Question regarding centripetal acceleration/period

In summary: You already have the basic ingredients there for calculating it.In summary, a satellite with a radius of 70,000 km orbits the Earth every 6.0 hours. The period of rotation is 21600 seconds or 6 hours. The acceleration of the satellite can be calculated using the equation a = v2/r or a = mv2/r, where v is the velocity and r is the radius. However, the numbers in the problem statement do not appear to be accurate and suggest that the satellite may not be in a free orbit, but instead under some form of propulsion.
  • #1
LifeMushroom
3
0

Homework Statement


A satellite orbits the Earth every 6.0 hours in a circle. Radius = 70,000 km
a) What is the period of rotation?
b) What is the acceleration of the satellite?

Homework Equations


a = v2/r
Fc = mv2/r
v = 2pir/T

The Attempt at a Solution


For a, I converted 6 hours into 21600 seconds. Then I divided 70000 km by 21600 to get 3.24 km/sec as the period, but I'm not sure if this is right.
For b, I used the velocity equation: v = 2pi*70000/3.24 = 135679 m/s. Then a = 1356792/70000? Not sure if this is correct, because it's a really high number.
Thanks!
 
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  • #2
I don't know that I agree with the numbers in the statement of the problem. A satellite in low Earth orbit, with radius ## R=6,400 ## km, takes about 90 minutes=## \frac{3}{2} ## hours for one orbit. Since, by Kepler's 3rd Law, ## \frac{R^3}{T^2}=constant ##,## \frac{( 70,000)^3}{6^2} =\frac{(6,400)^3}{1.5^2} ## if their numbers are accurate. If I did the arithmetic correctly, it's not even close to a match. The "satellite" in the problem could not be orbiting in a free orbit. It is covering the distance much too quickly, and would need to be under some form of propulsion. ## \\ ## (Just a rough estimate, ## T \approx 52 ## hours at ## R=70,000 ## km).
 
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  • #3
LifeMushroom said:
Then I divided 70000 km by 21600 to get 3.24 km/sec as the period, but I'm not sure if this is right.

For future reference, the period ##T## is always in terms of unit time, I believe. Do you remember the definition of a mathematical period? I think you already found ##T## in your post somewhere, though.

LifeMushroom said:
For b, I used the velocity equation: v = 2pi*70000/3.24 = 135679 m/s. Then a = 13567922/70000? Not sure if this is correct, because it's a really high number.

You have the right idea; just solve for the correct value of ##T## and I think you should be good.
 
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  • #4
LifeMushroom said:
I divided 70000 km by 21600 to get 3.24 km/sec as the period,
A few things wrong there.
As @Eclair_de_XII points out, the period is a time, not a speed. Can you state what is meant by the period of an oscillation (or orbit)?
Even then, what speed do you think you are calculating by dividing the radius of the orbit by a time? Is the satellite moving along a radius?!

LifeMushroom said:
v = 2pi*70000/3.24 = 135679 m/s. Then a = 1356792/70000
Now you have divided a distance by a speed.
 
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  • #5
Charles Link said:
I don't know that I agree with the numbers in the statement of the problem. A satellite in low Earth orbit, with radius ## R=6,400 ## km, takes about 90 minutes=## \frac{3}{2} ## hours for one orbit. Since, by Kepler's 3rd Law, ## \frac{R^3}{T^2}=constant ##,## \frac{( 70,000)^3}{6^2} =\frac{(6,400)^3}{1.5^2} ## if their numbers are accurate. If I did the arithmetic correctly, it's not even close to a match. The "satellite" in the problem could not be orbiting in a free orbit. It is covering the distance much too quickly, and would need to be under some form of propulsion. ## \\ ## (Just a rough estimate, ## T \approx 52 ## hours at ## R=70,000 ## km).
It comes out a lot closer if the 70,000km is the circumference instead of the radius.
 
  • #6
Wait, since it orbits once every 6 hours, wouldn't the period just be 21600 seconds then, if it's just a unit of time?
 
  • #7
LifeMushroom said:
Wait, since it orbits once every 6 hours, wouldn't the period just be 21600 seconds then, if it's just a unit of time?
Yes.
 
  • #8
Or just 6 hours. I see no need to convert to seconds unless the problem statement asks for it in seconds.
 
  • #9
For b what is the equation for centripetal acceleration? Start with that.
 

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle.

2. How is centripetal acceleration calculated?

Centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circle.

3. What is the relationship between centripetal acceleration and period?

The period of a circular motion is the time it takes for the object to complete one full revolution around the circle. The period is inversely proportional to the centripetal acceleration, meaning that as the period increases, the centripetal acceleration decreases, and vice versa.

4. How does centripetal acceleration affect the motion of an object?

Centripetal acceleration is responsible for continuously changing the direction of an object's velocity, keeping it in a circular path. Without it, the object would continue moving in a straight line tangent to the circle.

5. Can centripetal acceleration be negative?

Yes, centripetal acceleration can be negative if the object is slowing down or changing direction in the opposite direction of the center of the circle. However, the magnitude of the acceleration is always positive.

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