question: kepler's law of periods (1 Viewer)

asl2

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Use Kepler's law of periods to determine the time
required for Venus to complete an orbit around the sun
where the radius is (58.5*10^6) KM and the mass of
Venus is (4.9*10^24) KG
--------------------------------------
where am i going wrong?
----------------------------------------

R^3 Gm
------ = --------
T^2 4pi^2

4pi^2* R^3 T^2
------ = --------
Gm

T= Under root (4pi^2* R^3 / Gm)

= root (4pi^2*(58.5*10^6*1000)^3)/(6.67*10^-11)*(5.92*10^24)

= WRONG ANSWER !!!
 

kini mini

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Assuming the formula is correct ( I can't be completely sure, looks right though), there is an error on your 2nd line. You divided LHS by R^3 and multiplied RHS by R^3. There might be more errors but I haven't checked.
 

asl2

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huh?

the formula is....

R^3 Gm
____ = _________

T^2 4pi^2



r^3 * 4pi^2
_______________ = T^2

Gm
 

Weisy

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your formula doesn't make sense, unless it equals 1.

using the relationship between mass, period and radius:

M = 4*pi^2R^3
__________

G*T^2

I'm a little rusty...but isn't the mass here the mass of the central object, not the one in orbit? ok, you've done that.
 
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Weisy

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don't you need the mass of the sun, not the mass of venus?
 

sukiyaki

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Originally posted by Weisy
don't you need the mass of the sun, not the mass of venus?
oh is it?

note to myself - read the question probably
 

kini mini

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Originally posted by Weisy
don't you need the mass of the sun, not the mass of venus?
That would be error #2 - the whole idea of Kepler's law of periods is you only need the mass of the object being orbited. Sorry, I didn't bother checking your substitutions after the first error.

On your second line, you rearranged wrongly:


R^3 Gm
------ = --------
T^2 4pi^2

4pi^2* R^3 T^2
------ = --------
Gm

is clearly wrong, as I said before. How the heck did you end up with R^3 x T^2 on the RHS?? You want

R^3 = G.M(sun) T^2 = 4pi^2 x R^3
------- ------------ >>>>>> ----------------
T^2 4pi^2 G.M(sun)

You then make another error and get back on track with the right form. This error would be classed as a simplifying error in a maths exam and get you no marks from then on. What probably got you after that was the substitution.
 
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kini mini

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Actually, forget what I just said. I see now that the problem is the formatting that is done after you've arranged everything nicely in the post window. It was probably the substitution that was wrong then.
 

Weisy

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Originally posted by asl2
i thought it was the mass of the earth and not the sun ?:argue:
what has the earth got to do with it? you are talking about an orbital system which assumes a direct relationship between the planet and the body it is orbiting. From your initial question, you said this was Venus around the sun.

From Kepler's law of periods, the orbital radius cubed is proportional to the period squared. The constant value this proportionality is equal to is GM/4pi^2, where M is the mass of the central object. So the mass of Venus has nothing to do with it, only the mass of the sun affects period and radius of rotation.
 

asl2

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hmm... yeah.. thanks guyz... i think thats where the problem was...

its just that in the text book it said M(earth) so thats all..


thx
 
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