number 2
I divided the converted mass (kg) of the "new planet" by the mass of the Earth.
Therefore the answer is: 5.9 M(lowered E)? Is that what is meant?

tony873004
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You could have also converted the mass of the Earth into grams. That's easy. Just add 3 to your exponent.
5.97x10^24 kg =
5.97x10^27 g

#3, Does that seem like a reasonable answer? That means Phobos would go around Mars millions of times a second.

You're doing it right, but what are the units of G? And what are the units for Semi-major axis? Are they consistent? Also, watch out how you enter this into your calculator. It's a^3/(GM), not (a^3/G)*M which is how my calculator would want to do this problem if I entered it the way you typed it.

For problem number 3: I checked the units. If I convert the semi-major axis to meters than everything cancels out but seconds, because G's units are m^3/kg^-1/s^-2. The kilograms cancels out with the mass of the planet..

Thus after making the following changes: T = 122932.522 seconds. It wants the answer in hours so you convert T seconds ------> hours

T= 34.1479228

Is that right?

For number 4: I am still lost. I have that one answer, but I am lost as what to do with the distances and how to apply your hint to the problem.

SpaceTiger
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astronomystudent said:
T= 34.1479228
Is that right?
Yep.

For number 4: I am still lost. I have that one answer, but I am lost as what to do with the distances and how to apply your hint to the problem.
This one's pretty tricky for an intro astronomy course. Try answering these two questions:

1) How is a planet's total reflected light related to its distance from the sun?
2) How is a planet's brightness related to its total reflected light and its distance from earth?

SpaceTiger said:
1) How is a planet's total reflected light related to its distance from the sun?
2) How is a planet's brightness related to its total reflected light and its distance from earth?
Still, problem 4. For Jupiter, the mean distance from the Sun is 5.2028 A.U., this is already given. The distance from the Earth to the Sun is 1 A.U. Does this have anything to do with the albedo? Is that a measurement of "brightness" or does that not have anything to do with it. Magnitude? Otherwise, I have no idea. Also, I am still unclear about what I have already solved for.
I did (7.52/5.20)^2 = 2.091360947 which is 2.0 X10^9.

SpaceTiger
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astronomystudent said:
Still, problem 4. For Jupiter, the mean distance from the Sun is 5.2028 A.U., this is already given. The distance from the Earth to the Sun is 1 A.U. Does this have anything to do with the albedo?
The brightness of a planet does depend on albedo, but does the albedo change when you change its distance from the sun?

I did (7.52/5.20)^2 = 2.091360947 which is 2.0 X10^9.
2.09 is not equal to 2.0 x 109. The latter is exponential notation for two billion.

So 2.09 is just 2.09 no exponents involved.

I think the albedo changes with temperature and it is dependent on the "surface." I don't think it changes with distance? But, I'm not sure. Am I on the right track? I Just don't know where I am going with this?

SpaceTiger
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astronomystudent said:
So 2.09 is just 2.09 no exponents involved.
I think the albedo changes with temperature and it is dependent on the "surface." I don't think it changes with distance? But, I'm not sure. Am I on the right track? I Just don't know where I am going with this?
If you let Jupiter settle into a new equilibrium at a larger distance from the sun, you might see a change in the albedo, but those effects will be negligible for this calculation. Only concern yourself with brightness and distance. Did you study "flux"?

I am just lost on this problem. I don't even know what the 2.09 is in reference to. Are the units on that A.U.? Or do they cancel out or what?

No, we have not studided "flux"

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..

It is the brightness I am trying to solve for right?

SpaceTiger
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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?

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?

SpaceTiger
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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?

Textbook: Astronomy Today
Chaisson - McMillan

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.

SpaceTiger
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astronomystudent said:
Textbook: Astronomy Today
Chaisson - McMillan

How does the sun's apparent brightness change if you double the earth's distance from the sun?

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 alot of stuff.

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..

SpaceTiger
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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.

I just left you a note about that.

SpaceTiger
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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:

Read it and see if you can answer the question in my last post.

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tony873004
Gold Member
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.

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....