## Main Question or Discussion Point

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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chroot
Staff Emeritus
Gold Member
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

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

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

SpaceTiger
Staff Emeritus
Gold Member
astronomystudent said:
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
This answer is way too small because your units are inconsistent between density and volume. Remember, g/cc means grams per centimeter cubed.

tony873004
Gold Member
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.

Edit

** ST beat me to it :) . He and several others were quite helpful last year when I had similar questions.**

Last edited:
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?

tony873004
Gold Member
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.

SpaceTiger
Staff Emeritus
Gold Member
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.

tony873004
Gold Member
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.

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.

SpaceTiger
Staff Emeritus
Gold Member
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.

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.

tony873004
Gold Member
#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.

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

Is that right?

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?

SpaceTiger
Staff Emeritus
Gold Member
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.

SpaceTiger
Staff Emeritus
Gold Member
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?

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...

tony873004
Gold Member
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

How do you convert km to cm?

tony873004
Gold Member
That looks good for number 2. But read the question again. They don't want you to express your answer in grams.

tony873004
Gold Member
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.

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?

tony873004
Gold Member
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.

tony873004
Gold Member
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?

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.