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?Note: Write your answer in normal format up to 2 decimal points. Example: 12.34. Please DO NOT write the units in the electronic format. Definitely, write the units on paper. If I don't find the units in your hand-written notes, then marks will be deducted.
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?Note: Write your answer in normal format up to 2 decimal points. Example: 12.34. Please DO NOT write the units in the electronic format. Definitely, write the units on paper. If I don't find the units in your hand-written notes, then marks will be deducted.
Solution
To find the acceleration of the satellite, we can use the formula for the acceleration of an object moving in a circular orbit, which is given by:
a = v² / r
where:
- v is the velocity of the satellite
- r is the radius of the orbit
First, we need to find the radius of the orbit. This is given by the distance of the satellite above the Earth's surface plus the Earth's radius:
r = 16,405,682 m + 6,380,000 m = 22,785,682 m
Next, we need to find the velocity of the satellite. This is given by the circumference of the orbit divided by the time taken for one revolution:
v = 2πr / T
where:
- T is the period of the orbit
Converting the period from hours to seconds gives:
T = 5.05 hours * 3600 seconds/hour = 18,180 seconds
So:
v = 2π * 22,785,682 m / 18,180 seconds = 7,884.67 m/s
Finally, we can find the acceleration:
a = (7,884.67 m/s)² / 22,785,682 m = 2.73 m/s²
So, the acceleration of the satellite is 2.73 m/s².
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