# Excercise about a solar-like system

## Homework Statement

An Earth-like planet orbits around a Sun-like star with a circular orbit of period T=1 year. The system is very far from us. Assuming that our Earth is on orbit plane of the Earth-like planet, calculate:

i) the period of the partial occultation of the star from the planet;

ii) the resulting apparent magnitude variation of the star.

iii)Repeat the exercise for a giant planet, Jupiter-like, with an orbital period of T=4332 days, placed in the same system;

iv) Are the above mentioned phenomena observable with a ground-telescope?

[Data: Earth radius = 6370 km; Sun radius = 694000 km; 1 U.A. = 1.49 x 108 km; Jupiter orbit mean radius = 7.80 x 108 km; Jupiter mean radius = 69000 km]

## Homework Equations

This is a sketch of the eclipse: http://www.webalice.it/mizar02/figure/StarEcl.gif

So I've found the angle AOB as: α= 2 arcsin (REarth + RSun / 1 U.A. ).

the Stefan-Boltzmann law: Lstar=4πRstar2 σ T4

## The Attempt at a Solution

After the angle α is found, the eclipse period can be found as: Δt = (α/2π) T = 12.9 hours.

ii) Since luminosity can be written through the Stefan-Boltzmann law, I write the luminosity of the star before the occultation as: L1=4πRSun2 σ T4, and the luminosity during the eclipse as:

L2=4π(RSun2 - REarth2) σ T4.

Then, doing the ratio between these two luminosities, the magnitude difference can be found, and it results in a Δm of about 10-4.

Questions:

Is this part right (before I proceed to the other points)?
In particular, I'm not sure about the equation for L2: is the radiative surface equal to that I have reported?

i) 1 year
ii)Explain what is Rsun and Rearth .
you need apparent mag variation not luminos variation i think the ratio of fluxes σΤ^4 is equal to the ratio of appar magnitudes.
luminosity of star is always the same,but we can suppose it emits ration from a smaller surface
appar mag before/appar mag after = (L1/4πd ) / (L2/4πd)

You have to find the percentage of covered disk and say L2= L1/y

would be good approximation to subtract πRsun - πRearth = S' sun

and say S' sun /S sun = L 2 /L1 hence the ratio L2/L1 is known which means ratio of appar magnitudes is also known.

The question until now was extremely stupid which shows that the person who created is unlikely knows from astronomy ,amnd the last part requires knowledge that only kepler space telescope can detect earty like planets.Resolutions of ground based telescopes i think (with or withourt adaptive optics i cant remewmber) are 1'' .

when you subtract the apparent diam of the planet maybe you should take into account the distance of the planet from the star but then again if we suppose that the star is ligh years away from us ?i fthe exercise has given the distances of the planets from the stars you should find way to calculate the new apparent surfaces of the planets and subtr the from the Sun surface area.maybe you can use inverse square law

I'm sorry, i can not understand grammatic neither explanation.
As I said:
Δm = -2.5 log (L1/L2)

You say: You have to find the percentage of covered disk and say L2= L1/y
and I did that, and in fact this percentage depends on the covering planet radius.
You say: would be good approximation to subtract πRsun - πRearth = S' sun
but I really don't understand the general meaning of you explanation...

PS: what REarth and RSun can be if not the radii of Earth and Sun?

so what you did in the initial formula you gave you were subtracting the luminosity of the sun form the luminosity o fhte earth ....

S' is the new surface o fthe sun if we put the planet in fron t of ir

so what you did in the initial formula you gave you were subtracting the luminosity of the sun form the luminosity o fhte earth ....

S' is the new surface o fthe sun if we put the planet in fron t of ir

Your S' is not an area... so i don't understand.

D H
Staff Emeritus
Let's start from the start with your expression α= 2 arcsin (REarth + RSun / 1 U.A. ). Obviously there are some parentheses missing here. What I don't see is how you got the value 12.9 hours. Is that the value you know is correct? Using your numbers (Earth radius = 6370 km; Sun radius = 694000 km; 1 U.A. = 1.49 x 108 km) results in 13.1 hours. A value of 12.9 hours corresponds to using (Rsun-Rearth)/1 AU.

