3.4 Motion in space (Page 7/17)

Calculus volume 3 Page 7 / 17

{‖ C ‖}^{2} = G M ‖ r ‖ + ‖ r ‖ ‖ D ‖ cos θ .

Solving for $‖ r ‖,$

‖ r ‖ = \frac{{‖ C ‖}^{2}}{G M + ‖ D ‖ cos θ} = \frac{{‖ C ‖}^{2}}{G M} (\frac{1}{1 + e cos θ}),

where $e = ‖ D ‖ / G M .$ This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if $e > 1,$ a parabola if $e = 1,$ or an ellipse if $e < 1 .$ Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.

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Using kepler’s third law for nonheliocentric orbits

Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes

P^{2} = \frac{4 π^{2} a^{3}}{G (m + M)},

where m is the mass of the Moon and M is the mass of Earth, a represents the length of the major axis of the elliptical orbit, and P represents the period.

Given that the mass of the Moon is $7.35 \times 10^{22} kg,$ the mass of Earth is $5.97 \times 10^{24} kg,$ $G = 6.67 \times 10^{−11} m^{3} / kg \cdot {sec}^{2},$ and the period of the moon is 27.3 days, let’s find the length of the major axis of the orbit of the Moon around Earth.

It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:

27.3 days \times \frac{24 hr}{1 day} \times \frac{3600 sec}{1 hour} = 2,358,720 sec .

Substitute all the data into [link] and solve for a:

\begin{matrix} {(2,358,720 sec)}^{2} & = & \frac{4 π^{2} a^{3}}{(6.67 \times 10^{−11} \frac{m^{3}}{kg \cdot {sec}^{2}}) (7.35 x 10^{22} kg + 5.97 x 10^{24} kg)} \\ 5.563 \times 10^{12} & = & \frac{4 π^{2} a^{3}}{(6.67 \times 10^{−11} m^{3}) (6.04 x 10^{24})} \\ (5.563 \times 10^{12}) (6.67 \times 10^{−11} m^{3}) (6.04 \times 10^{24}) & = & 4 π^{2} a^{3} \\ a^{3} & = & \frac{2.241 \times 10^{27}}{4 π^{2}} m^{3} \\ a & = & 3.84 \times 10^{8} m \\ \approx & 384,000 km . \end{matrix}

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Titan is the largest moon of Saturn. The mass of Titan is approximately $1.35 \times 10^{23}$ kg. The mass of Saturn is approximately $5.68 \times 10^{26}$ kg. Titan takes approximately 16 days to orbit Saturn. Use this information, along with the universal gravitation constant $G = 6.67 \times 10^{−11} m^{3} / kg \cdot {sec}^{2}$ to estimate the distance from Titan to Saturn.

$a = 1.224 \times 10^{9} m \approx 1,224,000 km$

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Chapter opener: halley’s comet

This is a picture of Halley’s Comet. It is a bright ball of light towards the right of the picture with a tail of trailing light. There are also stars throughout the picture.

We now return to the chapter opener, which discusses the motion of Halley’s comet around the Sun. Kepler’s first law states that Halley’s comet follows an elliptical path around the Sun, with the Sun as one focus of the ellipse. The period of Halley’s comet is approximately 76.1 years, depending on how closely it passes by Jupiter and Saturn as it passes through the outer solar system. Let’s use $T = 76.1$ years. What is the average distance of Halley’s comet from the Sun?

Using the equation $T^{2} = D^{3}$ with $T = 76.1,$ we obtain $D^{3} = 5791.21,$ so $D \approx 17.96$ A.U. This comes out to approximately $1.67 \times 10^{9}$ mi.

A natural question to ask is: What are the maximum (aphelion) and minimum (perihelion) distances from Halley’s Comet to the Sun? The eccentricity of the orbit of Halley’s Comet is 0.967 (Source: http://nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html). Recall that the formula for the eccentricity of an ellipse is $e = c / a,$ where a is the length of the semimajor axis and c is the distance from the center to either focus. Therefore, $0.967 = c / 17.96$ and $c \approx 17.37$ A.U. Subtracting this from a gives the perihelion distance $p = a - c = 17.96 - 17.37 = 0.59$ A.U. According to the National Space Science Data Center (Source: http://nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html), the perihelion distance for Halley’s comet is 0.587 A.U. To calculate the aphelion distance, we add

P = a + c = 17.96 + 17.37 = 35.33 A .U .

This is approximately $3.3 \times 10^{9}$ mi. The average distance from Pluto to the Sun is 39.5 A.U. (Source: http://www.oarval.org/furthest.htm), so it would appear that Halley’s Comet stays just within the orbit of Pluto.

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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