PLEASE HELP! I am extremely confused. I have to finish this by today and there is no one to help me! Please I am begging you! I need help!

The planets orbit the Sun not in circles but in ellipses, which are “flattened” circles. The Earth’s orbit, however, isn’t flattened much. Its greatest distance from the Sun, 94.5 million miles, differs from its least distance, 91.4 million miles, by only about 3%.

To answer the following questions, make the following assumptions:

a. Summer includes all of the days from June 21 through September 21. During that time Earth travels in a circular orbit with a radius of 94.5 million miles. A year lasts 365 days.

b. Winter includes all of the days from December 21 through March 20. During that time Earth travels in a circular orbit with a radius of 91.4 million miles. The year you will consider lasts 365 days and is not a leap year. Solve. Show your work. Use 3.14 for π.

1. Find the distances that the Earth travels in summer and in winter. Give your answers in millions of miles rounded to the nearest tenth.

2. Find the Earth’s average rate of speed in summer and in winter. Give your answers in millions of miles per day rounded to the nearest hundredth.

3. Find the areas of the sectors that the Earth traverses in summer and in winter. Give your answers in millions of miles squared rounded to the nearest tenth.

Under the given assumptions, summer lasts 93 days (10 days for Jun, 31 days each for Jul and Aug, and 21 days for Sept), and winter lasts 90 days (11 days for Dec, 31 days for Jan, 28 days for Feb, and 20 days for Mar).

1. In the summer, we assume the Earth moves in a circular path with radius 94.5 million mi. In 365 days, the planet would cover a distance equal to the circumference of this circle, 2π * (94.5 million mi), or approximately (using π = 3.14) 593.46 million mi. Summer last 93 days, which is roughly 93/365 = 25.48% of one year. So during the summer, the Earth will traverse about 25.48% of the circumference of its circular path, or

0.2548 * (593.46 million mi) = 151.2 million mi

In the winter, the Earth's path is taken to be a circle of radius 91.4 million mi, so its circumference is 2π * (91.4 million mi) = 573.99 million mi. Winter has a duration of 90/365 = 24.68% of a year, so during winter the Earth travels

0.2468 * (91.4 million mi) = 22.6 million mi

2. The average speed of a body over some time interval is the distance the body travels divided by the duration of time.

During the summer, the Earth moves at an average speed of

(151.2 million mi) / (93 days) = 1.63 million mi / day

During the winter, it moves at a speed of

(22.6 million mi) / (90 days) = 0.251 million mi / day

3. Use the distances/arc lengths found in (1) to determine the measure of the central angle θ subtended by the summer and winter arcs. In either case, the ratio of arc length to circumference is equal to the ratio between the central angle and one complete revolution (2π radians).

By the same token, the ratio of sector area A to the area of the whole circle (πr^2) is proportional to the ratio of the central angle to 2π rad.

Summer:

(151.2 million mi) / (2π * (94.5 million mi)) = θ / (2π rad)

==>  θ = 1.6 rad

A / (π * (94.5 million mi)^2) = (1.6 rad) / (2π rad)

==>  A = 7144.2 million square mi

Winter:

(22.6 million mi) / (2π * (91.4 million mi)) = θ / (2π rad)

==>  θ = 0.2473 rad

A / (π * (91.4 million mi)^2) = (0.2473 rad) / (2π rad)

==>  A = 1033.0 million square mi

the answer is not negative reinforcement.