Question

A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?

A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.

Answer 1

The linear speed of the Earth in its orbit around the Sun is approximately 29.8 km/s.

Step-by-step explanation:

The linear speed of the Earth in its orbit around the Sun can be calculated using the formula:

linear speed = 2 * π * radius / time

Given that the Earth's distance from the Sun is 150,000,000 km and its orbital period is one year (365 days), we can plug these values into the formula:

linear speed = 2 * π * 150,000,000 km / (365 days * 24 hours * 60 minutes * 60 seconds)

Calculating this value gives us:

linear speed ≈ 29.8 km/s

Therefore, the linear speed of the Earth in its orbit around the Sun is approximately 29.8 km/s, which is closest to the option:

D. 30 m/s.

Answer 2

The answer is "Option B".

Step-by-step explanation:

`r=15* 10^(7)\ km\ = 15* 10^(10)\ m\\\\w=(2\pi)/(1\ year)\\\\=(2\pi)/(1* 365.24 * 24 * 60 * 60\ sec)\\\\a=w^2r\\\\=((2\pi)/(1* 365.24 * 24 * 60 * 60\ sec))^2 * 15 * 10^(10)\ (m)/(s^2)\\\\`

`=5.940 * 10^(-3) \ (m)/(s^2)\\\\=6 * 10^(-3) \ (m)/(s^2)\\\\=0.006\ (m)/(s^2)\\\\`