Space Math: The Dawn Mission -- Ion Rockets and Spiral Orbits
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Space Math: The Dawn Mission -- Ion Rockets and Spiral Orbits

Topic: Math

Body: Asteroids

Mission: Dawn (Dwarf Planets)

Science Education Standards: Content Standard E -- Science and Technology: Understandings about Science and Technology

Short Description: Students determine the shape of the trajectory taken by a spacecraft using a constant-thrust ion motor using differential and integral calculus for arc lengths.

Source: Space Math (GSFC)

Ion rocket motors provide a small but steady thrust, which causes a spacecraft to accelerate. The shape of the orbit for the spacecraft as it undergoes constant acceleration is a spiral path. The length of this path can be computed using calculus.

The arc length integral can be written in polar coordinates where the function, y = F(x) is replaced by the polar function r(Θ).

Because the integrand is generally a messy one for most realistic cases, in the following problems, we will explore some simpler approximations.

The Dawn spacecraft was launched on September 27, 2007, and will take a spiral journey to visit the asteroid Vesta in February 2015. Earth is located at a distance of 1.0 Astronomical Units from the Sun (1 AU = 150 million kilometers) and Vesta is located 2.36 AU from the Sun. The journey will take about 66,000 hours and make about 3 loops around Earth's orbit in its outward spiral as shown in the figure to the left.

Problem 1 - Suppose that the Dawn spacecraft travels at a constant outward speed from Earth's orbit. If we approximate the motion of the spacecraft by X = R cosΘ, Y=RsinΘ and R = 1 + 0.08 Θ, where the angular measure is in radians, show that the path taken by Dawn is a simple spiral.

Problem 2 - From the equation for R( Θ), compute the total path length of the spiral from R=1.0 to R = 2.36 AU, and give the answer in kilometers. About what is the spacecraft's average speed during the journey in kilometers/hour? [Note: Feel free to use a Table of Integrals!]

Problem 3 - The previous two problems were purely 'kinematic' which means that the spiral path was determined, not by the action of physical forces, but by employing a mathematical approximation. The equation for R(Θ) is based on constant-speed motion, and not upon actual accelerations caused by gravity or the action of ion engine itself. Let's improve this kinematic model by approximating the radial motion by a uniform acceleration given by R(Θ) = 1/2 A Θ2 where we will approximate the net acceleration of the spacecraft in its journey as A = 0.009. What is the total distance traveled by Dawn in kilometers, and its average speed in kilometers/hour?

Problem 2: R = 1.0 + 0.08 Θ and so dR/dΘ = 0.08 and dΘ/dR = 12.5. The integrand becomes ( 1+ 156R2 )1/2 dR .

If we use the substitution U = 12.5R dU = 12.5 dR and the integrand becomes 0.08 (1 + U2)1/2 dU. A table of integrals yields the answer

1/2 [ U (1 + U2)1/2 + ln (U + (1+ U2)1/2).

The limits to the integral are Ui = 12.5 x 1.0 = 12.5 and Uf = 12.5 x 2.36 = 29.5, and when the integral is evaluated we get 1/25 [ 29.5 (29.5) + ln (29.5 + (29.5) ) - 12.5 (12.5) - ln(12.5 + (12.5))] = 1/25 (870 + 4.1 - 156 - 3.2) = 28.6 Astronomical Units or 28.6 x 150 million km = 4.3 billion kilometers! The averages speed would be about 4.3 billion/66000 hrs = 65,100 kilometers/hour.

Problem 3 - dR/dΘ = A Θ so that dΘ/dR = 1/(A Θ).

From R(Θ), we can re-write dΘ/dR solely in terms of R as dΘ/dR = (1/(2Ar))1/2 so that the integrand becomes (1 + R/(2A))1/2 dR.

Unlike the integral in Problem 1 ,this integral can be easily performed by noting that if we substitute

U = 1 + R/(2A), and dU = dR/2A,

we get the integrand 2A U1/2 dU and so S = (4A/3) U3/2 + C.

The limits to this integral are Ui = 1 + 1.0/2A = 56. and Uf = 1 + 2.36/2A = 132.

Then the definite integral becomes S = (4 x 0.009/3) [ 1323/2 - 563/2] = 0.012 [ 1516 - 419] = 13.2 AU .Since 1 AU = 150 million km, the spiral path has a length of 2.0 billion kilometers. The averages speed would be about 2.0 billion km/66000 hours = 30,300 km/hour. The trip takes less time because the 'kinematic' motion is speeded up towards the end of the journey.