We can use Kepler's Third Law, which relates the orbital period and the orbital radius of a planet around its star.
Kepler's Third Law is given by the equation:
T^2 = (4π^2/GM) * R^3
Where T is the orbital period, G is the gravitational constant, M is the mass of the central star, and R is the orbital radius.
Given:
M = mass of the Sun = 1.989 × 10^30 kg
T = 10.1 years = 10.1 × 365.25 × 24 × 60 × 60 seconds (converted to seconds)
Substituting these values into the equation, we have:
(10.1 × 365.25 × 24 × 60 × 60)^2 = (4π^2/(6.673×10^-11))*(1.989 × 10^30) * R^3
Simplifying the equation, we get:
(10.1 × 365.25 × 24 × 60 × 60)^2 = (25.1707 × 6.673×10^-11) * R^3
Solving for R^3:
R^3 = [(10.1 × 365.25 × 24 × 60 × 60)^2]/[(25.1707 × 6.673×10^-11)]
Taking the cube root of both sides to solve for R, we have:
R ≈ ∛[(10.1 × 365.25 × 24 × 60 × 60)^2]/[(25.1707 × 6.673×10^-11)]
Calculating this value, the orbital radius, R, is approximately 2.831 × 10^11 meters.
If a planet with mass of 4.35 × 10^23kg is rotating around the sun with an orbital period of 10.1 years, what is the orbital radius of the planet? (The constant of universal gravitation is G=6.673×10^-11) Assume the orbit is circular
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