Bode's Law In 1772, Johann Bode published the following...

Bode's Law In $1772,$ Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2}, n \geq 2 $$ where $n$ is the number of the planet from the sun. (a) Determine the first eight terms of this sequence. (b) At the time of Bode's publication, the known planets were Mercury $(0.39 \mathrm{AU}),$ Venus $(0.72 \mathrm{AU}),$ Earth $(1 \mathrm{AU})$ Mars (1.52 AU), Jupiter (5.20 AU), and Saturn (9.54 AU). How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in 1781 and the asteroid Ceres was discovered in $1801 .$ The mean orbital distances from the sun to Uranus and Ceres" are $19.2 \mathrm{AU}$ and 2.77 AU, respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (c) The planets Neptune and Pluto* were discovered in 1846 and $1930,$ respectively. Their mean orbital distances from the sun are $30.07 \mathrm{AU}$ and $39.44 \mathrm{AU},$ respectively. How do these actual distances compare to the terms of the sequence? (f) On July $29,2005,$ NASA announced the discovery of a dwarf planet* $(n=11),$ which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

Bode's Law In $1772,$ Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun:
$$
a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2}, n \geq 2
$$
where $n$ is the number of the planet from the sun.
(a) Determine the first eight terms of this sequence.
(b) At the time of Bode's publication, the known planets were Mercury $(0.39 \mathrm{AU}),$ Venus $(0.72 \mathrm{AU}),$ Earth $(1 \mathrm{AU})$ Mars (1.52 AU), Jupiter (5.20 AU), and Saturn (9.54 AU). How do the actual distances compare to the terms of the sequence?
(c) The planet Uranus was discovered in 1781 and the asteroid Ceres was discovered in $1801 .$ The mean orbital distances from the sun to Uranus and Ceres" are $19.2 \mathrm{AU}$ and 2.77 AU, respectively. How well do these values fit within the sequence?
(d) Determine the ninth and tenth terms of Bode's sequence.
(c) The planets Neptune and Pluto* were discovered in 1846 and $1930,$ respectively. Their mean orbital distances from the sun are $30.07 \mathrm{AU}$ and $39.44 \mathrm{AU},$ respectively. How do these actual distances compare to the terms of the sequence?
(f) On July $29,2005,$ NASA announced the discovery of a dwarf planet* $(n=11),$ which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

Key Concepts

Empirical Modeling and Model Fitting

Empirical modeling involves constructing models based on observed data rather than derived laws. Bode's Law is an example of an empirical model where the formula was tested against known planetary distances and later used to predict additional ones. This leads to discussions on how well such models fit actual observations and their limitations in forecasting data outside the original sample.

Geometric Sequences

A geometric sequence is a series in which each term after the first is found by multiplying the previous term by a constant, known as the common ratio. In the formula discussed, the exponential component involving powers of 2 exemplifies a geometric progression within the sequence, which is essential for understanding the rapid increase in the predicted distances.

Bode's Law

Bode's Law is an empirical rule from the 18th century that proposes a simple numerical relationship to predict the distances of the planets from the Sun in astronomical units. Although not derived from fundamental physics, it historically provided a framework for comparing observed planetary distances with a mathematical pattern, sparking debate about the underlying order in the solar system.

Explicit Sequence Formulas

An explicit sequence formula defines each term of a sequence directly in terms of its index without requiring the computation of previous terms. This approach allows for immediate calculation of any term in the sequence, and in this context, it is used to generate predicted astronomical distances based on a simple mathematical expression.

Transcript

00:01 A1 is equal to 0 .4 and a .n is equal to 0 .4 plus 0 .3 multiplied by 2 to the power n minus 2 where n is greater than equal to 2.

00:13 In first part we have to find the first 8 terms of the given sequence.

00:17 A1 first term is given equal to 0 .4.

00:20 A2 will be equal to 0 .4 plus 0 .3 multiplied by 2 to the power 2 minus 2 that is equal to 0 .4.

00:30 To 0 .4 plus 0 .3 that is equal to 0 .7.

00:35 A3 will be equal to 0 .4 plus 0 .3 2 to the power 3 minus 2.

00:42 This will be equal to 0 .4 plus 2 multiplied by 0 .3 that is equal to 1.

00:49 A4 will be equal to 0 .4 plus 0 .3 multiplied by 2 to the power 4 minus 2.

00:56 We simplify this.

00:58 We get this is equal to 1 .1.

01:00 Point six a five will be equal to 0 .4 plus 0 .3 multiplied by 2 to the power 5 minus 2.

01:10 After simplification we get this is equal to 2 .8 a 6 will be equal to 0 .4 plus 0 .3 multiplied by 2 to the power 6 minus 2 after simplification we see that this is equal to 5 .2 7th term will be equal to 0 .4 plus 0 .3 multiplied by 2 to the power 7 minus 2.

01:36 After simplification we get this is equal to 10 and 8 term will be equal to 0 .4 plus 0 .3 multiplied by 2 to the power 8 minus 2 we simplify this and we get this is equal to 19 .6 these are the first 8 term of this sequence.

01:56 In second in second part we have given distance of mercury is equal to 0 .39 a u distance of venus is equal to 0 .72 a u distance of earth is given equal to 1 a u distance of mars is given equal to 1 .52 a u distance of jupiter is given equal to 5 .20 a u and distance of saturn is given equal to 9 .54 a u.

02:40 Now we have to compare the distances of the given planets to the sun to the terms of the sequence a1 minus d mercury is equal to 0 .4 minus 0 .39 that is equal to 0 .0 .0 .0 1.

02:59 A2 minus d2 is equal to 0 .7 minus distance of venice that is equal to 0 .7 minus 0 .72 that is equal to minus 0 .0 .0 .2...

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