Roche Limit: Approximating x/d

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In summary, the student attempted to solve an equation of motion for a test particle that is situated at a distance x from the center of a secondary (on a line that goes from the center of the secondary to the center of the primary) with a mass M and a mass of the secondary FD. He attempted to use a formula to express the force the primary exerts on the particle X, but got stuck. He then attempted to use the binomial approximation for calculating FO, but was not able to get close to the term he was looking for. He eventually solved for d using the equation of motion of the particle X.
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1. Homework Statement

Show that d=(9M/(4*pi*p))^1/3 is an approximation for the roche limit. Note that x/d <<1
with M = mass of the primary
p = density of the secondary
x= distance of a test particle from the center of the secondary (in part a) of the task one should give the motion equation for X. X is a particle that is situated at distance x from the center of the secondary (on a line that goes from the center of the secondary to the center of the primary)
m = mass of the secondary

Homework Equations

The Attempt at a Solution


In the critical point the tidal force excerted from the primary and the gravitational force that hold the secondary together are equal and therefore the secondary breaks. So I tried to put these forces equal and to solve the equation for d, but it did not work out.
 
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  • #2
Hi physicstudent_B, Welcome to Physics Forums.

Can you show us the details of what you tried and point out where you think it went wrong or failed?
 
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  • #3
Thank you for your message and for welcoming me to the Physics Forum.
So the first part of this question was to write the equation of motion for X:
I think that there are two forces, who effect particle X: the gravitational force from the secondary FD and the gravitational force from the primary FO
FO+FD = mx * w2d (D is orbiting around O with constant velocity and D is not moving)
FO = (G*mx*M)/(d-x)^2
FD= (here I am not so sure, because obviously X is situated somewhere inside of D if I understood the question right, but from another exercise I remember a formula that you can use to calculate the gravitational force the Earth exerts on a body of mass m that is inside the Earth somewhere at distance r from the center of mass: F=mgr/R (R= radius of the Earth) --> so I tried to use this formula to express the force the primary exerts on the particle X :
FD=(mx*G*m*x) / R^3
Now I tried to put FO=FD and resolve for d, but I got stuck
I think I should use the binominal approximation for calculatin FO (so that 1(d-x)^2 = 1/d^2 * (1+x/d), also I think that I should express the mass in terms of volume (4/3*pi*r^3) * density, but I don't get close to the term I should get, so I wonder if there is something wrong with how I calculate the two forces.
 
  • #4
You can assume that the small test mass is on the surface of the object. While it does not matter for the calculation, the object cannot fly away if it is inside.

Be careful with the signs. The gravitational forces point in opposite directions. If you want to add them as in your first equation one has to be negative (which one?).

In the first equation: The radius of the orbit of x is not d.

You are missing two equations:
w as function of the given parameters.
A formula relating m/R3 and the density.

Once you have those and fixed the sign issue and the radius, you should be able to solve for d.
 
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  • #5
Thank you. So with the signs right it should be:
FO-FD= mx*w2*d
and the formula relating m/R3 and density would be m/R^3 = p * 4/3 * pi
I have not yet an idea for w as function of the given parameters.
And if X is on the surface of O, the x would be the same as the radius of O?
 
  • #6
physicstudent_B said:
I have not yet an idea for w as function of the given parameters.
Your mass m is in a circular orbit around the bigger mass.
physicstudent_B said:
And if X is on the surface of O, the x would be the same as the radius of O?
Right.
 
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  • #7
OK, w2=GMm/d3
Then I could solve for d using the equation of motion of the particle X.
But I did not yet completely understand: I thought that FO and FD would be the same at distance d, so that FO-FD=0, but that is obviously not the case?
 
  • #8
You'll see if you can neglect the orbital motion later. You cannot just assume it to be negligible.
 
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Related to Roche Limit: Approximating x/d

1. What is the Roche Limit and why is it important for approximating x/d?

The Roche Limit is the distance at which a celestial body, such as a planet or moon, will be pulled apart by the tidal forces of another celestial body. It is important for approximating x/d because it helps determine the point at which a satellite or object will be destroyed by the gravitational pull of a larger body.

2. How is the Roche Limit calculated?

The Roche Limit is calculated using the formula d = 2.44R(Mm/M), where d is the distance between the two bodies, R is the radius of the larger body, M is the mass of the larger body, and m is the mass of the smaller body.

3. What factors can affect the Roche Limit?

The main factors that can affect the Roche Limit are the masses and sizes of the two bodies involved, as well as the distance between them. Other factors such as the density and composition of the bodies may also play a role.

4. Can the Roche Limit be used to predict the formation of planetary rings?

Yes, the Roche Limit can be used to predict the formation of planetary rings. When a celestial body, such as a moon, passes within its Roche Limit of a larger body, the tidal forces can cause the smaller body to break apart and form a ring of debris around the larger body.

5. Are there any exceptions to the Roche Limit?

Yes, there are some exceptions to the Roche Limit. For example, the Roche Limit does not take into account the strength of a celestial body's surface. If a body has a strong enough surface, it may not be pulled apart even if it passes within its Roche Limit. Additionally, the presence of other celestial bodies can also affect the Roche Limit calculation.

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