How Do You Calculate the Mass of a Star Based on Its Planet's Orbit?

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Homework Help Overview

The problem involves calculating the mass of a star based on the orbital characteristics of a planet. The planet orbits in a circular path with a specified radius and period, and the gravitational constant is provided for calculations.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to begin the calculation. Another participant suggests using the relationship between orbital velocity and gravitational force, indicating a method to derive the mass of the star.

Discussion Status

One participant claims to have found a solution and presents a specific mass value for the star. However, there is no indication of consensus on the approach or the correctness of the solution, as other participants have not confirmed or challenged this finding.

Contextual Notes

Participants reference Kepler's laws and discuss their relevance to the problem, but there is no agreement on their necessity for the solution. The original poster's initial confusion highlights the complexity of the topic.

od943
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Homework Statement


A distant star has a single planet circling it in
a circular orbit of radius 3.33 × 1011 m. The
period of the planet’s motion about the star
is 836 days.
What is the mass of the star? The
value of the universal gravitational constant
is 6.67259 × 10−11 N · m2/kg2.
Answer in units of kg.


Homework Equations


v = 2 π r / T
ac = v2 / r
Fc = mac
Kepler's laws?

The Attempt at a Solution


uhhhh...have i have no idea where to start
 
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wait i figured it out...you use v=(gm/r)^(1/2) and substitute in 2(pi)r/T (in seconds) for v...then solve for M
the answer i ended up with was 4.187530248*10^30
 
Nice.
 
od943 said:
wait i figured it out...you use v=(gm/r)^(1/2) and substitute in 2(pi)r/T (in seconds) for v...then solve for M
the answer i ended up with was 4.187530248*10^30

Well done! And your method is exactly right, Kepler's laws being unnecessary in this case, though just as useful if you know how.

You can, of course, combined the equations to get...

T2 = 4.π2.r3/GM

...and manipulated that as needed. You can see where Kepler's law comes from then, too.
 

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