Given that
• The distance from the earth to the sun is 1.5*10^11m, and
• The mass of the sun is 1.9 *10^30 kg ,
1. use Kepler’s Law of Periods to find the period of the earth’s orbit, recorded in seconds:;
2. show how to express your answer in years; and
3. if your answer does not agree with the accepted earth orbital period of 365.25 days, explain the discrepancy.
You have a problem to solve. You may need to type the problem or question here.
Identify the units on x in each of the following equations.
(a) x = yz (both y and z are measured in feet)
(b) x = 3 y^2 (y is measured in inches)
(c) x = t^3 (t is measured in seconds)
(d) x = 4F2 d (both F and d are measured in meters)
4. In an exponential expression such as A = be -(t/RC) , we cannot attach units to the exponent -(t/RC) because a base, such as e, can only be raised to a pure number. Consequently, if the variable t in this equation is measured in seconds, what does this imply about the units on the product RC?
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 and the mass of the Sun.
1. Kepler's Law of Periods states that the square of the period of an orbit is proportional to the cube of the semi-major axis of the orbit. The semi-major axis is half of the longest diameter of the elliptical orbit. In this case, the distance from the Earth to the Sun is the semi-major axis of the Earth's orbit.
The formula for Kepler's Law of Periods is:
T^2 = (4π^2 / GM) * a^3
where T is the period of the orbit, G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2), M is the mass of the central body (the Sun in this case), and a is the semi-major axis.
Plugging in the given values, we have:
T^2 = (4π^2 / (6.67430 x 10^-11 m^3 kg^-1 s^-2)) * (1.5 x 10^11 m)^3
Now we can solve for T by taking the square root of both sides:
T = sqrt((4π^2 / (6.67430 x 10^-11 m^3 kg^-1 s^-2)) * (1.5 x 10^11 m)^3)
Calculating this gives us the period of the Earth's orbit in seconds.
2. To express the answer in years, we need to convert the period from seconds to years. There are 365.25 days in a year since it takes approximately 365.25 days for the Earth to complete one orbit around the Sun.
To convert the period from seconds to years, we can use the following conversion factors:
1 year = 365.25 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
So, the conversion from seconds to years is:
T_years = T_seconds / (365.25 days * 24 hours * 60 minutes * 60 seconds)
Calculate this value to express the period of the Earth's orbit in years.
3. If the calculated period does not agree with the accepted Earth orbital period of 365.25 days, it may be due to various factors such as rounding errors, approximations made during calculations, or inaccuracies in the given values for the distance and mass of the Sun. Additionally, Kepler's Law of Periods assumes that the orbits are perfectly elliptical, but in reality, there are other factors like gravitational interactions with other planets that can affect the exact period of the Earth's orbit.