Calculating Orbital Radius from Period: Solving for r without Velocity

[joel amos]

Homework Statement

If a new planet has a year of 3.44578 Earth years, what is its radius of orbit in meters if Earth's orbital radius is 149,597,870,700 meters?

Formula: r = (vt)/(2pi)

How can I use this without velocity?

All help is appreciated as I'm not sure how to go about this: forumlas, explanations, answers, etc.

 

[gneill]

Does the name Kepler ring a bell?

 

[joel amos]

Does the name Kepler ring a bell?

I'm not sure whether answering yes or no will yield better results. The name is familiar, but it doesn't tell me much.

Is it this:

If so, I don't have the planet's mass...

 

[SteamKing]

You have a defective problem statement. The radius of the Earth itself is 6.4*10^6 meters. The radius of Earth's orbit around the sun is considerably larger: 149,600,000 kilometers.

 

[gneill]

I'm not sure whether answering yes or no will yield better results. The name is familiar, but it doesn't tell me much.

You should have been introduced to Kepler's three laws of planetary motion by now, if the content of this current problem is any indication. If not, a quick search of your text or the web will turn up the information -- it very well known.

Is it this:

If so, I don't have the planet's mass...

That's a rearrangement of Newton's version of Kepler's third law, derived from his law of universal gravitation. Yes, it involves actual distances and masses and Newton's gravitational constant and so on. But there's an easier way...

Look to the original Kepler version; the early astronomers had no idea what the masses of the planets were (nor the actual scale of the solar system), so everything was worked out by ratios.

 

[joel amos]

You have a defective problem statement. The radius of the Earth itself is 6.4*10^6 meters. The radius of Earth's orbit around the sun is considerably larger: 149,600,000 kilometers.

That's probably a transcription error on my part. How would I attempt this if the Earth's radius was given?

 

[joel amos]

You should have been introduced to Kepler's three laws of planetary motion by now, if the content of this current problem is any indication. If not, a quick search of your text or the web will turn up the information -- it very well known.

That's a rearrangement of Newton's version of Kepler's third law, derived from his law of universal gravitation. Yes, it involves actual distances and masses and Newton's gravitational constant and so on. But there's an easier way...

Look to the original Kepler version; the early astronomers had no idea what the masses of the planets were (nor the actual scale of the solar system), so everything was worked out by ratios.

Oo, could it be period squared = radius cubed?

 

[gneill]

You have a defective problem statement. The radius of the Earth itself is 6.4*10^6 meters. The radius of Earth's orbit around the sun is considerably larger: 149,600,000 kilometers.

Ha! Well done SteamKing. I missed that problem statement slip-up.

Probably easiest to just call the Earth's orbital radius 1AU to make the ratios easy to deal with. Convert results to meters afterwards, if required.

 

[gneill]

Oo, could it be period squared = radius cubed?

Indeed

Can you set up the appropriate ratios to solve your problem?

 

[gneill]

That's probably a transcription error on my part. How would I attempt this if the Earth's radius was given?

I'd ignore the value (since it's obviously specious, having nothing to do with the orbit size), and employ "common knowledge" of the Earth's orbital radius. Since I know that ratios will be involved, I'd first use the Earth's orbital radius as a unit of measure. Thus for the Earth's orbit, R = 1AU (the Astronomical Unit is the common term used).

Once the dust settles on the ratio math and you have the required distance of the unknown in terms of AU, you can convert AU units to meters or kilometers or whatever afterwards as required.

 

[joel amos]

I'd ignore the value (since it's obviously specious, having nothing to do with the orbit size), and employ "common knowledge" of the Earth's orbital radius. Since I know that ratios will be involved, I'd first use the Earth's orbital radius as a unit of measure. Thus for the Earth's orbit, R = 1AU (the Astronomical Unit is the common term used).

Once the dust settles on the ratio math and you have the required distance of the unknown in terms of AU, you can convert AU units to meters or kilometers or whatever afterwards as required.

Alright, so here's what I did:

T^2 = r^3
(3.44578 years)^2 =
11.873 years^2 = 11.873 AU^3
³√(11.873 AU^3) = 2.28 AU
2.28 AU * (149,597,870,700 m /1 AU) =

3.41 x 10^11 m

Is this correct?

 

[gneill]

Sure, that'll do it.

 

[joel amos]

Sure, that'll do it.

Thanks for all the help!