Answer: 4 AU
Kepler's third law states that square of orbital period of a planet is proportional to the cube of the average distance from the Sun.
[tex]T^2 \propto r^3[/tex]
[tex]\Rightarrow T^2=\frac {4\pi}{GM}r^3[/tex]
where, T is the orbital period, r is the average distance from Sun, G is the gravitational constant and M is the mass of the Sun.
If T is in Earth years, M is in Solar mass (1 Solar mass = mass of sun), r is in AU (astronomical unit) then:
[tex]T^2=r^3[/tex]
[tex]\Rightarrow r=T^{2/3}[/tex]
[tex]\Rightarrow r=8^{2/3}=4 AU[/tex]
According to Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.
According to Kepler's third law, the square of a planet's orbital period (T) is proportional to the cube of the length of the semi-major axis of its orbit (r).
We can express it through the following expression.
[tex]T^{2} = r^{3}[/tex]
where,
If an asteroid has an orbital period of 8 years, its average distance from the Sun is:
[tex]T^{2} = r^{3}\\\\(8y)^{2} = r^{3}\\\\r = 4 AU[/tex]
According to Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.
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