given at
. the distance from the earth to the sun is 1.5 * 10^11, and the mass of the sun is 1.9 *10^30 kg
1. use kepler's law of period to find the period of the earth's orbit recorded in seconds.
show how to express your answer in years: and if your answer does not agree with the accepted earh orbital period of 365.25 days, explain the discepency.
I am not certain what version of Kepler's law you are exposed to, I suspect the law of Harmonics.
What is your question about this?
To find the period of the Earth's orbit using Kepler's law of periods, we need to know the distance from the Earth to the Sun (r) and the mass of the Sun (M).
1. Start with Kepler's third law of periods:
T^2 = (4π^2) / GM * r^3
2. Substitute the given values into the equation:
T^2 = (4π^2) / ((6.67 * 10^-11 N m^2/kg^2) * (1.9 * 10^30 kg)) * (1.5 * 10^11 m)^3
3. Simplify the equation:
T^2 = (4π^2 * (1.5 * 10^11 m)^3) / ((6.67 * 10^-11 N m^2/kg^2) * (1.9 * 10^30 kg))
4. Calculate the value on the right side of the equation:
T^2 ≈ 9.92 * 10^34 s^2
5. Take the square root of both sides to find the period (T):
T ≈ sqrt(9.92 * 10^34 s^2)
≈ 3.15 * 10^17 seconds
6. To express the answer in years, divide by the number of seconds in a year:
T ≈ (3.15 * 10^17 seconds) / (365.25 * 24 * 60 * 60 seconds)
≈ 9.95 years
The period calculated using Kepler's law of periods is approximately 9.95 years. This is slightly different from the accepted Earth orbital period of 365.25 days (equivalent to 1 year). The discrepancy arises because the Earth's orbit is not perfectly circular but elliptical, with the distance from the Sun varying throughout the year. Kepler's law assumes a perfectly circular orbit, leading to a small deviation from the actual observed period.
To find the period of the Earth's orbit using Kepler's Law of Period, we need to know the distance between the Earth and the Sun. You have mentioned that the distance from the Earth to the Sun is 1.5 * 10^11 meters.
Kepler's Law of Period states that the square of the period (T) of an orbiting object is directly proportional to the cube of its average distance (r) from the central mass. Mathematically, it can be expressed as T^2 = k * r^3, where k is a constant.
1. Calculate the period:
Given the distance from the Earth to the Sun (r) = 1.5 * 10^11 meters.
Let's assume that the average distance between the Earth and the Sun (r) remains constant throughout the Earth's orbit. Therefore, k can be calculated using the known period of the Earth's orbit (365.25 days or 3.156 × 10^7 seconds) and distance (r).
T^2 = k * r^3
(3.156 × 10^7 seconds)^2 = k * (1.5 * 10^11 meters)^3
Now, solve for k:
k = (3.156 × 10^7 seconds)^2 / (1.5 * 10^11 meters)^3
Substitute the values and calculate k.
2. Calculate the period in seconds:
Now, with the value of k, we can find the period of the Earth's orbit in seconds using the known distance (r = 1.5 * 10^11 meters).
T^2 = k * r^3
T^2 = calculated value of k * (1.5 * 10^11 meters)^3
Now, solve for T:
T = sqrt(calculated value of k * (1.5 * 10^11 meters)^3)
Calculate the square root of the expression and express the result in seconds.
3. Convert the period to years:
To convert the period from seconds to years, divide the obtained time in seconds by the number of seconds in a year (365.25 days or 31,557,600 seconds).
Period in years = Period in seconds / 31,557,600
If the calculated period (in years) doesn't agree with the accepted Earth orbital period of 365.25 days, it might be due to rounding errors or the approximation that was used for the value of the Earth-Sun distance or the Earth orbital period.