Escape Through Functions

Nice problem. Simulation code is attached. Note that the net graviational potential energy is negative, which means that the particle has to have positive initial kinetic energy in order to escape the system.

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import math

# Constants

m = 1.0
G = 1.0

Nx = 2.0
Ny = 0.0

dx = 10.0**(-5.0)

lam1 = -1.0
lam2 = 1.0

#################################

# Integrate gravitational potential energy

U = 0.0

x = 0.0

while x <= 2.0*math.pi:

    x1 = x
    y1 = math.sin(x)

    x2 = x
    y2 = math.cos(x)

    dx1 = dx
    dx2 = dx

    dy1 = math.cos(x)*dx
    dy2 = -math.sin(x)*dx

    dL1 = math.hypot(dx1,dy1)
    dL2 = math.hypot(dx2,dy2)

    dM1 = dL1 * lam1
    dM2 = dL2 * lam2

    r1 = math.hypot(x1-Nx,y1-Ny)
    r2 = math.hypot(x2-Nx,y2-Ny)

    dU = -G*m*dM1/r1 - G*m*dM2/r2

    U = U + dU

    x = x + dx

#################################

# Verify that net U is negative
# Calculate escape velocity

print dx
print U

U = math.fabs(U)


# U = 0.5*m*(v0**2.0)

v0 = math.sqrt(2.0*U/m)

print v0

#################################

#>>> 
#0.0001
#-1.54850452534
#1.75983210866
#>>> ================================ RESTART ================================
#>>> 
#1e-05
#-1.54851262164
#1.75983670927
#>>> 

Nice problem. However, I find the notion of 'negative mass' difficult to interpret.

Consider the curve $y_1=\sin{x}$ . Mass of an elementary arc length near a point $(x,y_1)$ is:

$dm_1 = - dx\sqrt{1+\cos^2{x}}$

Gravitational potential energy due to this element is:

$dV_1 = -\frac{Gm \ dm_1}{\sqrt{(x-2)^2 + y_1^2}}$

Same for the curve $y_2=\sin{x}$ . Mass of an elementary arc length near a point $(x,y_2)$ is:

$dm_2 = dx\sqrt{1+\sin^2{x}}$

Gravitational potential energy due to this element is:

$dV_2 = -\frac{Gm \ dm_2}{\sqrt{(x-2)^2 + y_2^2}}$

Potential energy due to both elements:

$dV = dV_1 + dV_2 = \left(\sqrt{\frac{1+\cos^2{x}}{(x-2)^2 +\sin^2{x}}} -\sqrt{\frac{1+\sin^2{x}}{(x-2)^2 +\cos^2{x}}}\right)dx$

Total potential energy due to both curves is:

$V = \int_{0}^{2\pi} \left(\sqrt{\frac{1+\cos^2{x}}{(x-2)^2 +\sin^2{x}}} -\sqrt{\frac{1+\sin^2{x}}{(x-2)^2 +\cos^2{x}}}\right)dx$

Now, for the particle to just escape the gravitational field, it must be projected with a speed such that its final energy becomes zero. Applying the energy conservation principle gives:

$V + \frac{v^2}{2} = 0$ $\boxed{v = \sqrt{-2\int_{0}^{2\pi} \left(\sqrt{\frac{1+\cos^2{x}}{(x-2)^2 +\sin^2{x}}} -\sqrt{\frac{1+\sin^2{x}}{(x-2)^2 +\cos^2{x}}}\right)dx} \approx 1.76}$