Need help with an orbital mechanic question!

[Der Anfang]

I am running into huge inconsistencies every time I do this, nothing ever matches up when I try to confirm one or the other. I am trying to calculate the semi-major axis of an object based on it's orbital period. Every time I do so, I get a sma WAY shorter than it actually is supposed to be.

Let's work with the numbers, here:

The Earth-Moon system have a total mass of ~6.045847e24 kg (5.97237e24 kg + 7.3477e22 kg).

The moon orbits around the Earth at a semimajor axis of ~384399000 m, at an orbital speed of 1022 m/s, totalling at an orbital period of 27 d (2360584.7 s).

Let's say I don't know the semi major axis, but I have the orbital period of the object. How do I know the seminajor axis of the object in orbit? Please help!!

[kurja]

So you'd have the masses, orbital period and average orbital velocity and you want to know the sma? I don't know really. Can you even? I think you'd need to know Ap or Pe, otherwise you'd be assuming a zero eccentricity orbit so your sma would be just that... Here's a spreadsheet where I doodled with some orbital parameters, maybe playing around with it helps you.

[Der Anfang]

So you'd have the masses, orbital period and average orbital velocity and you want to know the sma? I don't know really. Can you even? I think you'd need to know Ap or Pe, otherwise you'd be assuming a zero eccentricity orbit so your sma would be just that... Here's a spreadsheet where I doodled with some orbital parameters, maybe playing around with it helps you.

I'm... not sure if you were being sarcastic or not. I know it doesn't matter if you know the PE or AP because orbital period and sma are corelated, aren't they? The sma is what determines the orbital period, so how do you determine the sma from knowing only the orbital period?

[GeneralVeers]

Wikipedia has the following formula:

where T is the orbital period, a is the semi-major axis, and u is the standard gravitational parameter (gravitational constant times mass) of the central body.

[kurja]

Not sarcastic, maybe a bit of a language barrier here. And it looks like you're right and I was wrong, Ap or Pe should not be needed when you have the orbital period.

Is there something wrong with this solution?

edit - ^which is the same as given in the ninja post above

[Der Anfang]

Wikipedia has the following formula:

https://upload.wikimedia.org/math/0/e/7/0e7cf81b92bf0236f35342d10700f8d9.png

where T is the orbital period, a is the semi-major axis, and u is the standard gravitational parameter (gravitational constant times mass) of the central body.

I have seen that formula and it doesn't help... I'm not sure how you use it if you don't know a. I know you can use that to calculate the time of the orbit, but what I am looking for is how to calculate a with only the other parametres that I do know. I've seen that formula many times, but it makes no sense how you use it if you don't know the sma. Does nobody understand what I am asking or is there something I am not understanding?

[kurja]

are you looking for how to rearrange this

to solve for a?

[Der Anfang]

are you looking for how to rearrange this

https://upload.wikimedia.org/math/2/6/b/26b9ed71b1c8dbce008d396135ff6070.png

to solve for a?

Yes.

[kurja]

Yes.

when in doubt, wolfram alpha

Just plug in the equation above and ask how to solve for a.

[FleshJeb]

when in doubt, wolfram alpha

Just plug in the equation above and ask how to solve for a.

I grieve for our children's problem-solving skills...

Der Anfang, you have to do the same thing to both sides of the equation to isolate the term you want. I'll do it in steps.

Pardon my formatting:

Original equation: T = (2*pi)*[a^3/G(M1+M2)]^2

Step 1, divide by 2*pi: T/(2*pi) = [a^3/G*(M1+M2)]^(1/2)

Step 2, square both sides: [T/(2*pi)]^2 = a^3/[G*(M1+M2)]

Step 3, multiply by denominator: [G*(M1+M2)]*[T/(2*pi)]^2 = a^3

Step 4, take cube root: {[G*(M1+M2)]*[T/(2*pi)]^2}^(1/3) = a

EDIT: I'll solve it so you can check (with units):

G = 6.67*10^-11 (m^3 kg^-1 s^-2) (gravitational constant of the universe) *EDIT negative exponent

[T/(2*pi)]^2 = 1.41x10^13 (s^2)

G*(M1+M2) = 4.04*10^14 (m^3 s^-2)(the kg cancels)

[G*(M1+M2)]*[T/(2*pi)]^2 = 5.70*10^25 (m^3)(the s cancels) *EDIT typo in exponent

cube root that big nasty mess above = 3.85*10^8 (m)(uncubed the meters) *deleted yet another typo

That matches the listed value, so it looks like I did it right, within rounding errors.

