Using Kepler's Laws to find the mass of a star

  • Thread starter Thread starter jamesc5
  • Start date Start date
  • Tags Tags
    Laws Mass Star
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 3K views
jamesc5
Messages
2
Reaction score
0

Homework Statement


In recent years, a number of nearby stars have been found to possesses planets. Suppose, the orbital radius of such a planet is found to be 4.9 1011 m, with a period of 1280 days. Find the mass of the star.


Homework Equations


T^2=4pi√((r^3)/Gm))


The Attempt at a Solution


(1280)^2=4pi√((4.9E11)^3/(6.67E-11)m)
1638400=4pi√((1.17649E35)/(6.67E-11)m)
1286796.351=√((1.17649E35)/(6.67E-11)m)
1.655844849E12=(1.17649E35)/(6.67E-11)m)
110.4448514m=6.67E-11
m=6.039E-13

I thought this was the right solution; however, I am not getting the correct answer as I am doing it online and it tells me I am doing it wrong. Am I using the wrong equation?
 
Physics news on Phys.org
I see at least two mistakes. The equation is wrong (c.f. Wikipedia); the period must be expressed in seconds, not days.
 
welcome to pf!

hi james! welcome to pf! :smile:
jamesc5 said:
… with a period of 1280 days
…
(1280)^2=4pi√((4.9E11)^3/(6.67E-11)m)

days ! :wink:
 
voko said:
I see at least two mistakes. The equation is wrong (c.f. Wikipedia); the period must be expressed in seconds, not days.

I tried switching days to seconds and i got T=110592000 seconds so i changed my equation...

(11059200)^2=4pi√((4.9E11)^3)/(6.67E-11)m)
9.733E14=√((4.9E11)^3)/(6.67E-11)m)
9.473E29=((4.9E11)^3)/(6.67E-11)m)
(1.42E40)m=(4.9E11)^3
m=8.234E-6

I changed it to seconds, but I still don't seem to be getting the right answer.
 
The equation is start with is wrong. I mean this one: T^2=4pi√((r^3)/Gm)) As I said. look up the correct equation.