Orbital velocity help

  • Thread starter tinksy
  • Start date
  • #1
5
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hi, could someone please help me with this problem??

If a small planet were discovered with a distance from the sun eight times that of the Earth, what would you predict for its period in (Earth) years. (i.e. how many times longer would it take to go round the sun than the Earth does.)
 
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Answers and Replies

  • #2
18
0
i will not give you the answer straight away, but here's the hint---
get orbital velocity v as a function of radius r

dynamics of the system(*) :

from newton's second law and law of gravitation,
(m*v^2)/R = (G*M*m)/R^2


get v(r) and substitute in the kinemetical relation which you got correct,that is, T=2(pi)r/v

this will give you T(r) which is usually called "kepler's third law".


*assumption: all orbits are circular

justification : though the orbits that actually elliptical ,the eccentricity is very small.(dont worry if you don't understand this,take this as a side remark).

cheers :smile:
 
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  • #3
5
0
thanks teddy...

could u tell me if i'm right?

i've used kepler's law T^2 (directly proportional to) R^3
so if R = 8R (as for Earth, R = 1AU so for the planet, R = 8AU)
kepler's law: R^3/T^2 = 1
therefore, T^2 = 8^3/1
so T = (root)512 = 16(root)2 .....???? is that correct???
 
  • #4
18
0
yup,its right.
bye :smile:
 
  • #5
5
0
yaaaaaay!!! thanks
 

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