Oscillation in Space: Derive an Equation to Determine Period in Years

[xgoodtimesx]

In the middle of empty space two equal masses, M = 2.1 solar masses, lie fixed on a horizontal line separated by a distance D = 8.4 AU. Halfway between the masses lies a third mass. The third mass is displaced vertically a distance y where y << D. Due to the gravitational attraction from the other two masses, the third mass will be pulled back toward the line. The subsequent motion of the third mass will be approximately simple harmonic. Determine the period of the motion in years.


Please derive a equation to solve this problem

 

[tiny-tim]

Hi xgoodtimesx!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[nlightner]

, and provide a solution.

To determine the period of the oscillation in this scenario, we can use the equation for the period of a simple harmonic motion, T = 2π√(m/k), where m is the mass of the object and k is the spring constant.

In this case, the third mass is experiencing a restoring force due to the gravitational attraction from the other two masses. This force can be approximated using Newton's law of universal gravitation, F = Gm1m2/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two fixed masses, and r is the distance between them.

Since the third mass is only displaced a small distance y from the line, we can assume that the force acting on it is proportional to y. This can be expressed as F = -ky, where k is the spring constant equivalent for gravitational force.

Equating these two equations, we get -ky = Gm1m2/r^2. Solving for k, we get k = -Gm1m2/r^3.

Substituting this value of k into the equation for period, T = 2π√(m/k), we get T = 2π√(m/(-Gm1m2/r^3)). Simplifying, we get T = 2π√(-r^3/(Gm1m2)).

Now, we can substitute the given values of m1, m2, r, and the mass of the third object (m) into this equation to solve for the period in seconds.

T = 2π√(-8.4 AU)^3/((6.67x10^-11 Nm^2/kg^2)(2.1 solar masses)(2.1 solar masses))

Converting AU to meters and solar masses to kilograms, we get T = 2π√(-1.256x10^14 m^3/(1.4x10^30 kg)^2)

Simplifying, we get T = 2π√(-1.256x10^-2 m^3/1.96x10^60 kg^2)

Taking the square root and converting to years, we get T = 2π√(3.21x10^-63) years

Simplifying, we get T = 2π(