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
  • #1
astronomystudent
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I need help identifying equations and trying to start these problems. Thanks...

1.) If Jupiter has an average temperature of 125 K, at what wavelength is it emitting the most energy according to Wien's Law?

2.) If a new planet is discovered beyond the orbit of Neptune which has a radius three times Earth's radius and a densiy of 1.2 g/cc, what is the mass of the new planet in terms of Earth's mass?

3.) If Phobos had a semimajor axis of 29,000 km and Mars has a mass of 0.16 M(lowered E), what would be its period around Mars in hours? Be careful with your units here!

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!

5.) 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 think it is composed of?
 
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  • #2
astronomystudent said:
1.) If Jupiter has an average temperature of 125 K, at what wavelength is it emitting the most energy according to Wien's Law?
Wein's law is an equation. Look it up in your book's index, and solve it.
2.) If a new planet is discovered beyond the orbit of Neptune which has a radius three times Earth's radius and a densiy of 1.2 g/cc, what is the mass of the new planet in terms of Earth's mass?
Do you know how to find the volume of a sphere of radius r?

I'll continue with further problems once you start making an attempt. We're not going to do your homework for you.

- Warren
 
  • #3
Answers To Numbers 1 & 2

I tried to do the first two problems. I looked up the formulas and stuff. I am still going to need help with the others. I just don't know where to begin.

1.) Wien's Law = (wavelength)(Temperature)= 2.898x10^7(constant)
(wavelength)(125 K) = 2.898X10^7
wavelength= 2.898x10^7/125 K
wavelength= 231,840 Angstroms
ANSWER = 231,840 Angstroms

2.) V=4/3(pi)r^3
density= mass/volume
New planet radius = 19134.45 km
V = 4/3(pi)(19134.45)^3
V= 2.93 X10^13 g/cc

density = mass/volume
1.2g/cc = mass/2.93 x10^13
(1.2g/cc)(2.93 x10^13) = mass
mass = 3.58 X10^13 kg
ANSWER = 3.58 x10^13 kg

Please let me know about these answers, and any other help you can offer on the other problems.
 
  • #4
astronomystudent said:
ANSWER = 231,840 Angstroms

Right


2.) V=4/3(pi)r^3
density= mass/volume
New planet radius = 19134.45 km
V = 4/3(pi)(19134.45)^3
V= 2.93 X10^13 g/cc

With the units you put in, this is the volume in km^3. g/cc is a unit of density.


density = mass/volume
1.2g/cc = mass/2.93 x10^13
(1.2g/cc)(2.93 x10^13) = mass
mass = 3.58 X10^13 kg
ANSWER = 3.58 x10^13 kg

This answer is way too small because your units are inconsistent between density and volume. Remember, g/cc means grams per centimeter cubed.
 
  • #5
You got #1 right. But keep in mind that K does not mean constant. It means Kelvins. It's there to make the units dimentionally consistant.

Also, is Angstroms the best unit to express the answer?

#2. Wrong.
Becareful with your units. To get volume, you cubed kilometers, so you should end up with km^3, not g/cc.

Always do your math with your units, and never plug in your numbers until you have your formulas written to solve for the unknown. Then if your units cancel properly, you can be more confident with your answer.

Edit

** ST beat me to it :) . He and several others were quite helpful last year when I had similar questions.**
 
Last edited:
  • #6
So is the answer to number 2 wrong or do I just have it labeled with the wrong units? Also, can I get some help on the other problems by chance?
 
  • #7
The answer is wrong.
People here will be happy to help you with the others provided that you at least start the problem so we can see where you went wrong.

#3, do you have a formula for period?
And just a clue, you can Google for the answer to check yourself.
 
  • #8
astronomystudent said:
So is the answer to number 2 wrong or do I just have it labeled with the wrong units?

As tony said, you didn't get the right answer. I was just saying that you would get the right answer if you used a set of consistent units -- your equations are correct.
 
  • #9
Forget about my hint to Google for the correct answer to check yourself. The problem says [if]. They changed both Mars' mass and Phobos' Semi-major axis.
 
  • #10
For problem #2 I am still confused as to what I need to do. How do I keep consistent units, do I need to convert to different units.
 
  • #11
astronomystudent said:
For problem #2 I am still confused as to what I need to do. How do I keep consistent units, do I need to convert to different units.

