# Saturn makes one complete orbit of the Sun every 29.4 years. Calculate the radius of the orbit of...

## Question:

Saturn makes one complete orbit of the Sun every 29.4 years. Calculate the radius of the orbit of Saturn. Hint: It is a very good approximation to assume this orbit is circular.

## Period in a circular orbit

The period {eq}T {/eq} of a body in a circular orbit with radius {eq}r {/eq} about a planet of mass {eq}M {/eq} is

{eq}T = \dfrac{2\pi r^{3/2}}{\sqrt{G M}}, {/eq}

where {eq}G = 6.67 \times 10^{-11} \ \text{N} \ \text{m}^2/\text{kg}^2 {/eq} is the gravitation constant.

## Answer and Explanation:

The mass of the Sun is {eq}M=1.99 \times 10^{30} \ \text{kg} {/eq}. Given that the period of the Saturn is {eq}T = 29.4 \text{ years} = 9.28 \times 10^8 \ \text{s} {/eq}, then the radius of its orbit is

{eq}T = \dfrac{2\pi r^{3/2}}{\sqrt{G M}} \quad \Rightarrow \quad r^{3/2} = \dfrac{T\sqrt{GM}}{2\pi} \quad \Rightarrow \quad r = \bigg(\dfrac{T}{2\pi} \bigg)^{2/3} (GM)^{1/3} {/eq}

{eq}\begin{align} r &= \bigg(\dfrac{9.28 \times 10^8 \ \text{s}}{2\pi} \bigg)^{2/3} \big( (6.67 \times 10^{-11} \ \text{N} \ \text{m}^2/\text{kg}^2)(1.99 \times 10^{30} \ \text{kg}) \big)^{1/3} \\ &= 1.43 \times 10^{12} \ \text{m} . \end{align} {/eq}

Therefore, the radius of the circular orbit of Saturn is {eq}1.42 \times 10^{12} \ \text{m} {/eq}.

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