Edited October 1, 2015 by FleshJeb
solution / typos

[YNM]

Earth - Moon system is quite complex - Moon's mass is within one percent of Earth's (means that it's significant - use within Kepler's 3rd law requires all the mass in the system actually, not just the larger one), and the Sun plays a fair role too. Also, the Moon's orbital period is a bit more closer to 27+(1/3) day than 27 (that's a difference of ~24000 seconds !), so having your answer close to 5000 km away from the actual is quite a good calculation.

That being said, the equation you use to compute period comes from Kepler's 3rd law :

Where :

a : Semi-major Axis

T : Orbital period (of both bodies actually, orbiting the barycenter)

G : Gravitational constant. Using one that goes to the fourth decimal might helps a bit.

M : Mass of larger body

m : Mass of smaller body, often omitted when falls below 1% of M

pi : you should know this.

There's more convenience here to rearrange them algebraically because no roots yet. You only need to move/leave the (a^3) to one side and let everything else move to the other side. Cubic root then can be taken.

Edited September 29, 2015 by YNM

[PB666]

I am running into huge inconsistencies every time I do this, nothing ever matches up when I try to confirm one or the other. I am trying to calculate the semi-major axis of an object based on it's orbital period. Every time I do so, I get a sma WAY shorter than it actually is supposed to be.

Let's work with the numbers, here:

The Earth-Moon system have a total mass of ~6.045847e24 kg (5.97237e24 kg + 7.3477e22 kg).

The moon orbits around the Earth at a semimajor axis of ~384399000 m, at an orbital speed of 1022 m/s, totalling at an orbital period of 27 d (2360584.7 s).

Let's say I don't know the semi major axis, but I have the orbital period of the object. How do I know the seminajor axis of the object in orbit? Please help!!

http://www.masteringphysicssolutions.net/chapter-6-6-the-center-of-mass-of-the-earth-moon-sun-system/

[Der Anfang]

I grieve for our children's problem-solving skills...

Der Anfang, you have to do the same thing to both sides of the equation to isolate the term you want. I'll do it in steps.

Pardon my formatting:

Original equation: T = (2*pi)*[a^3/G(M1+M2)]^2

Step 1, divide by 2*pi: T/(2*pi) = [a^3/G*(M1+M2)]^(1/2)

Step 2, square both sides: [T/(2*pi)]^2 = a^3/[G*(M1+M2)]

Step 3, multiply by denominator: [G*(M1+M2)]*[T/(2*pi)]^2 = a^3

Step 4, take cube root: {[G*(M1+M2)]*[T/(2*pi)]^2}^(1/3) = a

EDIT: I'll solve it so you can check (with units):

G = 6.67*10^11 (m^3 kg^-1 s^-2) (gravitational constant of the universe)

[T/(2*pi)]^2 = 1.41x10^13 (s^2)

G*(M1+M2) = 4.04*10^14 (m^3 s^-2)(the kg cancels)

[G*(M1+M2)]*[T/(2*pi)]^2 = 5.70*10^27 (m^3)(the s cancels)

cube root that big nasty mess above = 1.77*10^9 (m)(uncubed the meters) = 3.85*10^8 (m)

That matches the listed value, so it looks like I did it right, within rounding errors.

Wow, yes, thank you so much! I figured it out with your help. However, you equition step "[G*(M1+M2)]*[T/(2*pi)]^2 = 5.70*10^27 (m^3)(the s cancels)" was a little off. My results came down to 5.694e25, instead of e27. Not sure why, but my end result came close to yours, but I ultimately got a=384,727,403.1 m. Maybe it is because I used the exact precise results of solving each part of the equation without rounding?

[FleshJeb]

Gah! Thanks for checking my work!

I didn't run the calc again, but I estimated with the exponents, and it should be 10^-11 * 10^24 * 10^6^2 = 10^25.

Also the 1.77*10^9 was from the bad 10^27 number, ignore it. I fixed my typos.

Speaking of which, I had listed the exponent of G as positive and not negative, so I fixed that too.

Yes, I'm sure it was the rounding, I think I only used four or five significant figures.

Clearly, I should not do science on 40 hours without sleep.

Did the colors help? I wasn't sure if they'd add to the confusion or not.

Edited October 1, 2015 by FleshJeb

[GoSlash27]

Der Anfang,

rearranging these motion equations is simply a matter of straight algebra.

((t/2pi)^2*GM)^(1/3)=a

Best,

-Slashy