How Do You Calculate the Mass of a Star Based on Planetary Data?

In summary, the planet has a diameter of 1.8*10^7 m and a rotation period of 22.3 hours. Its mass is 1.48*10^25 kg and the mass of the star is 1.39*10^64 kg.
  • #1
tibessiba
18
0
Gravitational Force---Need Help Fast!

Homework Statement



You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8*10^7 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits from its star with a period of 402 Earth days. Once on the surface you find that the acceleration due to gravity is 12.2 m/s^2.

What is the mass of the planet? in kg

What is the mass of the star? in kg

Equations:
Mass(of planet)= g*R^2/G

The Attempt at a Solution



Ok, so I calculated the mass of the planet to be 1.48*10^25 kg using the above equation. M=[(12.2m/s^2)*(9*10^6m)^2]/6.67*10^-11 Nm^2/kg^2.

Now I am stuck with how to find the mass of the star and which equation to use to get it..

Thank you
 
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  • #2
Anybody?
 
  • #4
It states that the planet orbits 2.2*10^11 m from its star. This would them be the distance, correct??
 
  • #5
So, then I use T^2=4pi^2/GM somehow? I guess I'm not exactly sure how...
 
  • #6
oops, I forget r^3 in the equation..
so:

T^2=(4pi^2/GM)r^3
 
  • #7
T should = 34732800 sec
r should be 2.2*10^11 m

correct??
 
  • #8
Using that I calculated the mass of the star to be 1.39 * 10^64...
but that is wrong...
so where did I go wrong??
 
  • #9
I calculated it again and this time I got 1.44*10^37, but that is also wrong...

What am I doing wrong?
 
  • #10
If I rearrange the equation to find the mass of the star this is what I get:

M=[(2pir^3/2)^2]/GT

Is this right? It is what I have been using, but I can't come up with the correct answer.
 
  • #11
tibessiba said:
It states that the planet orbits 2.2*10^11 m from its star. This would them be the distance, correct??
Yes, that is a or r in Kepler's forumula relating period with distance for one mass orbiting another mass, e.g. moon around a planet or planet about a star.

In the simplest form, one may assume that the mass of the star greatly exceeds the mass of the planet.

So [tex]T^2\,=\,\frac{4{\pi^2}{a^3}}{GM}[/tex], which can be rearranged to get

[tex]M\,=\,\frac{4{\pi^2}{a^3}}{GT^2}[/tex]

So substitute in the appropriate numbers

G = 6.67 x 10-11 N m2/kg2, the universal gravitational constant,

a = 2.2 x 1011 m

T = 402 d * 24 h/d * 3600 s/h = 3.47328 x 107 s

Now compare the mass of the star with the mass of the planet. The mass calculated for the star might need adjusting for the mass of the planet since M is the sum of the masses, but if M >> m(planet), the M is approximately the mass of the star.

M=[(2pir^3/2)^2]/GT
This is not correct. Be careful about moving terms and exponents. T should be squared and if one brings r^3 inside the parentheses and squares those terms, then one must use r^(3/2) within the parentheses.
 
Last edited:
  • #12
Ok... I see.

Thank you very much!
 

What is gravitational force?

Gravitational force is a fundamental force of nature that describes the attraction between two objects with mass. It is responsible for keeping planets in orbit around the sun and objects on Earth's surface.

How is gravitational force calculated?

Gravitational force is calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

What factors affect the strength of gravitational force?

The strength of gravitational force is affected by the mass of the objects and the distance between them. The greater the mass of the objects, the stronger the force, and the farther apart they are, the weaker the force.

Can gravitational force be shielded or canceled?

Gravitational force cannot be shielded or canceled. It is a fundamental force and is always present between objects with mass, regardless of any barriers or obstacles between them.

What are some real-world applications of gravitational force?

Gravitational force has many real-world applications, including keeping the planets in our solar system in orbit, allowing us to walk on Earth's surface, and enabling the operation of devices like scales and elevators. It is also crucial in fields such as astronomy, engineering, and physics.

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