Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is [tex]T_R = 11.8 \ days[/tex]
Explanation:
From the question we are told that
The semi - major axis of the rocky debris [tex]a_R = 45.0\ AU[/tex]
The semi - major axis of Planet D is [tex]a_D = 60 \ AU[/tex]
The orbital period of planet D is [tex]T_D = 18.164 \ days[/tex]
Generally from Kepler third law
[tex]T \ \ \alpha \ \ a^{\frac{3}{2} }[/tex]
Here T is the orbital period while a is the semi major axis
So
[tex]\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}[/tex]
=> [tex]T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }[/tex]
=> [tex]T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }[/tex]
=> [tex]T_R = 11.8 \ days[/tex]
The orbital period of the rocky debris in space is 12 days.
From Kepler's law, we know that the square of the period of a planet in years is equal to the cube of its mean distance from the sun in astronomical units (AU).
Mathematically;
[tex]T^2 = r^3[/tex]
Now, we have the mean distance as 45.0 AU, we need to obtain its period.
semi major axis of debris rR= 45.0 AU
semi major axis of planet D rD = 60 AU
Period of planet D TD = 18.164 days
Period of debris TR = ?
[tex]\frac{TR^2}{TD^2} = \frac{rR^3}{rD^3}[/tex]
[tex]\frac{rR^3}{rD^3} * TD^2[/tex]
[tex]TR^2 = 45^3/60^3 * 18.164^2[/tex]
TR = 12 days
Missing parts:
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
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