A planets orbital period distance from the sun

[buc4933]

If a planet were found with an orbital period of 64 years. how might you estimate its distance from the sun? If its orbit is circular, what is it radius?<br /><br />No clue on how I would answer this <br /><br />Any help would be great thanks

 

[buc4933]

I think something with 90 degrees and multiplying that by like 360 degrees or something.

 

[buc4933]

Is this the formula that I'm looking for p^2=a^3

 

[MeteorWayne]

That's not right.<br /><br />Approximately, The period equals the square root of the cube of the semimajor axis (a)-for a circle, that's the radius)<br /><br />P= SQRT(a^3)<br /><br />Now with my head cold, I'm going to have to go offline and do the algebraic conversion to get "a" on the left side of the equation, and get everything else on the right. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[MeteorWayne]

Yeah, that's the same as what I posted, and looks easier to move the "a" than the way I wrote it <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[buc4933]

thats what i was thinking...

 

[MeteorWayne]

so the radius of a circular orbit is the cube root of the period squared.<br /><br />I won't even attempt to write that without symbols <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[buc4933]

something like this maybe.......p^2=a^3 (64)^2=a^3 so 4096 and taking the cube root of that gives us 15.9999 a=15.9999?????

 

[MeteorWayne]

Looks like you did it right.<br />For example; <br />Saturn 9.6 AU. P=29.7 Years<br />Yours 16 AU ... P=64 Years<br />Uranus 19.1 AU. P=83.8 Years <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[buc4933]

ok so know that I have the distance of that planet 15.99au what how would i find the radius of this planet

 

[nexium]

Since your planet is 64 years it is closer most of the time than Pluto, likely about the orbit of Neptune. Neil

 

[buc4933]

that may be true, but there has to be some calculation to figure its radius

 

[MeteorWayne]

No, it would be in between the orbits of Saturn and Uranus, as I showed above. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[MeteorWayne]

You mean the size of the planet itself?<br />At that distance, with a telescope you should be able to measure the angular size of the image and then calculate the actual size (which would be the diameter) at that distance.<br /><br />That formula I don't have handy <img src="/images/icons/frown.gif" /> <br /><br />It is however, basic trigonometry....which I haven't used much lately. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[buc4933]

ok but i still dont understand the actual radius

 

[buc4933]

no i think they are asking for the radius of its orbit

 

[MeteorWayne]

Well, the radius of it's orbit is what you've already figured out; 15.99 AU. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[buc4933]

I thought what i had figured out was that planets distance from the sun

 

[buc4933]

sorry i didnt really follow all that

 

[MeteorWayne]

That's the definition of the orbit's radius! <br /><br />What do you think the radius is? <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[buc4933]

so would i take 15.999 and multiply it by 2 or am i wrong

 

[MeteorWayne]

Since his question referenced the sun, I didn't go into all the other gory details.<br />Yes there are a lot of other factors, including the mass of the planet, but the effect of that is so small it can be ignored for a general question.<br />Certainly the mass of the star (or planet) makes a big difference, but I didn't get the sense that's what was being asked for here.<br /><br />The simplified formula seemed adequate. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[MeteorWayne]

No, times 2 would be the diameter of the orbit. The radius is 15.99.<br /><br />Do you understand the difference between radius and diameter? <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[yevaud]

This formula is Kepler's formula, otherwise known as "Vis Visa."<br /><br />Here's a fairly good site for an explanation: <a target="_blank" href="http://www.Orbitsexplained.com">www.Orbitsexplained.com </a> <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>

 

[buc4933]

Ok thanks guys for all your help