# How far would a planet be from the Earth, when its....

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1. Mar 7, 2016

### wiegetz

1. The problem statement, all variables and given/known data
How far would be a planet from the earth, when its period would be 2 years?
T = 2 years/730 days
a = 150*106km
2. Relevant equations
a3/T2 = C
(C is the Kepler-Constant)
3. The attempt at a solution
I tried inserting T in days and years, but I always get a wrong solution, since C (The distance from earth) does need to increse if it needs longer for its period. Any idea why?

2. Mar 7, 2016

### drvrm

i think usually ( T^2/ R^3) is a constant and R is the average distance from the mother planet- say Sun
for
Earth

T = 3.156 x 10^7 s ; R = 1.4957 x 10^11 m ; Constant( T^2 /R^3 ) = 2.977 x 10-19
so place the values properly and see!

3. Mar 7, 2016

4. Mar 7, 2016

### wiegetz

I cannot follow you how you got the values for T and R. The teacher has given us a, a = 150*106km and T = 2 years/730 days. I can follow your T which is just re-calculated to seconds. Do I need to use seconds in this term to get out a correct answer as well? And why would the teacher give us a then?

(TBH, the tasks of this teaching can be quite annoying)

5. Mar 7, 2016

### Staff: Mentor

C is not a distance from Earth. It's a constant of proportionality and has units of distance3 / time2 as you've written the ratio.

According to Kepler's laws the square of the period is proportional to cube of the distance, as you've noted. That makes it convenient to set up these sorts of problems as ratios so that the actual constant of proportionality cancels (drops out).

In short:
$$\frac{T_2^2}{a_2^3} = \frac{T_1^2}{a_1^3}$$
You can work in any convenient units so long as you remain consistent. Since you know values for the Earth, and the problem is stated in terms of Earth years, why not work with "Earth units"? So the time unit is the year and the distance unit is the AU. Just plug in the known values in those units and solve for the unknown quantity. Once you have the result in "Earth units", then you can convert to whatever units are convenient for you. For example, if you get a value in AU you can convert that to meters or miles or whatever else you require.

6. Mar 7, 2016

### wiegetz

Thank you for your response. Sadly I still do not know what to do, if I use the formula you gave to me, I get approx. 1,18-24 which can be only wrong. I do not know what to do with this "simple" forumla. Frustrating :( I need to get a formula with this one, where the results increses if I increase a in that case, which I simply cannot can get.

BTW, is that a rather simple or complex task (So in which forum it should belong to?) I'd say simple.

7. Mar 7, 2016

### SammyS

Staff Emeritus
.Show how you put things together when you got that answer.

8. Mar 7, 2016

### wiegetz

If you need to, why not? I doubt it will help you though solving my problem.

I tried out following things: T2/a3
and a3/T2
and a/T
and T/a

some are sensless, but trying does not harm. Still I got a bad result: What I need is, if I increase T, the result does need to increse, not decrease (since a planet wont be closer if it is rotating slower to its host star). Maybe a mistake from the teacher? Or is the solution so simple that I cannot see it.

9. Mar 7, 2016

### Staff: Mentor

Take the formula as it was shown in post #5 and assign values to the variables. Hint: If you choose $T_1$ and $a_1$ to represent the period and orbit radius for the Earth, then their values are 1 yr and 1 AU. You can just assume the units and plug in 1 and 1 for their values. Plug in what you are given for the mystery planet. What's left that doesn't have a value?

10. Mar 8, 2016

### wiegetz

In short:
$$\frac{T_2^2}{a_2^3} = \frac{T_1^2}{a_1^3}$$

I see know how easy it is. Could you say if my result is correct? : I inserted for T22 2 years, and for a 1 AU (Since how you said, but 150000000km equals 1,0026880684005897 AU (1AU) ) . Then I became 4 (I think 4 AU?) As a result, which should be 600000000km, so 4 times the distance that earth is away from the sun. The formula makes sence, since if I increase T the value increases as well.

If my result is correct, how would I need to proceed if I would like to use 150*106km, unconverted for a?

Until here, great help, I like the fact that you give only advice and not a proper result (no joke) So I can learn more. The old sites I visit always gave a result, which is in my opinion bad, since you won't learn anything then.

11. Mar 8, 2016

### SteamKing

Staff Emeritus
You can't say the answer is 4 AU. You have to look at the ratio of period to distance.

If you make the substitution T2 = 2 ⋅ T1, that can simplify things a bit, since T1 = 1 year

Then you would have $\frac{(2⋅T_1)^2}{a_2^3} = \frac{T_1^2}{a_1^3}$

By cross-multiplication, you can solve for a2 in terms of a1 and T1 cancels out.

Remember, you are trying to find the distance of this planet from earth, or a2 - a1 essentially.

12. Mar 8, 2016

### Staff: Mentor

wiegetz, I'm not sure that I'm following the logic in your attempt. If you start with the relationship:
$$\frac{T_2^2}{a_2^3} = \frac{T_1^2}{a_1^3}$$
and choose $T_1$ and $a_1$ to represent known values for the period and orbit radius of Earth, then taking all time units to be years and all distance units to be AU you can write:

$\frac{T_2^2}{a_2^3} = \frac{1^2}{1^3}$

$\frac{T_2^2}{a_2^3} = 1$

You can see how choosing the units to be years and AU's is convenient.
Rearranging:

$a_2^3 =T_2^2$

This then represents the situation for your mystery planet (or any planet in the solar system for that matter) so long as the units you use are years and AU. If you plug in "2" (2 years) for the period of the planet then you can find its orbit radius in AU. Take note that the "a" in the equation is cubed, so you need to deal with that when you solve for its value.

13. Mar 8, 2016

### wiegetz

Finally, I hope:

I continued from where you stopped:

T22/a23 = 1 (Inserting 2 years for T2)
22/a23 = 1 (equivalence transformation with *22 to the right)
a23 = 1*22 ( pull 3rd root to the right)
a2 = 3√1*22
a2 ≈ 1,5874

Sentence = Planet is ≈ 1,5874 AU away from Earth, which equals ≈ 237471660km.

Seems to be correct, since when inserted for a a number close to 1 comes out.

Is that finally correct now? Please be so ))

Last question: How to proceed, if I would use (a) unconverted, as 150*106 this time?

14. Mar 8, 2016

### Staff: Mentor

Yes that looks fine, but you need to include the units on the result: 1.5874 AU.

If you want to use other units (like seconds or days for time, or meters or kilometers for distances), then go back to the ratio formula in post #5 and plug in the given values using the desired units. Your ratio for the Earth values won't turn out so nice (it won't be a simple "1"), but the equation remains valid. It will simply take a bit more work to resolve a (it's just algebra at that point).

15. Mar 8, 2016

### SteamKing

Staff Emeritus
No, you still haven't realized that both the earth and this other planet are orbiting the sun, which is why you can apply the Kepler period relation in the first place.

It takes the earth 1 year (by definition) to orbit the sun at a distance of 1 AU, and this other planet takes 2 years to orbit the same body at some greater distance which you have just calculated.

But the problem asks for the distance between the earth and this other planet, not the distance between that planet and the sun.

16. Mar 8, 2016

### Staff: Mentor

That's true. But that distance would vary with time so there'd be no one answer. I suspect that the original problem has been paraphrased or misquoted. It likely meant for the student to find the distance from the Sun, not the Earth.

17. Mar 9, 2016

### wiegetz

I will use my result to present, since the teacher did not say anything about the sun, and since we started the topic recently, I do not think that he expected us to know it. If I have more questions I will ask, thanks for the help!