For example, if you want the mass in kilograms, you have:

Mass (in kg) = Density (in kg per Liter) x Volume (in Liters)

Notice how the units cancel one another on the right side of the equation: kg -> kg/L*L. You can do this other ways:

Mass (in Earth masses) = Density (in Earth masses per centimeters cubed) x Volume (in centimeters cubed)

If the units you're given in the problem don't work out like in the above equations, then you need to convert them before plugging them into the equations.
 
  • #12
2.) I understand what you are saying. I will try again after I convert my units and post my new work.

3.) I am trying to find the period in "hours". I found one equation for period and it reads as follows: P(years)=R(A.U.)^3/2
I have the semimajor axis which I think I plug in for P. Years is a gravitational constant, but then I have the mass and I don't know what to do with that. Is there another formula/equation for the period?

4.) I assume I use the equation for energy, but the energy equations don't include A.U. So I am lost? Also once you solve energy how do you begin to fid the "size and brightness" from Earth? Would that include telescope calculations?

5.) No idea, this problem is very difficult.
 
  • #13
#3. The way you wrote it is basically Kepler's 3rd law. P is period, not semi-major axis. R is semi-major axis in A.U. (1 AU = Earth Sun average distance).

Look on Wikipedia for "Orbital Period". They've re-written that equation you give for more general use, with standard SI units.

#4. This one is easier than you think. You don't need energy equations, just the inverse square law. Basically, it says that if planet 2 is three times as far from the sun as planet 1, then it will receive 1/3^2 the amount of energy as planet 1.

#5. For part 1, look up small angle approximations. Part 2 is similar to #2 and #3 combined.
 
  • #14
Okay, I think maybe I have the correct answer to number 2.

I converted the radius to cc.
So the new radius = 1.931445 x 10^13 cc
Thus the new volume = 2.93 x 10^40 cc
And the new mass = 3.52 x 10^40 g
ANSWER = 3.52 x10^40 g

Is that right?
 
  • #15
For number four:
(apparent brightness (near)/apparent brightness (far)) = (distance (far)/distance (near))^2

so i...
(apparent brightness (near)/apparent brightness (far))= (7.52 AU/5.20 AU)2
answer = 2.09 x 10^9 (not sure of units)

I didn't understand your other notes about "the planet is 3x as far from the sun as planet 1, then it will receive 1/3^2 amount of energy as planet 1. Is what I did all I need to do? Am I missing another step and I am confused about which units go with my answer?
 
  • #16
astronomystudent said:
I converted the radius to cc.
So the new radius = 1.931445 x 10^13 cc

cc is a unit of volume. It stands for "centimeters cubed". I don't know how you got the above number, but you want the radius to be in centimeters.
 
  • #17
astronomystudent said:
I didn't understand your other notes about "the planet is 3x as far from the sun as planet 1, then it will receive 1/3^2 amount of energy as planet 1. Is what I did all I need to do? Am I missing another step and I am confused about which units go with my answer?

The inverse square law will give you the relative amount of sunlight received at each planet. However, the brightness of a planet will depend not just on the amount of sunlight it receives and reflects, but also on how much of that light we get after the reflection. Can you see how these combined effects might scale with distance?
 
  • #18
Okay, let's try number 2.

I converted the radius to cm: 1,913,445,00 cm
new volume = 2.93 x 10^28 cm

new mass = 3.52 x 10^28 g

How does that look?

I looked at your notes on number 4 and I am still confused. What exactly did I solve for, and how does that relate to what I need to do to solve the rest of the problem? Am I done or not? And in terms of distance, I am lost..

Thank you...
 
  • #19
#2, no. This is where you made your error?
V = 4/3(pi)(19134.45)^3

Do it with units:

V = 4/3(pi)(19134.45 km)^3
V = 4/3(pi)19134.45^3 km^3
V = 9491269039275.3 km^3

But since you're given density in cm^3 (also called cc), you don't want your volume to be in km^3. You want it to be in cm^3. What could you do to the original formula
V = 4/3(pi)(19134.45)^3
to ensure that your answer comes out in cm^3?

How do you convert km to cm?
 
  • #20
That looks good for number 2. But read the question again. They don't want you to express your answer in grams.
 
  • #21
astronomystudent said:
(apparent brightness (near)/apparent brightness (far))= (7.52 AU/5.20 AU)2
answer = 2.09 x 10^9 (not sure of units)

There are no units here. You're simply stating that the closer planet receives 2.09 x 10^9 times as much energy as the farther planet.

Does that answer seem reasonable? You did it right, but check your math.
 
