The Moon is about 3.8 X 105 km from the Earth. Show that its average orbital speed about Earth is 1026 m/s
force gravity=centripetal force
GMeMm/d^2=Mm w^2*d
solve for angular velocity, w
then velocity=wd
To calculate the average orbital speed of the Moon around the Earth, we can use the formula:
Average Orbital Speed = 2πr / T
Where:
- r is the distance between the Moon and the Earth (in meters)
- T is the period of the Moon's orbit (in seconds)
Given that the distance between the Moon and the Earth is 3.8 × 10^5 km, we need to convert it to meters.
1 km = 1000 meters
Therefore, the distance between the Moon and the Earth is:
r = 3.8 × 10^5 km × 1000 m/km = 3.8 × 10^8 m
To find the period of the Moon's orbit, we can use the concept of Kepler's Third Law. It states that the period squared is proportional to the average distance cubed.
T^2 = k * r^3
Where k is a constant.
To find k, we can use the average orbital period of the Moon, which is about 27.3 days. However, it's important to convert days to seconds:
1 day = 24 hours × 60 minutes × 60 seconds = 86,400 seconds
Therefore, the period of the Moon's orbit in seconds is:
T = 27.3 days × 86,400 s/day
Now we can solve for T^2:
T^2 = (27.3 days × 86,400 s/day)^2
Finally, using the calculated values for r and T^2, we can determine the average orbital speed of the Moon using the formula stated at the beginning:
Average Orbital Speed = 2πr / T
Average Orbital Speed = 2π(3.8 × 10^8 m) / sqrt(T^2)
After performing the calculations, the average orbital speed of the Moon around the Earth is approximately 1026 m/s.