Which value is correct depends on what you are supposed to be measuring.

There is a slightly easier way to do this. You don't need arcsin. The small angle approximation is valid to more than four decimal places here.

The same concept of approximation applies to the variation in magnitude. You could use the Stefan-Boltzmann law in full to get the answer. The Earth-like (or Jupiter-like) planet is also radiating, so the "correct" answer should account for this radiation. However, since radiation is proportional to T4, you can ignore the radiation from the planet itself (the relative error that results from this simplification is on the order of 10-6 or so). That leaves you with a straightforward calculation that does not involve temperature at all for the reduction in intensity.

There's one problem here: The problem did not ask for the reduction in intensity. It asked for the "resulting apparent magnitude variation of the star". You need to convert that reduction in intensity to a reduction in apparent magnitude.

Thanks.
I repeated the calculation about the eclipse period, and it results in 12.9 hours again.
The α angle is equal to 9.3 x 10-3 radiants for you too?

However, I never thought about the possibility to "adding" the planet emission to the star emission.
In my opinion, it's natural to think about a decrease in luminosity caused from the planet transit.
(When we have a solar eclipse here on the earth we see the dark and not a "powerized sun" ).

You say that the calculus for the intensity is straightforward, but it's just what i can not understand, as you can see from my formula about the luminosity...

D H
Staff Emeritus
Regarding the ellipse period: Show your calculations. It appears you might be losing precision due to some intermediate result.

Regarding luminosity: You missed the point. To be pedantically correct one should incorporate the planetary emissions. However, this pedantic correctness will only affect the result in the sixth decimal place. Why bother? A real telescope, earth-based or space-based, will not see that error. One important aspect of physics is learning to distinguish between that which is important and that which isn't.

Finally, regarding apparent magnitude versus luminosity, your text or lectures must have something to say about magnitude and how it relates to luminosity. How do you calculate apparent magnitude?

I'm sure that is not an approximation error because i do a "unique" calculation with a calculator.
So i miss not decimals.
As I said: α=9.3 x 10-3 radiants.
then Δt=(α/2π)T=12.9 hours.

Hower, about the magnitude, I consider: m1-m2= - 2.5 log (L1/L2).
Is this not right?

EDIT: maybe the difference is in the A.U. definition: the exercise give 1.49 x 108 km, while i used the value that i always remember, that is 150 x 106 km.

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D H
Staff Emeritus
Show your calculation of that angle. Also note that 9.3 is not a very precise value. It is a good idea to use a couple of extra digits in your intermediate calculations. You are losing precision.

All your values are off by a bit from published values. I used your stated values. You should not get 12.9 hours using your calculation and those values. You should get 13.1.

α/2π=1.48 x 10-3
(α/2π) x T = 46 624,44 sec (/ 60sec / 60 min) = 12.95 hours

where α= 9.3 x 10-3 radiants and T=1 year = 31.5 Msec

I'm not loosing precision.
I calculated the eclipse period with a calculator so I use until 10 decimals from the first operation to the last.

Well, if i suppose A.U.=1.5 x 108 km the result is 13.02 hours.
So your result is surely more correct.
I apologize.

D H
Staff Emeritus
Your magnitude calculation looks good, more or less.

Could a ground-based telescope see this small a change?

Of course It can not.
But I don't know the limit, or the law, that governs the magnitude limit parameter.
I only know that 10-4 is a really low difference in magnitude.
Maybe we can say that it equals to a difference in luminosity equal to 1%, but this can not help me neither...

D H
Staff Emeritus
The Kepler mission can see variations as small as these. What's the difference? (Big hint: What is a telescope looking through that the Kepler satellite isn't?)

Of course the atmosphere is the problem.
But how can this be translated in numerical (quantitative) hints?

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D H
Staff Emeritus
Twinkle, twinkle little star ...

What is it?

D H
Staff Emeritus
Why do stars twinkle? Can modern astronomers overcome this, and if so, to what extent?

I'm sorry, I can not found out what is the order of magnitude scintillation, and so I can not asnwer your question...
Can you give me a link?

D H
Staff Emeritus