Chapter 1.2, Problem 102E

Distance from the Earth to the Sun It follows from Kepler’s Third Law of planetary motion that the average distance from a planet to the sun (in meters) is

$d={\left(\frac{GM}{4{\pi }^2}\right)}^{1/3}{T}^{2/3}$

where M = 1.99 × 10³⁰ kg is the mass of the sun, G = 6.67 × 10⁻¹¹ N · m²/kg² is the gravitational constant, and T is the period of the planet’s orbit (in seconds). Use the fact that the period of the earth’s orbit is about 365.25 days to find the distance from the earth to the sun.

To determine

The distance from the earth to the sun.

Answer to Problem 102E

The distance from the earth to the sun is $\underset{\_}{1.496\times {10}^{11}\text{m}}$.

Explanation of Solution

Formula used:

The distance is calculated using the formula $d={\left(\frac{GM}{4{\pi }^2}\right)}^{\frac{1}{3}}{T}^{\frac{2}{3}}$.

Where, d is the distance from the earth to the sun, G is the gravitational constant, M is the mass of the sun, T is the period of the planet’s orbit in sec.

Division of powers:

“The division of two powers with the same base a and different exponents m and n is given by, $\frac{a^m}{a^n}=a^{m-n}$”.

That is, while dividing two powers with same base, the exponents are subtracted and the base will remain the same.

Law used:

Laws of exponents:

“To raise a power to a new power, multiply the exponents”.

${\left(a^m\right)}^n=a^{mn}$

Calculation:

Given that, the mass of the sun is $1.99\times {10}^{30}\text{kg}$ and the gravitational constant is $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ and the period of earth’s orbit is about $365.25\text{days}$.

It is known that, $1\text{day}=86400\text{sec}$.

Use the above unit conversion to convert $365.25\text{days}$ into sec as follows.

$\begin{array}{c}365.25\text{days}=365.25\times 1\text{day}\\ \text{=365}\text{.25}\times 86400\text{sec}\\ \text{=31557600}\text{sec}\end{array}$

Note that, the period of earth’s orbit is about $31557600\text{sec}$.

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G, $1.99\times {10}^{30}\text{kg}$ for M, $31557600\text{sec}$ for T to calculate d as follows.

$\begin{array}{c}d={\left(\frac{6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 1.99\times {10}^{30}\text{kg}}{4{\pi }^2}\right)}^{\frac{1}{3}}{\left(31557600\sec \right)}^{\frac{2}{3}}\text{m}\\ ={\left(\frac{13.2733\times {10}^{30-11}}{4{\pi }^2}\right)}^{\frac{1}{3}}\left(99862.5484\right)\text{m}\\ ={\left(0.3362\times {10}^{19}\right)}^{\frac{1}{3}}\left(99862.5484\right)\text{m}\end{array}$

Now, move decimal 1 place to the right and subtract 1 from the power of 10 in the first bracket of the above equation to simplify the expression.

$\begin{array}{c}d={\left(3.362\times {10}^{18}\right)}^{\frac{1}{3}}\left(99862.5484\right)\text{m}\\ =\left(1.498\times {10}^6\right)\left(99862.5484\right)\text{m}\\ =149594.0975\times {10}^6\text{m}\end{array}$

Similarly, move decimal 5 places to the left, add 5 to the power of 10 to express the above quantity in proper scientific notation as follows.

$\begin{array}{c}d=1.495940975\times {10}^{6+5}\text{m}\\ =1.\text{495940975}\times {10}^{11}\text{m}\end{array}$

Round off the above quantity up to three decimal places.

$d=\text{1}\text{.496}\times {10}^{11}\text{m}$

Therefore, the distance from the earth to the sun is $\underset{\_}{1.496\times {10}^{11}\text{m}}$.