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Gravitation: Orbits


Terms (orbits)

Problems on Orbits

Equations of motion

We can write expressions for both the angular momentum and the total energy. If p θ is the magnitude of the momentum in the tangential direction, then since this perpendicular to , L = rp θ . But p θ = mv θ = m = mr = mr . Hence L = r(mr ) = mr 2 . Hence:

L = mr 2    

We can also write an expression for the total energy as a sum of the radial kinetic energy term, the angular kinetic energy term and the potential term:

E = 1/2m + -    

Rearranging and dividing through the left side by m 2 r 4 and the right by L 2 , and canceling factors of dt 2 we find:

= - -    

To find the equations of motion we want to find r in terms of θ . In principle we could take the square root of both sides of the above equation, separate the variables, integrate to find θ(r) , and then invert to find r(θ) . This involves a lot of messy algebra which is not very enlightening, so we will just state the result:

= (1 + εcosθ)    

where ε is the eccentricity that we saw in Ellipses and Foci and is now given by:

ε =    

This equation determines the motion of all orbital systems in the universe.

We can also find the maximum and minimum values of r . The minimum occurs where the expression for 1/r is a maximum. This is when cosθ = 1 and the maximum is therefore . Thus:

r min =    

The maximum occurs for the minimum of 1/r . There are two cases: first, when ε < 1 , the minimum of 1/r is . Thus:

r max =    

When ε≥1 , the expression for 1/r can take on the value zero when cosθ = - 1/ε . Hence the maximum value r can take on is infinite in this case.

We can also take the equation and using r 2 = x 2 + y 2 , and cosθ = x/r , we can write:

x 2 + y 2 = -    

Solving the orbits

The orbits are determined by the various values that ε can take.

Circular orbits

When ε = 0 , the expression for ε tells us that E = - . The negative value of the energy just means that the potential energy is more negative than the kinetic energy is positive. In this case we have r min = r max = . The particle is trapped at the very bottom of a potential well, and the radius does not change as it goes around the orbit, hence forming a circle. Substituting this value for r into the energy we have E = - . Note that we could have derived this directly by summing the potential energy we found for a circular orbit with the kinetic energy (Gravitational Potential Energy).

E = 1/2mv 2 + U = - = -    

Figure %: Potential well for a circular orbit.
In the case ε = 0 we can see that this equation simplifies to x 2 + y 2 = . This describes a circle with radius .

Elliptical Orbits

Elliptical orbits occur when 0 < ε < 1 . This means that - < E < 0 . Again the particle is trapped in a potential well, oscillating now between r min and r max .

Figure %: Potential well for a elliptical orbit.
We can also solve the equation for ε in this range. The algebra turns out to be complicated, but we find that:

+ = 1    

where a = and b = . This is an ellipse with its center at (- L 2 ε/GMm 2(1 - ε 2), 0) , and with semimajor and semiminor axis length a and b respectively. It can also be shown that one focus of this ellipse is at the origin.

Parabolic and Hyperbolic Orbits

Parabolic orbits occur when ε = 1 , which means that E = 0 .

Figure %: Potential graph for parabolic and hyperbolic orbits.
Hyperbolic orbits occur when ε > 1 , which means that E > 0 . In both cases the particle will go off to infinity. In the parabolic case the particle barely has enough energy to make it to infinity, but the hyperbolic orbit makes it with energy to spare. In the parabolic orbit this equation simplifies to y 2 = ±2x . This is an equation for a parabola with its vertex at (, 0) .

Figure %: Shapes of various orbits.
Shows the shapes and locations of the circular, elliptical and parabolic orbits.

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