• Support PF! Buy your school textbooks, materials and every day products Here!

Kepler's Law of planetary motion

  • Thread starter Tulatalu
  • Start date
  • #1
29
1

Homework Statement


Two stars of masses M and m, separated by a distance d, revolve in circular orbits around their center of mass.Show that each star has a period given by
T^2= (4π^2)(d^3)/ G(M+m)

Homework Equations


[/B]


The Attempt at a Solution


I[/B] know the Kepler's Laws can be expressed as T^2= (4π^2)(d^3)/ G(M) but i don't know how it is applied when 2 planets interact with each other (both in circular motion)
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
It's a given that the orbits are circular. Simple kinematics, therefore.
 
  • #3
29
1
It's a given that the orbits are circular. Simple kinematics, therefore.
Can you explain a bit further please, how does it relate to the Kepler's law. I thought their must be something to do with the gravitational force
 
  • #4
gneill
Mentor
20,792
2,770
I know the Kepler's Laws can be expressed as T^2= (4π^2)(d^3)/ G(M) but i don't know how it is applied when 2 planets interact with each other (both in circular motion)
Kepler's 3rd Law is a special case of Newton's more general formula relating the period to the size of the orbit (the semi-major axis for elliptical orbits, the orbit radius for circular orbits). Kepler's Law makes the unstated assumption that the mass M of the Sun is much, much greater than that of the planet m in orbit so that (M + m) ≈ M. Newton's laws can be applied without making such an assumption.

In this problem you are to show that the individual periods of the orbits of the objects about their common center of mass are as given. You should be able to determine the radius of each orbit by locating the center of mass. Hint: use the angular motion form for centripetal acceleration (involving ω) since there's a simple relationship between ω and period.
 
  • #5
29
1
Kepler's 3rd Law is a special case of Newton's more general formula relating the period to the size of the orbit (the semi-major axis for elliptical orbits, the orbit radius for circular orbits). Kepler's Law makes the unstated assumption that the mass M of the Sun is much, much greater than that of the planet m in orbit so that (M + m) ≈ M. Newton's laws can be applied without making such an assumption.

In this problem you are to show that the individual periods of the orbits of the objects about their common center of mass are as given. You should be able to determine the radius of each orbit by locating the center of mass. Hint: use the angular motion form for centripetal acceleration (involving ω) since there's a simple relationship between ω and period.
I still don't knowhow to apply Kepler's Law to this problem. I can mathematically prove the Kepler's Law in case of satellite moving around earth but with 2 planet in circular orbit I have no idea. Can you please explain a bit further and use the equation so that it would be easier for me to follow.
 
  • #6
DEvens
Education Advisor
Gold Member
1,203
451
You don't need Kepler's law to derive the period of he orbits. You are given that the orbits are circular, the masses, and the radius. You should be able to work out the rest fairly easily. What speed must the stars be moving at for the circular orbits to be stable? And once you have the speed, what is the period?
 
  • #7
gneill
Mentor
20,792
2,770
I still don't knowhow to apply Kepler's Law to this problem. I can mathematically prove the Kepler's Law in case of satellite moving around earth but with 2 planet in circular orbit I have no idea. Can you please explain a bit further and use the equation so that it would be easier for me to follow.
You use the same methodology: Pick one of the planets. Determine the radius of the orbit. It's moving in a circle so equate the gravitational acceleration (due to the other planet, which takes the role of the "Sun" in this case) to the centripetal acceleration.
 
  • #8
29
1
Thanks everybody. It turn out to be not as complicated as I think it is :D
 
  • #9
29
1
You use the same methodology: Pick one of the planets. Determine the radius of the orbit. It's moving in a circle so equate the gravitational acceleration (due to the other planet, which takes the role of the "Sun" in this case) to the centripetal acceleration.
Coukd I ask one more question : how about the period of 3 identical planets moving around another planet and they are positioned one third of a revolution apart from each other? Do we have consider the force between the three planets or just simply the force between them and the planet in the centre?
 
  • #10
gneill
Mentor
20,792
2,770
Coukd I ask one more question : how about the period of 3 identical planets moving around another planet and they are positioned one third of a revolution apart from each other? Do we have consider the force between the three planets or just simply the force between them and the planet in the centre?
Um, there would be no planet at the center if they are spaced equally around the orbit. Like the two-planet scenario, the center of the orbits is empty.

But yes, you must take into account all the forces acting and resolve them (vectors!) into the net force that provides the centripetal force for each planet.

Note that many configurations of multiple objects in mutual orbit are not stable over long periods of time. If you're interested in the topic, I might suggest starting with a search on "Klemperer rosette" :)
 
  • #11
29
1
Um, there would be no planet at the center if they are spaced equally around the orbit. Like the two-planet scenario, the center of the orbits is empty.

But yes, you must take into account all the forces acting and resolve them (vectors!) into the net force that provides the centripetal force for each planet.

Note that many configurations of multiple objects in mutual orbit are not stable over long periods of time. If you're interested in the topic, I might suggest starting with a search on "Klemperer rosette" :)
Now it's too much for me :D but thanks anyway.
 

Related Threads on Kepler's Law of planetary motion

Replies
16
Views
1K
Replies
1
Views
3K
  • Last Post
Replies
2
Views
931
Replies
2
Views
4K
Replies
6
Views
2K
Replies
8
Views
468
Replies
3
Views
840
Replies
4
Views
1K
  • Last Post
Replies
5
Views
649
Top