Physics Problem: Velocity of Satellite 230km Above Earth?

  • Thread starter wikidrox
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In summary, the junior physicist's calculated velocity of 30,000 m/s for an artificial satellite traveling 230 km above Earth's surface is significantly higher than what is expected for an object in Earth's orbit. A rough calculation suggests that the speed should be around 7700 m/s for a circular orbit at that distance. The possibility of a highly eccentric elliptical orbit is also mentioned. Another calculation suggests a speed of approximately 7.8 km/s for a parabolic orbit, emphasizing the importance of considering the appropriate equations for different types of orbits.
  • #1
wikidrox
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
Last edited:

Related to Physics Problem: Velocity of Satellite 230km Above Earth?

1. What is the formula for calculating the velocity of a satellite 230km above Earth?

The formula for calculating the velocity of a satellite is V = √(GM/R), where G is the gravitational constant (6.67 x 10^-11), M is the mass of the Earth (5.97 x 10^24 kg), and R is the distance between the satellite and the center of the Earth (230km + radius of the Earth).

2. How is the velocity of a satellite affected by its distance from Earth?

The velocity of a satellite is affected by its distance from Earth because of the gravitational force between the two objects. The further the satellite is from Earth, the weaker the gravitational force and the slower its velocity. This is why satellites in higher orbits have slower velocities than those in lower orbits.

3. What units are typically used to measure the velocity of a satellite?

The velocity of a satellite is typically measured in meters per second (m/s) or kilometers per hour (km/h).

4. Can the velocity of a satellite change once it is in orbit?

Yes, the velocity of a satellite can change while it is in orbit. This can happen due to various factors such as atmospheric drag, gravitational forces from other objects, and adjustments made by ground control.

5. How does the velocity of a satellite affect its orbit around Earth?

The velocity of a satellite affects its orbit around Earth because it determines the balance between the centrifugal force of the satellite and the gravitational force of Earth. If the velocity is too high, the satellite may escape Earth's gravity and enter a new orbit around the sun. If the velocity is too low, the satellite may crash back to Earth. The correct velocity will result in a stable circular or elliptical orbit around Earth.

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