## Astronomy C

Test your knowledge of various Science Olympiad events.
nobodynobody
Member
Posts: 36
Joined: January 8th, 2020, 5:41 pm
Division: C
State: OH
Pronouns: He/Him/His
Has thanked: 9 times
Been thanked: 7 times

### Re: Astronomy C

Idk if this question even has a solution but here goes:

Imagine 3 masses where M2>>M1 and the mass of M3 is negligible. M1 is orbiting around M2 at distance R with a period of P. M3 is also is orbiting M2 with the same period of P. What distance(s) could M3 be orbiting M2 from? Assume ideal circular orbits.
Class of '23
2021 events: Astro, Digi, SOM, WICI

"No." - Marie Curie

0sm0sis
Member
Posts: 6
Joined: July 25th, 2020, 3:30 pm
Division: C
State: OH
Has thanked: 2 times
Been thanked: 4 times

### Re: Astronomy C

nobodynobody wrote:
July 25th, 2020, 3:08 pm
Idk if this question even has a solution but here goes:

Imagine 3 masses where M2>>M1 and the mass of M3 is negligible. M1 is orbiting around M2 at distance R with a period of P. M3 is also is orbiting M2 with the same period of P. What distance(s) could M3 be orbiting M2 from? Assume ideal circular orbits.
By equating forces in a non-inertial frame, we have the gravitational forces = mass * angular velocity ^2 * radius. Replacing the angular velocity on RHS with the period of the orbit of M1 gives us M1M2 / R^2 + M2M3 / r^2 = M2M3 r / R^2 (where r is the desired distance). Then we can taylor expand and simplify to:

r = R ( 1 ± cbrt( M1 / 3*M2 ) )
2020-21 Events:

Circuit Lab, Detector, Machines, SOM

"All science is either physics or stamp collecting." - Ernest Rutherford

hippo9
Member
Posts: 270
Joined: March 12th, 2018, 9:35 am
Division: C
State: IN
Has thanked: 1 time
Been thanked: 6 times

### Re: Astronomy C

0sm0sis wrote:
July 25th, 2020, 3:56 pm
nobodynobody wrote:
July 25th, 2020, 3:08 pm
Idk if this question even has a solution but here goes:

Imagine 3 masses where M2>>M1 and the mass of M3 is negligible. M1 is orbiting around M2 at distance R with a period of P. M3 is also is orbiting M2 with the same period of P. What distance(s) could M3 be orbiting M2 from? Assume ideal circular orbits.
By equating forces in a non-inertial frame, we have the gravitational forces = mass * angular velocity ^2 * radius. Replacing the angular velocity on RHS with the period of the orbit of M1 gives us M1M2 / R^2 + M2M3 / r^2 = M2M3 r / R^2 (where r is the desired distance). Then we can taylor expand and simplify to:

r = R ( 1 ± cbrt( M1 / 3*M2 ) )
I haven't done this in a while so I could be completely off, but shouldnt the distances be the same? By kepler's 3rd P^2=(4((pi)^2)a^3)/GM. No where in kepler's 3rd is the mass of the orbiting object included. Therefore if the periods are the same, they should be orbiting at the same distance.
"If we knew what we were doing, it wouldn’t be called research." - Albert Einstein

2018: Battery Buggy, Road Scholar, Roller Coaster
2019: Chem Lab, Code, Disease, Fossils, Geo Maps, Sounds
2020 and 2021: Astro, Chem Lab, Code, Fossils, Geo Maps, Sounds

RiverWalker88
Exalted Member
Posts: 103
Joined: February 24th, 2020, 7:14 pm
Division: C
State: NM
Pronouns: He/Him/His
Has thanked: 79 times
Been thanked: 143 times
Contact:

### Re: Astronomy C

hippo9 wrote:
July 25th, 2020, 7:26 pm
0sm0sis wrote:
July 25th, 2020, 3:56 pm
nobodynobody wrote:
July 25th, 2020, 3:08 pm
Idk if this question even has a solution but here goes:

Imagine 3 masses where M2>>M1 and the mass of M3 is negligible. M1 is orbiting around M2 at distance R with a period of P. M3 is also is orbiting M2 with the same period of P. What distance(s) could M3 be orbiting M2 from? Assume ideal circular orbits.
By equating forces in a non-inertial frame, we have the gravitational forces = mass * angular velocity ^2 * radius. Replacing the angular velocity on RHS with the period of the orbit of M1 gives us M1M2 / R^2 + M2M3 / r^2 = M2M3 r / R^2 (where r is the desired distance). Then we can taylor expand and simplify to:

r = R ( 1 ± cbrt( M1 / 3*M2 ) )
I haven't done this in a while so I could be completely off, but shouldnt the distances be the same? By kepler's 3rd P^2=(4((pi)^2)a^3)/GM. No where in kepler's 3rd is the mass of the orbiting object included. Therefore if the periods are the same, they should be orbiting at the same distance.
p^2=\frac{4 \pi^2 a^3}{GM} I'm pretty sure M, in this case, is mass. And, as the semimajor axis is being divided by mass, mass and semimajor axis cubed are inversely proportional. Note: No crosschecking or other research was done for this post, so I could be the incorrect one here.

