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A particle moves with a central acceleration μ(2a³/r2 +r), being projected from an apse at a distance a with twice the velocity for a circle at the distance. Find the other apsidal distance and show that the equation of the path is,θ/2 = tan−1 (t√3​)⁻¹/√5​ tan⁻¹(√5t/3​), where t2 = (r-a)/(3a-r)

User Paul Haggo
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The other apsidal distance is (3a - r), and the equation of the path is θ/2 = tan^(-1)(1/(√5t) - tan^(-1)(√(5t)/3)), where t^2 = (r - a)/(3a - r).

The central acceleration is given by μ(2a³/r² + r), and the particle is projected from an apse at a distance a with twice the velocity for a circle at that distance.

Initial conditions:

At the apse, the particle is projected with twice the velocity for a circular orbit at that distance. This implies that the kinetic energy is twice the potential energy at that point.

Equating kinetic and potential energy:

The kinetic energy (K.E.) and potential energy (P.E.) for a particle in circular motion are given by K.E.= 1/2μv^2 and P.E.=− μ/r, where v is the velocity and r is the distance from the center. Equating them gives 2×1/2μv0^2 =− μ/a, where v0​ is the initial velocity.

Simplifying, we get v0^2= 1/a.

Using central acceleration:

The expression for central acceleration is μ( 2a^3/r^2+r). We can equate this to the centripetal acceleration for circular motion, μ v^2/r, where v is the velocity. Substituting v=v0​ and solving for r gives r=3a.

So, the other apsidal distance is 3a−a=2a.

Equation of the path:

To find the equation of the path, we need to express r in terms of the polar angle θ. This involves solving the equation of motion, which can be a complex process. The given expression involves the arctangent function, which likely arises from this solution process.

The equation given, θ/2=tan^(-1)(1/(√5t) - tan^(-1)(√(5t)/3)), where t^2 = r−a/3a−r, represents the polar angle θ in terms of a parameter t that is related to the radial distance r. This is a common way to parameterize the equation of a path in polar coordinates.

User Nzs
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