Orbital Period of the Earth where the Sun's Mass Changes

In summary, the conversation discusses a homework problem involving the speed of a planet in orbit around a star and the effects of the star losing mass. The equations used are stated, and the individual's attempts to solve the problem are described. Ultimately, the mistake is identified as using the same symbol for two different distances, and the correct solution is discussed.
  • #1
Maik
4
0
Member advised to use the provided formatting template when starting a new thread in a homework forum.
Hey,
this is going to be my first post here so I'm not sure how it all works, so just tell me if I do something out of order please. Anyway I have been given this homework assignment and part of it was the question stated below.
---------------------------------------------------------------------------------------------------------------------------------------
The speed (v) of a planet in orbit around a star of mass M is related to its orbital distance (r) and orbital semimajor axis (a) by

v2=GM((2/r)-(1/a))

Use this equation to show that if the Sun instantly lost a fraction f of its mass, reducing its mass from M to M(1 − f), then the Earth’s (originally circular) orbit would have a period of

T =((1 − f)/(1 − 2f))3/2 years

[You may assume Kepler’s Third Law: orbital period is proportional to orbital semimajor
axis to the power 3/2.]
---------------------------------------------------------------------------------------------------------------------------------------

The Equations that I am supposed to use are stated in the problem, however I also considered the equation for the angular momentum and energy relating the kinetic energy and the gravitational potential energy:
L=mvr and E=1/2mv2 -GMm/r

I started off by trying to find an expression for a and the using that in Keplars 3rd rule, however rather than simplifying the whole expression to what they asked it made it more of a mess. I then attempted to solve it for the point at which r=a (initially) and then the same for the new orbit. By then considering the conservation of angular momentum I found an expression on the new a (calling it c) as follows: c= a/(1-f) ( by saying that L2=m2 u2 a2 = m2 v2 c2 (where u is the initial velocity and v is the final velocity) and then putting the expression from the problem in as the v2 and then simplifying it down using the initial velocity of u as squrt.(GM/a)) I then put this in keplars 3rd rule but with no success.
I could have made some mistake here or forgot to consider something but I just can not get my head around it.
It would be great if someone could just hint me where I could have gone wrong or what I should/can consider to stay the same during the mass loss.

Thanks a lot in advance already!
 
Physics news on Phys.org
  • #2
I don't see how the angular momentum would help here. Consider v2=GM((2/r)-(1/a)). We know v and r of the Earth today satisfy this equation, and we know r=a as the orbit is supposed to be circular.
How does the equation look like directly after the Sun lost some mass? v and r will still be the same, but two other things change. You can write that as new equation. Set them equal to get rid of v and you should get the new semimajor axis a function of the old one and the mass loss.
 
  • #3
Hello. Welcome to PF.

Maik said:
I then attempted to solve it for the point at which r=a (initially) and then the same for the new orbit. By then considering the conservation of angular momentum I found an expression on the new a (calling it c) as follows: c= a/(1-f) ( by saying that L2=m2 u2 a2 = m2 v2 c2 (where u is the initial velocity and v is the final velocity) and then putting the expression from the problem in as the v2 and then simplifying it down using the initial velocity of u as squrt.(GM/a))
Your result c = a/(1-f) is not correct. I believe your mistake might be due to using the symbol "a" for two different distances. You use "a" to denote the distance of the planet from the star at the moment the star's mass changes. But you also use "a" do denote the semi-major axis in the formula v2 = GM[(2/r)-(1/a)]. But the radius of the initial circular orbit does not equal the semi-major axis of the new elliptical orbit.
 
  • #4
mfb said:
v and r will still be the same, but two other things change. You can write that as new equation. Set them equal to get rid of v and you should get the new semimajor axis a function of the old one and the mass loss.

Thank you so much, by considering that v and r stay the same I was able to solve it.
Just out of curiosity, how do I know that the v and r will stay the same even when the sun losses mass?
 
  • #5
TSny said:
Your result c = a/(1-f) is not correct. I believe your mistake might be due to using the symbol "a" for two different distances. You use "a" to denote the distance of the planet from the star at the moment the star's mass changes. But you also use "a" do denote the semi-major axis in the formula v2 = GM[(2/r)-(1/a)]. But the radius of the initial circular orbit does not equal the semi-major axis of the new elliptical orbit.
That's right, I just saw that in my working now, thank you for pointing it out to me.
I should really be more careful when it comes to things like this.
Thank you for your help!
 
  • #6
Maik said:
Thank you so much, by considering that v and r stay the same I was able to solve it.
Just out of curiosity, how do I know that the v and r will stay the same even when the sun losses mass?
At the moment the Sun loses mass, the Earth is a certain distance from the Sun (r) and has a given velocity(v). The Sun losing mass will not change these properties at that moment, just the future trajectory of the Earth from that moment on.
 
  • #7
Janus said:
At the moment the Sun loses mass, the Earth is a certain distance from the Sun (r) and has a given velocity(v). The Sun losing mass will not change these properties at that moment, just the future trajectory of the Earth from that moment on.
Ah I see that makes sense.
Thank you!
 

Related to Orbital Period of the Earth where the Sun's Mass Changes

What is the orbital period of the Earth when the Sun's mass changes?

The orbital period of the Earth is the time it takes for the Earth to complete one full orbit around the Sun. The Sun's mass changing does not affect the Earth's orbital period significantly. It remains approximately 365.24 days.

How does the Sun's mass affect the Earth's orbital period?

The Sun's mass has a gravitational pull on the Earth, which keeps it in its orbit. However, the Earth's orbit is primarily determined by its distance from the Sun, rather than the Sun's mass. Therefore, small changes in the Sun's mass do not have a significant impact on the Earth's orbital period.

Can the Sun's mass change? If so, how frequently does it change?

Yes, the Sun's mass does change over time, but these changes are very small and occur over long periods. The Sun's mass decreases as it converts hydrogen into helium through fusion. This process is gradual and takes place over millions of years.

What is the current mass of the Sun?

The current mass of the Sun is approximately 1.989 x 10^30 kilograms. It is estimated to lose about 5 million tonnes of mass per second through nuclear fusion.

How does the Earth's orbital period affect life on Earth?

The Earth's orbital period plays a crucial role in the Earth's climate and the changing of seasons. It also affects the length of days and nights, which impacts the growth and behavior of plants and animals. Any significant changes in the Earth's orbital period could have significant consequences for life on Earth.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
863
  • Introductory Physics Homework Help
Replies
30
Views
555
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
986
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
827
Replies
1
Views
996
Back
Top