Gravitation: Orbits

Problem : In Solving the Orbits we derived the equation:

= - +

= -

Solution for Problem 1 >>

Making the y substitution, we have:

= - y 2 + y +

We can then complete the square on the right-hand side and we have:

= - y - + +

Then we let p = y - and we have:

= - p 2 + 1 + = - p 2 + C

where we have just defined C in terms of the constant E, G, M, m, L . We can now separate variables:

= dθ'âá’cos -1(p'/)|p p1 = θ - θ 1âá’p = cos(θ - θ 1) - cos-1(p 1/) = cos(θ - θ 0)

Finally, recalling p = (1/r) - , and we can pick an axis such that θ ₀ = 0 :

= 1 + cosθ

The quantity under the radical is defined a ε , the eccentricity.

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Problem : Using the expression we derived for (1/r) , show that this reduces to x ² = y ² = k ² -2kεx + ε ² x ² , where k = , ε = , and cosθ = x/r .

Solution for Problem 2 >>

We have:

= (1 + εcosθ)âá’1 = (1 + ε )âá’k = r + εx

We can solve for r and then use r ² = x ² + y ² :

x 2 + y 2 = k 2–2kxε + x 2 ε 2

which is the result we wanted.

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Problem : For 0 < ε < 1 , use the above equation to derive the equation for an elliptical orbit. What are the semi-major and semi-minor axis lengths? Where are the foci?

Solution for Problem 3 >>

We can rearrange the equation to (1 - ε ²)x ² +2kεx + y ² = k ² . We can divide through by (1 - ε ²) and complete the square in x:

x - - - =

Rearranging this equation into the standard form for an ellipse we have:

+ = 1

This is an ellipse with one foci at the origin, the other at (, 0) , semi-major axis length a = and semi-minor axis length b = .

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Problem : What is the energy difference between a circular earth orbit of radius 7.0×10³ kilometers and an elliptical earth orbit with apogee 5.8×10³ kilometers and perigee 4.8×10³ kilometers. The mass of the satellite in question is 3500 kilograms and the mass of the earth is 5.98×10²⁴ kilograms.

Solution for Problem 4 >>

The energy of the circular orbit is given by E = - = 9.97×10¹⁰ Joules. The equation used here can also be applied to elliptical orbits with r replaced by the semimajor axis length a . The semimajor axis length is found from a = = 5.3×10⁶ meters. Then E = - = 1.32×10¹¹ Joules. The energy of the elliptical orbit is higher.

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Problem : If a comet of mass 6.0×10²² kilograms has a hyperbolic orbit around the sun of eccentricity ε = 1.5 , what is its closest distance of approach to the sun in terms of its angular momentum (the mass of the sun is 1.99×10³⁰ kilograms)?

Solution for Problem 5 >>

Its closest approach is just r _min , which is given by:

r min = = (6.44×10-67)L 2

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