EDIT: Removed math bbcode because apparently math doesn't like being jammed into a spoiler
Last edited by RiverWalker88 on July 25th, 2020, 10:52 pm, edited 1 time in total.
Socorro High School (2021 Events: Astro, Chem Lab, Circuit Lab, Codybusters, Detector, ExDes, Machines)
2021 Socorro High Invitational Director (Thanks to those who competed and volunteered, it was fun!)
RiverWalker88's Userpage (Mostly Complete)

Lemonism Forever

hippo9
Member
Posts: 270
Joined: March 12th, 2018, 9:35 am
Division: C
State: IN
Has thanked: 1 time
Been thanked: 6 times

### Re: Astronomy C

RiverWalker88 wrote:
July 25th, 2020, 10:50 pm
hippo9 wrote:
July 25th, 2020, 7:26 pm
0sm0sis wrote:
July 25th, 2020, 3:56 pm

By equating forces in a non-inertial frame, we have the gravitational forces = mass * angular velocity ^2 * radius. Replacing the angular velocity on RHS with the period of the orbit of M1 gives us M1M2 / R^2 + M2M3 / r^2 = M2M3 r / R^2 (where r is the desired distance). Then we can taylor expand and simplify to:

r = R ( 1 ± cbrt( M1 / 3*M2 ) )
I haven't done this in a while so I could be completely off, but shouldnt the distances be the same? By kepler's 3rd P^2=(4((pi)^2)a^3)/GM. No where in kepler's 3rd is the mass of the orbiting object included. Therefore if the periods are the same, they should be orbiting at the same distance.
p^2=\frac{4 \pi^2 a^3}{GM} I'm pretty sure M, in this case, is mass. And, as the semimajor axis is being divided by mass, mass and semimajor axis cubed are inversely proportional. Note: No crosschecking or other research was done for this post, so I could be the incorrect one here.

EDIT: Removed math bbcode because apparently math doesn't like being jammed into a spoiler
Once again correct me if I am wrong but M should be the mass of the body that the object is orbiting around, not the mass of the object that is orbiting. So in this problem, the mass is equal to M2 in both situations.
These users thanked the author hippo9 for the post:
RiverWalker88 (July 26th, 2020, 8:19 am)
"If we knew what we were doing, it wouldn’t be called research." - Albert Einstein

2018: Battery Buggy, Road Scholar, Roller Coaster
2019: Chem Lab, Code, Disease, Fossils, Geo Maps, Sounds
2020 and 2021: Astro, Chem Lab, Code, Fossils, Geo Maps, Sounds

Name
Member
Posts: 424
Joined: January 21st, 2018, 4:41 pm
Division: C
State: NY
Pronouns: He/Him/His
Has thanked: 39 times
Been thanked: 32 times

### Re: Astronomy C

hippo9 wrote:
July 25th, 2020, 11:11 pm
Once again correct me if I am wrong but M should be the mass of the body that the object is orbiting around, not the mass of the object that is orbiting. So in this problem, the mass is equal to M2 in both situations.
From what I understand, the mass in KTL is the combined mass of the entire system (M+m), in which the mass of the smaller object can usually, but not always be ignored.
Now to be honest I have no idea what 0sm0sis did, so I'll try to answer this myself? I'm don't think a equation though fully satisfies the question asked.
So m3's mass is negligible meaning that I can set a baseline with it. Let's just arbitrarily say the mass of m2 is 1 solar mass and has a period of 1 year, leaving us with a distance of 1 au. Now with m1's mass being much less then m2's mass, it should be effectively negligible as well meaning that the period should also be one year. But "much less then" is also arbitrary. So instead, lets just assume it means less then. We can set a upper bound of the period by saying that the mass of m2=m1, which means the combined mass is 2 solar mass. With 2 solar mass and a period of 1 year we get a distance of 1.26 au (cbrt2). So my attempt to answer the question "What distance(s) could m3 be orbiting m2 from" resulted in the answer of 1 to 1.26 times the period of m2
this is when i realized it asked for m3 around m2 and not m1 around m2 oops. so the answer for m3 should be between .79 and 1 times the period of m1 around m2.
South Woods MS, Syosset HS '21
Favorite Past Events: Microbe, Invasive, Matsci, Fermi
Events: Astro, Code, Fossils
My Userpage
UTA/MIT/H-B
Astro: 1/5/x
Code: 1/11/x
Fossils: 3/47/13
GeoMaps: x/5/4


0sm0sis
Member
Posts: 6
Joined: July 25th, 2020, 3:30 pm
Division: C
State: OH
Has thanked: 2 times
Been thanked: 4 times

### Re: Astronomy C

I think there's been some confusion about the question and my solution, so I'll try to better explain my thought process.

