Physics Problem: Velocity of Satellite 230km Above Earth?

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The discussion centers on the velocity of an artificial satellite orbiting 230 km above Earth's surface, with a calculated speed of 30,000 m/s deemed excessively high. Participants agree that this speed is unrealistic for Earth orbit, with a more accurate estimate for a circular orbit being around 7,700 m/s. Calculations suggest that at this altitude, gravitational acceleration is approximately 9.2 m/s². One contributor revisits the virial theorem, leading to a refined velocity estimate of about 7.8 km/s for a circular orbit. Overall, the consensus is that the initial calculation is significantly incorrect.
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I have been asked to review a junior physicist's calculations regarding an artificial satellite traveling 230 km above the Earth's surface where acceleration due to gravity is about 9.0m/s2. He calculated a velocity of 30000 m/s. And the radius of the Earth was measured at about 6370km. Does his speed seem right? I don't even know if I have enough information to figure this out.
 
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wikidrox,

30,000 m/s is WAY to fast for anything in Earth orbit. That's 30km/s, or about 20miles/sec, which is 72,000 miles/hr, gets you around the Earth every 20 minutes. Can't happen.
 
A quick calculation on the back of an envelope (literally) gives g at that distance to be about 9.2 m/s^2 (so 9 m/s^2 is OK if you're not fussy), and this gives a speed of about 7700 m/s for a circular orbit. It is entirely possible my math is off, especially when I'm being distracted by a yummy blueberry muffin.

Maybe your "junior physicist" was doing a highly eccentric elliptical orbit, which if true, is highly eccentric in itself. :)

Zz.
 
I calculated ~11 km/s. I used Etotal = 0 => KE = -Ugrav. That's a parabolic orbit, though, isn't it. Dammit! Take one semester off and you pay for it the whole time. I'll get back to this.

EDIT:
OK, now I remember from the virial (sp?) theorem (or something) that KE = -(1/2)Ugrav for a circular orbit. Then that would lead to the velocity I was smoking for the parabolic orbit divided by a factor of √2. This gives ~11√2 km/s ~ 7.8 km/s. I would like to take this opportunity to agree with ZapperZ.
 
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