Help with Calculating Planet's Orbital Period

In summary, the conversation is about predicting the period of a newly discovered planet based on its distance from the sun. The conversation includes equations and references to Kepler's third law, with the assumption that all orbits are circular. The final conclusion is that the predicted period is 16 times the Earth's period, represented as T = 16√2.
  • #1
tinksy
5
0
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.)
 
Last edited:
Physics news on Phys.org
  • #2
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:
 
Last edited:
  • #3
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
yup,its right.
bye :smile:
 
  • #5
yaaaaaay! thanks
 

Similar threads

Replies
1
Views
1K
Replies
10
Views
525
Replies
3
Views
2K
Replies
6
Views
978
Replies
11
Views
2K
Replies
13
Views
7K
Back
Top