Question regarding centripetal acceleration/period

[LifeMushroom]

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
F_c = 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!

 

[Charles Link]

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).

 

[Eclair_de_XII]

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.

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.

 

[haruspex]

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?!

v = 2pi*70000/3.24 = 135679 m/s. Then a = 135679²/70000

Now you have divided a distance by a speed.

 

[haruspex]

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.

 

[LifeMushroom]

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?

 

[haruspex]

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.

 

[CWatters]

Or just 6 hours. I see no need to convert to seconds unless the problem statement asks for it in seconds.

 

[CWatters]

For b what is the equation for centripetal acceleration? Start with that.