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
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.
To find the orbital radius of the planet, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the orbital period is proportional to the cube of the semi-major axis (orbital radius) of the planet's orbit.
Step 1: Convert the orbital period to seconds.
Given: Orbital period = 10.1 years
1 year = 365.25 days (considering leap years)
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
So, the orbital period in seconds is:
10.1 years * 365.25 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute = 3.18 x 10^8 seconds
Step 2: Apply Kepler's Third Law.
T^2 = (4π^2 / GM) * a^3
Where:
T = Orbital period (in seconds) = 3.18 x 10^8 seconds (calculated in Step 1)
G = Universal gravitational constant = 6.673 x 10^-11 N(m/kg)^2 (given)
M = Mass of the Sun = 1.989 x 10^30 kg (approximate value)
a = Orbital radius (semi-major axis) of the planet's orbit (to be found)
Now, rearrange the equation to solve for a:
a^3 = (T^2 * GM) / (4π^2)
a = cube root [(T^2 * GM) / (4π^2)]
First, calculate T^2 * GM:
T^2 * GM = (3.18 x 10^8 seconds)^2 * (6.673 x 10^-11 N(m/kg)^2 ) * (1.989 x 10^30 kg)
Now, substitute the values and solve for a:
a = cube root [(T^2 * GM) / (4π^2)]
a = cube root [(3.18 x 10^8 seconds)^2 * (6.673 x 10^-11 N(m/kg)^2 ) * (1.989 x 10^30 kg) / (4π^2)]
Performing the calculation, the orbital radius of the planet is approximately 2.96 x 10^11 meters.
To find the orbital radius of the planet, we can use Kepler's third law and the concept of centripetal force.
Kepler's third law states that the square of the orbital period of a planet is proportional to the cube of its orbital radius. Mathematically, this can be expressed as:
T^2 = k * R^3
where T is the orbital period, R is the orbital radius, and k is a constant.
In our case, the orbital period is given as 10.1 years. Therefore, we can rewrite the equation as:
(10.1 years)^2 = k * R^3
Now, let's solve for the constant k. Rearranging the equation, we have:
k = (10.1 years)^2 / R^3
Given that the constant of universal gravitation, G, is 6.673 x 10^-11, we can substitute it into the equation:
k = (10.1 years)^2 / R^3 = G * M
where M is the mass of the sun. Rearranging the equation, we have:
R^3 = (10.1 years)^2 / (G * M)
Now, substitute the mass of the sun, M, which is approximately 1.989 x 10^30 kg:
R^3 = (10.1 years)^2 / (G * 1.989 x 10^30 kg)
Substituting the values of G (6.673 x 10^-11) and the mass of the sun (1.989 x 10^30 kg), we can calculate the orbital radius:
R^3 = (10.1 years)^2 / (6.673 x 10^-11 * 1.989 x 10^30 kg)
Calculate the right side of the equation:
R^3 = (10.1 years)^2 / (1.331 x 10^19 m^3/kg)
Next, take the square root of both sides to solve for R:
R = [ (10.1 years)^2 / (1.331 x 10^19 m^3/kg) ]^(1/3)
Finally, substitute the values into the equation:
R = [ (10.1 years)^2 / (1.331 x 10^19 m^3/kg) ]^(1/3)
= [ (10.1 * 365.25 * 24 * 60 * 60 s)^2 / (1.331 x 10^19 m^3/kg) ]^(1/3)
After performing the calculation, the orbital radius of the planet is obtained.