A satellite orbits the earth a distance of 16,405,682 m above the planet's surface and takes 5.05 hours for each revolution about the earth. The earth's radius is 6,380,000 m. What is the acceleration of this satellite?
Question
A satellite orbits the earth a distance of 16,405,682 m above the planet's surface and takes 5.05 hours for each revolution about the earth. The earth's radius is 6,380,000 m. What is the acceleration of this satellite?
Solution
To find the acceleration of the satellite, we can use the formula for the acceleration of an object moving in a circular path, which is:
a = v² / r
where:
- a is the acceleration,
- v is the velocity, and
- r is the radius of the circular path.
However, we don't have the velocity of the satellite. We can find it using the formula:
v = 2πr / T
where:
- v is the velocity,
- r is the radius of the circular path, and
- T is the period of the motion (time for one revolution).
First, we need to find the total radius of the satellite's path. This is the radius of the Earth plus the distance of the satellite from the Earth's surface:
r = radius of the Earth + distance of the satellite from the Earth's surface r = 6,380,000 m + 16,405,682 m r = 22,785,682 m
Next, we need to convert the period from hours to seconds because the standard unit of time in physics is seconds:
T = 5.05 hours * 3600 seconds/hour T = 18,180 seconds
Now we can find the velocity:
v = 2πr / T v = 2π * 22,785,682 m / 18,180 s v = 7,884.9 m/s
Finally, we can find the acceleration:
a = v² / r a = (7,884.9 m/s)² / 22,785,682 m a = 2.73 m/s²
So, the acceleration of the satellite is approximately 2.73 m/s².
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