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
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.
To find the orbital radius of the planet, we can use Kepler's third law of planetary motion:
T^2 = (4π^2 / GM) × r^3
Where:
T = orbital period
G = gravitational constant
M = mass of the sun
r = orbital radius of the planet
Given:
M = mass of the planet = 4.35 × 10^23 kg
T = orbital period = 10.1 years = 10.1 × 365.25 × 24 × 60 × 60 seconds (since 1 year = 365.25 days, 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds)
G = gravitational constant = 6.673 × 10^-11 m^3/kg/s^2
Plugging in the values into the formula, we have:
T^2 = (4π^2 / GM) × r^3
(10.1 × 365.25 × 24 × 60 × 60 seconds)^2 = (4π^2 / (6.673 × 10^-11 m^3/kg/s^2) × (r^3)
Simplifying:
(10.1 × 365.25 × 24 × 60 × 60)^2 = (4π^2 / (6.673 × 10^-11)) × (r^3)
4783859440.25 = 4π^2 / (6.673 × 10^-11) × (r^3)
Rearranging the equation to solve for r^3:
r^3 = (4783859440.25 × 6.673 × 10^-11) / (4π^2)
Taking the cube root of both sides to find r:
r = (4783859440.25 × 6.673 × 10^-11) / (4π^2)^(1/3)
Calculating the expression:
r ≈ 1.496 × 10^11 meters
Therefore, the orbital radius of the planet is approximately 1.496 × 10^11 meters.
To find the orbital radius of the planet, we can use Kepler's Third Law of Planetary Motion, which relates the orbital period, orbital radius, and the mass of the central body (in this case, the Sun).
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the orbital radius (r):
T^2 ∝ r^3
We can rearrange the equation to solve for the orbital radius:
r = (T^2 * G * M_s) / (4π^2)
Where:
T is the orbital period in seconds
G is the gravitational constant (6.673×10^-11 N m^2 / kg^2)
M_s is the mass of the Sun (1.989 × 10^30 kg)
π is a mathematical constant approximately equal to 3.14159
Now, let's calculate the orbital radius:
Step 1: Convert the orbital period from years to seconds:
Given T = 10.1 years
1 year = 365.25 days (accounting for leap years)
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
T = 10.1 years * 365.25 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute
T ≈ 3.18 × 10^8 seconds
Step 2: Plug the values into the formula:
r = (T^2 * G * M_s) / (4π^2)
= (3.18 × 10^8)^2 * (6.673×10^-11) * (1.989 × 10^30) / (4 * 3.14159^2)
Perform the calculation using a calculator, and the result will be the orbital radius in meters.