Orbiting at the same distance R is a valid approximation if M3 is far enough away from M1 (where the gravitational force between the two is negligible) but if all three masses are collinear and M3 is very close to M1, there exist two other solutions to the problem, because the gravitational force is appreciable and must be factored in. I assumed this is what nobodynobody was asking for. In my solution, the fact that r and R were very close (r/R is approximately 1) was a very important part for the Taylor expansion.

I'm a little confused by Name's solution, but I think something important that was left out is that the orbit of M1 around M2 is approximately circular because M2 >> M1, and the center of mass of the system pretty much coincides with the center of mass of M2. Because of this, we can't set too high an upper bound. That being said, M1 being much much less than M2 doesn't mean that the gravitational force from it is negligible, especially if it's very close to M3.
These users thanked the author 0sm0sis for the post:
RiverWalker88 (July 26th, 2020, 8:18 am)
2020-21 Events:

Circuit Lab, Detector, Machines, SOM

"All science is either physics or stamp collecting." - Ernest Rutherford

nobodynobody
Member
Posts: 36
Joined: January 8th, 2020, 5:41 pm
Division: C
State: OH
Pronouns: He/Him/His
Has thanked: 9 times
Been thanked: 7 times

### Re: Astronomy C

0sm0sis wrote:
July 26th, 2020, 7:48 am
I think there's been some confusion about the question and my solution, so I'll try to better explain my thought process.

Orbiting at the same distance R is a valid approximation if M3 is far enough away from M1 (where the gravitational force between the two is negligible) but if all three masses are collinear and M3 is very close to M1, there exist two other solutions to the problem, because the gravitational force is appreciable and must be factored in. I assumed this is what nobodynobody was asking for. In my solution, the fact that r and R were very close (r/R is approximately 1) was a very important part for the Taylor expansion.

I'm a little confused by Name's solution, but I think something important that was left out is that the orbit of M1 around M2 is approximately circular because M2 >> M1, and the center of mass of the system pretty much coincides with the center of mass of M2. Because of this, we can't set too high an upper bound. That being said, M1 being much much less than M2 doesn't mean that the gravitational force from it is negligible, especially if it's very close to M3.
Yup! What I was going for was the lagrangian points. While it is true that a different radius of orbit will give a different period in a 2 body problem, M1 can change things in a restricted 3-body problem. At the lagrangian points, objects can orbit around M2 and essentially remain in the same place relative to M1 (thus having the same period). The equation for L1 and L2 is exactly what you answered, however there are still 3 more points. The other 3 points (L3, L4, and L5) are at the same distance R.

To reiterate, M1 also has mass and has an effect on M3.

Note: I realized after writing the question that we could take this further. If we stretch the question a bit, we could set up a scenario where M3 orbits around M1. We could define the period of M3 as the time it takes to return to the exact same location (when the M1 system orbits M2) . M3 techinically wouldn't be orbiting M2 in a circle, but the entire system does. In that case, every single distance (within the hill sphere of M1) that M3 could orbit M1 from that results in a period that divides evenly into P would result in M3 returning to the same location every P years. Unlike lagrangian points, it wouldn't stay at a constant distance from M2.

Edit: forgot spoilers, oops!
Last edited by nobodynobody on July 26th, 2020, 11:03 am, edited 1 time in total.
Class of '23
2021 events: Astro, Digi, SOM, WICI

"No." - Marie Curie

0sm0sis
Member
Posts: 6
Joined: July 25th, 2020, 3:30 pm
Division: C
State: OH
Has thanked: 2 times
Been thanked: 4 times

### Re: Astronomy C

1. What principle of quantum mechanics prevents white dwarfs and neutron stars from collapsing?
2. Explain how millisecond pulsars acquire the vast energy required to spin at up to 700 times per second.
3. What kind of emissions occur due to starquakes on a magnetar's surface?
These users thanked the author 0sm0sis for the post:
nobodynobody (July 26th, 2020, 11:49 am)
2020-21 Events:

Circuit Lab, Detector, Machines, SOM

"All science is either physics or stamp collecting." - Ernest Rutherford

RiverWalker88
Exalted Member
Posts: 103
Joined: February 24th, 2020, 7:14 pm
Division: C
State: NM
Pronouns: He/Him/His
Has thanked: 79 times
Been thanked: 143 times
Contact:

### Re: Astronomy C

1. Pauli exclusion principle
2. An neutron star draws in matter from a companion star, which strikes the surface parallel to the star's motion, providing a "push" on the star, and speeding it up.
3. Gamma ray emissions

Socorro High School (2021 Events: Astro, Chem Lab, Circuit Lab, Codybusters, Detector, ExDes, Machines)
2021 Socorro High Invitational Director (Thanks to those who competed and volunteered, it was fun!)
RiverWalker88's Userpage (Mostly Complete)

Lemonism Forever

### Who is online

Users browsing this forum: No registered users and 0 guests