The earth moves around the sun in a nearly circular orbit of radius 1.50*10^11m. During the three summer months (an elapsed time of 7.89*10^6s), the earth moves one-fourth of the distance around the sun. (a)what is the average speed of the earth? (b) what is the magnitude of the average velocity of the earth during this period?
I can't get the right answer for this problem again.Please give me some hints to do it.Thanks a lot.
What is the circumference of a circle? Divide that by 4, and you have the distance Mother Earth moved. You are given the time,
avg velocity= distance/time.
I get about2.99E4 m/s
To solve this problem, you need to understand the concepts of speed and velocity.
(a) Average speed is defined as the total distance traveled divided by the total time taken. In this case, we are given that the Earth moves one-fourth of the distance around the sun in 7.89*10^6 seconds. To find the average speed, you should divide this distance by the given time.
The total distance traveled can be calculated by multiplying the circumference of the circular orbit by one-fourth. The circumference of a circle is given by 2πr, where r is the radius of the circle.
So, you can calculate the average speed using the formula:
Average speed = (1/4) * (2π * r) / time
Substituting the given values:
Average speed = (1/4) * (2 * π * 1.50 * 10^11 m) / (7.89 * 10^6 s)
Calculate the average speed using the above formula, and make sure to convert units appropriately.
(b) The magnitude of average velocity is the total displacement divided by the total time taken. Displacement is a vector quantity, meaning it has both magnitude and direction. In this case, as the Earth moves in a circular orbit, its direction is constantly changing, and the total displacement is zero. Therefore, the magnitude of the average velocity will also be zero.
So, to answer part (b), the magnitude of the average velocity during this period is zero.
I hope these hints help you solve the problem correctly.