If a planet with mass of 4.35 x 10^23 kg 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 x 10^-11) Show all work and assume the orbit is circular

Answer 1

To find the orbital radius of the planet, we can use the following formula:

T^2 = (4π^2 / GM) × r^3

Where:
T = orbital period of the planet
G = constant of universal gravitation
M = mass of the sun
r = orbital radius of the planet

First, we need to convert the orbital period from years to seconds because the constant of universal gravitation, G, is given in SI units:

Orbital period in seconds = 10.1 years × 365.25 days/year × 24 hours/day × 60 minutes/hour × 60 seconds/minute

Orbital period in seconds = 318,156,360 seconds

Rearranging the formula to solve for r, we have:

r^3 = (GM / (4π^2)) × T^2

Substituting the given values:

r^3 = (6.673 × 10^-11 N(m/kg)^2 × (1.989 × 10^30 kg)) / (4π^2) × (318,156,360 s)^2

r^3 = 3.986× 10^14 m^3/s^2 × 2.194 × 10^16 s^2

r^3 = 8.756 × 10^30 m^3

Taking the cubic root of both sides to solve for r:

r = (8.756 × 10^30)^(1/3) m

r ≈ 2.784 × 10^10 m

Therefore, the orbital radius of the planet is approximately 2.784 × 10^10 meters.