  • #22
Okay, I am not sure whether my answer for number 2 is write or wrong. You wrote that it was wrong, and then that is looked good? I need to convert my answer from grams to kg right b/c that is the units for mass right?

For problem 4 is that all I have to do for the problem? That is what I am asking. There are two parts to this question but I think i have only done one part?
 
  • #23
You got it right. I said wrong before I noticed you posted your correct answer. 3.52 x 10^28 g is correct.

It's not asking for kg either. It wants you to express the new planet's mass in terms of Earth masses. You'll need to know the mass of the Earth to do this.
 
  • #24
Part 4 gets trickier. It tells you how far the planets are from the Sun, but then asks you how that affects their brightness from Earth. Hint: It gives you their Sun distances in AU. How far is the Earth from the Sun in AU?
 
  • #25
For number 2: the mass of the Earth is 5.97 x 10^24 kg.. so I do need to convert my answer to Kg but what dos "in terms of Earth's mass" do I divide

Also for number 3: I used the following equation and got an answer
T = 2(pi) (square root of a^3/u) where U = GM
T = 2(pi) (square root of (29,000^3)/(6.67x10^-11)(9.552x10^23)
T = 3.88746768x10^-11 hours
Answer = 3.9 x 10^-11 hours


I am still working on number 4 based on the information you gave me.
 
  • #26
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?
 
  • #27
Your answer for #2 is correct, 5.9Me.
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.
 
  • #28
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
ANSWER = 34 Hours

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.
 
  • #29
astronomystudent said:
T= 34.1479228
ANSWER = 34 Hours
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?
 
  • #30
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.
 
  • #31
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.
 
  • #32
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?
 
  • #33
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"?
 
  • #34
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?
 
  • #35
No, we have not studided "flux"
 
<h2>1. What are some common problems in astronomy?</h2><p>Some common problems in astronomy include understanding the origins and evolution of the universe, studying the behavior and composition of celestial objects, and predicting and tracking astronomical events such as eclipses and meteor showers.</p><h2>2. How can I solve problems in astronomy?</h2><p>Solving problems in astronomy often involves using mathematical equations and models to analyze and interpret data collected from observations and experiments. It also requires critical thinking and creativity to come up with new theories and explanations for observed phenomena.</p><h2>3. What are some resources for help with astronomy problems?</h2><p>There are many resources available for help with astronomy problems, including textbooks, online tutorials and courses, scientific journals and articles, and professional organizations such as the American Astronomical Society. Additionally, seeking guidance from experienced astronomers and participating in peer discussions can also be helpful.</p><h2>4. How can I improve my problem-solving skills in astronomy?</h2><p>To improve problem-solving skills in astronomy, it is important to have a strong foundation in mathematics and physics. Practicing with a variety of problems and seeking feedback from peers and mentors can also help refine problem-solving techniques and critical thinking abilities.</p><h2>5. Are there any common mistakes to avoid when solving astronomy problems?</h2><p>Some common mistakes to avoid when solving astronomy problems include incorrect use of mathematical equations, misinterpretation of data, and overlooking important details or assumptions. It is also important to critically evaluate solutions and consider alternative explanations before accepting a conclusion.</p>

1. What are some common problems in astronomy?

Some common problems in astronomy include understanding the origins and evolution of the universe, studying the behavior and composition of celestial objects, and predicting and tracking astronomical events such as eclipses and meteor showers.

2. How can I solve problems in astronomy?

Solving problems in astronomy often involves using mathematical equations and models to analyze and interpret data collected from observations and experiments. It also requires critical thinking and creativity to come up with new theories and explanations for observed phenomena.

3. What are some resources for help with astronomy problems?

There are many resources available for help with astronomy problems, including textbooks, online tutorials and courses, scientific journals and articles, and professional organizations such as the American Astronomical Society. Additionally, seeking guidance from experienced astronomers and participating in peer discussions can also be helpful.

4. How can I improve my problem-solving skills in astronomy?

To improve problem-solving skills in astronomy, it is important to have a strong foundation in mathematics and physics. Practicing with a variety of problems and seeking feedback from peers and mentors can also help refine problem-solving techniques and critical thinking abilities.

5. Are there any common mistakes to avoid when solving astronomy problems?

Some common mistakes to avoid when solving astronomy problems include incorrect use of mathematical equations, misinterpretation of data, and overlooking important details or assumptions. It is also important to critically evaluate solutions and consider alternative explanations before accepting a conclusion.

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