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September 18, 2013

Spacecraft missions to outer planets, minor planets or comets

Filed under: science — ckrao @ 11:50 am

After recently hearing news about Voyager 1 being the first human-made object to leave our solar system, I looked up other spacecraft that have visited the outer planets, minor planets (e.g. asteroids) and comets.

Outer Planets

 Spacecraft  Launch Date  Agency  Remarks
Pioneer 10 3-Mar-72 NASA / ARC flyby of Jupiter in Dec 1973, lost contact in Jan 2003 when 12b km from earth
Pioneer 11 6-Apr-73  NASA /ARC  flyby of Jupiter in Dec 1974, Saturn in May 1979, lost contact in 1995
Voyager 2 20-Aug-77  NASA / JPL flyby of all outer planets, the only one to fly by Uranus and Neptune
Voyager 1 5-Sep-77  NASA / JPL flyby of Jupiter in Sep 1977, Saturn in Mar 1979, now 19b km from earth
Galileo 18-Oct-89 NASA flyby of Venus, Earth, asteroids Gasra and Ida (both incidental) before detailed study of Jupiter and its moons, probe dropped into its atmosphere
Ulysses 6-Oct-90 NASA / ESA flyby of Jupiter (Feb 1992), orbiting sun, measuring solar wind and gamma ray bursts
Cassini–Huygens 15-Oct-97 NASA/ESA flyby of Jupiter & Saturn, Huygens deployed from Cassini, landed on Saturn’s moon Titan in Jan 2005
New Horizons 19-Jan-06  NASA expected to fly by Pluto in July 2015
Juno 5-Aug-11  NASA expected to reach Jupiter in Aug 2016

Minor Planets and comets (other than Halley)

 Spacecraft  Launch Date  Agency  Remarks
Dawn 27-Sep-07 NASA orbited Vesta July ’11-Sep ’12, expected to reach Ceres in Feb ’15
Hayabusa 9-May-03 JAXA visited asteroid Itokawa bringing back to earth tiny grains of asteroidal material in June 2010
NEAR Shoemaker 17-Feb-96 NASA studied near-earth asteroid Eros, touched down in Feb 2001
Deep Space 1 24-Oct-98 NASA/JPL flybys of asteroid Braille and Comet Borelly
Stardust 7-Feb-99 NASA/JPL returned dust samples from Comet Wild 2, also intercepted comet Tempel 1 in Feb 2011
Deep Impact 12-Jan-05 NASA/JPL impactor collided with nucleus of comet Tempel in July 2005, extended mission EPOXI flew by Comet 103P/Hartley, now on its way to asteroid 2002 GT
Rosetta 2-Mar-04 ESA flybys of asteroids Steins and Lutetia in 2008, 2010; expected to reach comet 67P/Churyumov-Gerasimenko in mid 2014 deploying the lander Philae
Chang’e-2 1-Oct-10 CNSA flew by asteroid 4179 Toutatis in Dec 2012 after orbiting moon
International Cometary Explorer (ICE) 12-Aug-78 NASA/ESA first spacecraft to pass through comet tail (Comet Giacobini-Zinner)

Halley’s Comet (during its most recent near-earth encounter in 1985-6)

 Spacecraft  Launch date  Agency  Remarks
Giotto 2-Jul-85 ESA passed within 600km of Halley’s comet, then flew by Comet Grigg-Skjellerup in Jul 1992
Vega 1 15-Dec-84 USSR flyby of Venus (descent craft surfaced on Venus) and Halley’s Comet (within 9,000km Mar ’86)
Vega 2 21-Dec-84 USSR flyby of Venus (descent craft surfaced on Venus) and Halley’s Comet (Mar ’86)
Suisei 18-Aug-85 ISAS (now part of JAXA) within 151,000km of Comet Halley in Mar ’86
Sakigake 7-Jan-85 ISAS (now part of JAXA) within 7m km of Comet Halley

Further reading:

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September 16, 2013

Polynomial extrapolation of finite exponential sequences

Filed under: mathematics — ckrao @ 11:40 am

Here is a cute problem I first encountered many years ago:

A polynomial P(x) of degree n satisfies P(k) = 2^k for k = 0, 1, 2, \ldots, n. Determine P(n+1).

Recall that a polynomial of degree n is uniquely determined by its value at n+1 points. Hence P(n+1) could have only one possible value. Is it 2^{n+1}? Let’s see: here is the solution I saw at the time.

Consider the polynomial \displaystyle P(x) = \binom{x}{0} + \binom{x}{1} + \ldots + \binom{x}{n}. Then P(x) is a polynomial of degree n (the k‘th term has degree k+1). Furthermore, since \binom{x}{k} = 0 for x = 0, 1, \ldots, k-1, we have for k = 0, 1, 2, \ldots, n the following:

\displaystyle P(k) = \binom{k}{0} + \binom{k}{1} + \ldots + \binom{k}{n} = \binom{k}{0} + \binom{k}{1} + \ldots + \binom{k}{k} = 2^k.

(Here we recall that the sum of a row of Pascal’s triangle is a power of 2.)

This shows that P(x) is our unique polynomial with the desired properties, and we find

\displaystyle P(n+1) = \binom{n+1}{0} + \binom{n+1}{1} + \ldots + \binom{n+1}{n} = 2^{n+1} - 1.

Neat isn’t it that the polynomial almost climbs up to 2^n but ends up one short. 🙂 Let’s see how this result generalises. Modify the above question so that P(k) = a^k for k = 0, 1, 2, \ldots, n (where a is a constant not equal to 1). What is P(n+1)? This time we let

\displaystyle P(x) := (a-1)^0 \binom{x}{0} + (a-1)^1 \binom{x}{1} + \ldots + (a-1)^n \binom{x}{n} = \sum_{i=0}^n (a-1)^i \binom{x}{i}.

As before P(x) is clearly a degree n polynomial. Furthermore, for k = 0, 1, 2, \ldots, n we apply the binomial theorem to obtain

\begin{aligned} P(k) &= \sum_{i=0}^n (a-1)^i \binom{k}{i}\\ &= \sum_{i=0}^k (a-1)^i \binom{k}{i}\\ &= (a-1+1)^k\\ &= a^k. \end{aligned}

This constructed polynomial has the desired properties, so by uniqueness we simply substitute x=n+1 to find P(n+1):

\begin{aligned} P(n+1) &= \sum_{i=0}^n (a-1)^i \binom{n+1}{i}\\ &= \sum_{i=0}^{n+1} (a-1)^i \binom{n+1}{i} - (a-1)^{n+1} \binom{n+1}{n+1}\\ &= a^{n+1} - (a-1)^{n+1}.\end{aligned}

The way such polynomials can be constructed (i.e. not pulled out of thin air!) is to write down a finite difference table – refer to an earlier blog post here: if the first diagonal of the table is a_0, a_1, \ldots, a_n where a_0 = P(0), then

\displaystyle p(x) = \sum_{i=0}^n a_i \binom{x}{i}.

In the first problem it is easy to verify that a_i = 1, in the generalisation a_i = (a-1)^i.

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